Solution Manual For Circuits and Networks Analysis and Synthesis 6th Edition A Sudhakar , Shyammoh

CHAPTER 1 (SOLUTION MANUAL) 1-2.1

R = 30 Ω

V = 500 V

500 I= = 16.66 A 30 1-2.3

(a) W = P × t;

P=

P = I2R = (16.66)2 × 30 = 8.33 kW

V 2 (12) 2 = = 14.4 watts R 10

∴ W = 14.4 × 2 × 60 = 1728 J (b) Decrease

1-2.4

Q = 50 × 10–6 C; V = 150 V

50´10 –6 1 = µF C = Q/V= 150 3 4 µF Total capacitance = 15 1 1 4 W = CV2 = ´ × 10–6 × (150)2 = 0.3 × 10–2 J 2 2 15 1-2.5

In the circuit shown in Fig Q.5 , the 20 Ω resistance is parallel with the 30 Ω resistance.

The equivalent resistance R = 20 || 30 = 12 Ω.

The 12 Ω resistance is parallel with the 10 Ω resistance. Hence, the current through the 5 Ω

12 10 +12 5´12 × 2.73 A = 22

resistance = IT ×

Power consumed by the 5 Ω resistance = I2R = 52 × 5 = 125 watts 1-2.6

Power absorbed by the 5 Ω resistance 5 + 5v = I3 + I5 v = 2 × I3 5 + 10I3 = I3 + I5 5 + 9I3 = I5 I3 = V : I 5 = V 3 5 9V V = 5+ 3 5 V 5 + 3V – = 0 5 25 + (15V – V) = 0 14V = – 25 25 V =– 14

Solution Manual (1) Page No. 1

= –1.79V V 2 1.792 P5Ω = = = 638mW R 5 1-2.7

The circuit is redrawn as shown below.

0.1 k 1 = 0.25 4 0.025 × 103 = 25 Ω

Current flowing through the 25 Ω resistance due to the 5 A source 100 × 5 I 25 = 100 + 25 100 × 5 = 125 Iy = I25 = 4 A Current flowing through the 25 Ω resistance due to the dependent source 100 × 0.8 ix I25d = 100 + 25 4 = × 0.8 ix 5 Total current flowing through the 25 Ω resistance Ix = Iy + I25d = 4 + 0.64 iy 1-2.8

i4 =

Þ Þ

∴

20´6 0.2i4 ´6 = 12 + 0.12i4 + 4+6 10 0.88i4 = 12 i4 = 13.63 A P4Ω = (13.63)2 × 4 = 743.8 W V = 13.63 × 4 = 54.54 V P20A = –1090.9 W

P0.2i4 = (54.54)(0.2 × 13.63) = –148.76 W

Solution Manual (1) Page No. 2

54.54 = 495.87 W 6 Ptotal = P4Ω + P20A + P0.2i4 + P6Ω

P6Ω = ∴

=0 1-4.1

I = 2 A; t = 15 s; W = 1000 J Q = It = 30 C V=

1-4.2

1000 = 33.33 V 30

V = 120 V;

I = 0.8 A

120 = 150 Ω 0.8 Q 75 Q 10 (i) (a) I = = 75 A (ii) t = =2s = = 1 5 t I Q 10 = 20 A (b) I = = t 0.5 Q 5 (c) I = = = 2.5 A t 2 1 W = CV2 2 1 0.3 = C(20)2 2 0.3´2 ∴ C= = 1.5 mF 400 R=

1-6.1

1-6.2

1-6.3

The rate of change of voltage across each capacitor and the combination is the same, i.e.,

1-8.1

dV = 4500 volts/second dt

In Fig Q.1 (a),

In Fig Q.1 (b),

50 VAB = VT 50 + 60 50 = 4.55 V = 10 × 110

30 30 +10 + 100 30 = 100 × 140

VAB = VT

1-8.2

VAB = 21.43 V Since each resistor is across the 100 V source, the voltage drop across each source is 100 V. Hence, each meter reads 100 V.

1-8.4

Actual resistance R =

20 = 0.2 kΩ 100 mA

Voltage applied V = 0.2 × 103 × 150 × 10–3 = 30 V ∴ the change in voltage = 10 V 1-8.5

Total voltage

V = 5.5 + 7.2 + 12.3 = 25 V

Solution Manual (1) Page No. 3

If R4 is added to the circuit, remaining voltage appearing across the resistor = 30 – 25 V = 5 V 1-9.1

V1 = –60 V

V3 = VA – VB = 60 – 260

Þ

V2 = VA = 60 V

V3 = –200 V

V4 = VS = 260 V i2 =

Þ

60 =3A 20

i3 =

–200 = – 40 5

i1 = i2 + i3 = –37 A

–V2 = –5 A 12 –35 i5 = i2 = –35 A 3 i4 =

1-9.3

Find Req.:

= Req = RAB = 22.5Ω 1-9.4

In the circuit shown in Fig Q.4, the 40 Ω resistor is parallel with the 60 Ω resistor. The equivalent is in series with the 76 Ω resistor.

The equivalent resistance R1 = {[40 || 60] + 76} = 100 Ω In the other branch, the 82 Ω resistor is in series with the 18 Ω resistor.

The equivalent resistance R2 = 82 + 18 = 100 Ω

Solution Manual (1) Page No. 4

The equivalent resistances R1 and R2 are in parallel.

R3 = 100 || 100 = 50 Ω The resistance R3 is in series with the 100 Ω resistance. Hence, the total resistance = 150 Ω.

1-9.5

a) Let R = 80 Ω Req = ?

Req = 60 Ω b) If Req = 80 Ω, R = ?

100 ( R + 20) 100 + R + 20 100 R + 2000 80 – 10 = 120 + R 70 (120 + R) = 100 R + 2000 8400 – 2000 = 100 R – 70R 30R = 6400 6400 R= 30 R = 213.33 Ω c) If Req = R; 10 + 100 R + 2000 R= 100 + R + 20 2 120R + R = 10 + 100R + 2000 R2 + 20R – 2010 = 0 80 Ω = 10 +

R=

=

1-9.6

Let

– 20 ± 400 + 4 × 2010 1 440 – 20 2

= 35.93 Ω

RP1 = 2.2 K || 3.3 K || 5.6 K = 1.08 K RP2 = 1 K || 5.6 K || 3.3 K = 0.68 K RP3 = 1 K || 2 K || 5.6 K = 0.61 K RP4 = 1 K || 3.3 K || 2.2 K = 0.57 K

Solution Manual (1) Page No. 5

1.08 = 5.19 mA 1 +1.08 0.68 Current in 2.2 kΩ, I2 = 10 mA × = 2.36 mA 2.2 + 0.68 0.61 Current in 3.3 kΩ, I3 = 10 mA × = 1.56 mA 3.3 + 0.61 0.57 Current in 5.6 kΩ, I4 = 10 mA × = 0.92 mA 5.6 + 0.57

Current in 1 kΩ, I1 = 10 mA ×

1-9.7

Power VI = 75 watts Voltage across each bulb = 110 V Current in each bulb =

75 = 0.682 A 110

Total current = 6 × 0.682 = 4.092 A

Solution Manual (1) Page No. 6

CHAPTER 2 (SOLUTION MANUAL) 2-5.1

2-5.2

The circuit can the be redrawn as shown below.

Loop equations are 5I1 – 3I2 = –120 (1) –3I1 + 7I2 = –60 (2) From the equations, we have I1 = –39.23 A, I2 = –25.4 A Power in the 4 Ω resistor = (25.4)2 4 = 2580 watts When no power is delivered to the 4 Ω resistor, the current I2 becomes zero. Then the equations are 5I1 – 3I2 = –120 (3) –3I1 – 7I2 = –V + 40 (4) Since I2 = 0, then I1 = –24 A V = –32 volts In the circuit shown in Fig Q.2, the three currents are taken as I1, I2, and I3.

Here, we use supermesh analysis. Taking I1 and I2 as a combined loop, 10I2 –4I2 – I3 = 10 V (1) –I2 + 5I3 = 10 V (2) I1 = 10 A (3) From the above equations, we have I2 = 5.3 A, I3 = 3.06 A VAB = V1Ω – 10 VAB = 1(5.3 – 3.06) – 10 = –7.76 volts

2-5.3

KVL to the loop 1: –100 + 0.1I1 + R1(I1 – I2) = 0 and (I2 – I3) = 30 A But (I1 – I2) = 20 A The above equations are (1) –100 + 0.1 I1 + 20 R1 = 0 and I1 – I3 = 50 A KVL to the loop 2: 0.3 I2 + R2(I2 – I3) + R1(I2 – I1) = 0 = 0.3(I1 – 20) + 30 R2 – 20 R1 = 0 (2) = 0.3I1 – 20 R1 + 30 R2 = 6 KVL to the loop 3: 0.2 I3 + 120 + R2(I3 – I2) = 0 0.2(I1 – 50) + 120 – 30 R2 = 0 0.2 I1 – 30R2 = –110 (3) From equations (1), (2) and (3),

Solution Manual (2)

1

I1 = –6.66 A 2-5.4

R1 = 4.88 Ω

R2 = 3.82 Ω

By redrawing the given circuit,

By KVL to the loop 1: 1000 I1 + 2000(I1 – I2) – 10 + 10000(I1 – I3) = 0 13000 I1 – 2000 I2 – 10000 I3 = 10

By KVL to the loop 2: 2000(I2 – I1) + 10000 I2 + 1000(I2 – I3) = 0 –2000 I1 + 13000 I2 – 1000 I3 = 0

By KVL to the loop 3: 10000(I3 – I1) + 10 + 1000(I3 – I2) + 2000 I3 = 0 –10000 I1 – 1000 I2 + 13000 I3 = –10 From equations (1), (2), (3), I2 = 0.265 mA I3 = – 0.416 mA I1 = 0.45 mA and VAB = 10000 I2 = 2.65 volts 2-5.5

(1) (2) (3)

Consider the two ends to the variable resistor R are A and B. When R is infinite, the circuit can be drawn as shown.

From the figure, VB =

100 ´ 4 = 50 V 8

VA = 0 Therefore, the voltage across R = VA – VB = – 50 V. When R is zero, the circuit can be drawn as shown.

From the nodal equation,

V – 100 V V – 100 V – 100 =0 + + + 4 4 4 4

Solution Manual (2)

2

V = 75 V I = I1 + I2 = 12.5 A From the equation I = a + bv, when voltage is zero, the current I becomes a.

When the current is zero, the voltage V = 2-5.6

–a b

Therefore, a = 12.5 and b = 0.25.

Here we use supermesh analysis.

From Fig Q.2, we have the supermesh equation.

–1 + i1 + i1 – i3 + i2 – 3 + i4 – 5 + i4 – i3 = 0 The other equations are i2 – i1 = 3 i2 – i4 = 1 i3 = 2 A From the above equations, we get i1 = 1.2 A, i2 = 4.2 A, i4 = 3.2 A

(1) (2) (3) (4)

2-5.7

I1 = 15 A

1 Vx 9 1 = [3(I3 – I2)] 9 I –I = 3 2 3 I I I3 – 15 = 3 – 2 3 3

and I3 – I1 =

0.33 I2 + 0.67 I3 = 15

By KVL to the loop 2: 2I2 + 3(I2 – I3) + 1(I2 – I1) = 0 But I1 = 15, 6I2 – 3I3 = 15 From equations (1) and (2), I2 = 11 A I3 = 17 A and I = I2 = 11 A. 2-5.8

(1)

(2)

By KVL to the loop 1: –31 + 5 I1 + 10 I1 + 1(I1 – I2) – 10 V1 = 0 But V1 = 31 – 5I1 –31 + 16 I1 – I2 – 10(31 – 5I1) = 0 66 I1 – I2 = 341 (1)

By KVL to the loop 2: 4I2 + 10 V1 + 1(I2 – I1) = 0

Solution Manual (2)

3

4I2 + 10(31 – 5I1) + I2 – I1 = 0 –51 I1 + 5 I2 = –310 (2) From equations (1) and (2), I1 = 5 A, I2 = –11 A Current supplied by the 31 V source = I1 = 5 A Current in the 4 Ω resistor = I2 = –11 A 2-5.9

I1 = 2 A, I2 = 3 A

From the figure,

10(I3 – I1) + 20(I3 – I4) = 24

(1)

30I3 – 20I4 = 44

(2)

Þ

But I3 – I4 = 0 for V20 = 0 From (1) and (2), I3 = 4.4 A, I4 = 4.4 A 5(I4 – I2) + 20(I4 – I3) = V2 V2 = –20I3 + 25I4 – 15

Þ

V2 = 7 V for V20 = 0 2-6.1

The circuit of Fig. Q.1 can be redrawn as follows:

Consider nodes at 1, 2, and 3 as shown and corresponding voltages are V1, V2, and V3.

Here, we use supernode analysis and taking combined single-node equation, 5A–

V V1 V2 – +4A– 3 –4=0 25 20 40

(1)

V1 – V1 = 100

(2)

V2 – V1 = 60

(3)

From the above equations, we have V3 = –60.87 V; V1 = 39.13 V Power supplied by the 5 A source, P5 = V1 × I5 = 39.13 × 5 = 195.7 watts

Solution Manual (2)

4

2-6.2

V1 = 20 V V10 = 0 is given and V2 = 0.

V2 – V1 V2 V2 – V3 + + + Ix = 0 2 10 3 0 – 20 0 0 – V3 + (2 – 2) = 0 + + 2 10 3

By KCL at the Node 2,

V3 = 30 V = V3 = 30 volts.

2-6.5

V1 = 8 V

V −V V2 – V1 –2+ 2 3 =0

By KCL at the node 2: 1 2 V2 – 8 V2 – V3 =0 + 1 2 1.5 V2 – 0.5 V3 = 10

(1)

V3 – V1 V3 – V2 V3 =0 + + 2 2 4 V3 – 8 V3 – V2 V3 =0 + + 2 2 4

By KCL at the node 3:

–0.5 V2 + 1.25 V3 = 4 From equations (1) and (2), V3 = 6.77 V V2 = 8.92 V,

Solution Manual (2)

(2)

5

V1 – V2 = – 0.92 A 1 V – V3 I2 = 1 = 0.615 A 2 V – V3 I3 = 2 = 1.075 A 2 V I4 = 3 = 1.69 A 4 I1 =

2-6.6

[Here, V1 is mentioned, so assume V1 as voltage at the node 1] Assume voltage at nodes 2 and 3 as V2 and V3 respectively.

By KCL at the node 1: –5 +

V1 – V2 V1 – V3 =0 + 20 50

0.07 V1 – 0.05 V2 – 0.02 V3 = 5

By KCL at the node 2: V2 = 0.4 V, 0.4 V1 – V2 = 0

By KCL at the node 3:

(1) (2)

V3 – V1 V3 – V2 – 0.01 V1 = 0 + 50 30 – 0.03 V1 – 0.033 V2 + 0.05 3V3 = 0 From equations (1), (2) and (3), V2 = 59.35 V, V1 = 148.3 V, 2-6.7

V3 = 120.9 V.

Assume Vx and Vy as node voltages at nodes x and y respectively.

Vy = –6 volts.

By KCL at the node x,

Vx – 10 Vx – 4V2 Vx –Vy + + =0 2 4 5 But V2 = 10 – Vx.

Vx – 10 Vx – 4(10 – Vx ) Vx – (–6 ) =0 + + 2 4 5 10 40 6 ⎡1 5 1⎤ ⎢ 2 + 4 + 5 ⎥ Vx – 2 − 4 + 5 = 0 ⎣ ⎦ and

1.95 Vx = 13.8 Vx = 7.07 volts V2 = 10 – Vx = 2.93 volts

Solution Manual (2)

6

(3)

CHAPTER 3 (SOLUTION MANUAL) 3-1.1

After converting star to delta transformation, the circuit can be drawn as shown.

3-1.2

Here, R1 = R2 = R3 = 3 Ω Resistances of 1.2 Ω and 0.75 Ω are in series, which are parallel to the 3 Ω resistance Hence, RAB = 1.18 Ω (a)

RA =

R1 R2 1× 2 = R1 + R2 + R3 1 + 2 + 1

= 0.5 Ω

RB =

R1 R3 R1 + R2 + R3

1× 2 2 = 1+ 2 +1 4 = 0.5 Ω =

RC =

R2 R3 R1 + R2 + R3

1×1 1 = 1+ 2 +1 4 = 0.25

=

Req = 1.25║1.5 1.25 × 1.5 = 1.25 + 1.5 1.875 = 2.75 = 0.6818 RAB = 1.1818 Ω

Solution Manual (3) - 1

(b)

RA = RB

R1 R2 3× 5 15 = = = 1.25 Ω R1 + R2 + R3 3 + 5 + 4 12

R1 R3 12 = = 1Ω R1 + R2 + R3 12

RC =

R2 R3 5×9 20 = = = 1.67 Ω R1 + R2 + R3 3 + 5 + 4 12

R1 =

RA RB + RB RC + RC RA RC

(325 × 367) + (367 × 3) + (3 × 725) 32.69 = 3Ω 3 = 10.89 Ω R R + RB RC + RC RA 32.69 R2 = A B = RB 367 = 8.91 Ω R R + RB RC + RC RA 32.69 R3 = A B = RA 3.25Ω = 10.06 Ω =

Solution Manual (3) - 2

0.924 + 1.8 0.924 + 1.8 RAB = 0.61 Ω RAB =

3-1.3

By source transformation,

Therefore, the network is reduced to

Combining the current source,

By current division, current through the 2.2k resistor is 10.5m × 6.8k I 2.2 = (6.8 + 2.2) k I2.2 = 7.9 mA Current through the 6.8k resistor is 10.5 – 7.9 I6.8 = 2.6 mA Voltage across each resistor = 17.38 V 3-2.1

The current I' in the 2 W resistor is due to the 2V1 source only.

Solution Manual (3) - 3

From the figure, the equivalent resistance Req = [(3|| 2 + 1) || 12] + 2 = 3.86 Ω Total current in the circuit IT =

2 = 0.52 A 3.86

Current passing through the 1 Ω resistor I1Ω =

0.52´12 = 0.44 A 12 + 2.2

Current in the 2 Ω resistor due to the 2 V source only I' =

0.44´3 = 0.265 A 5

Current I" in the 2 Ω resistor is due to the 4 V source only.

Equivalent resistance Req = [{[(2 || 12)] || 2} + 3] = 4.15 Ω Total current in the circuit =

4 = 0.96 A 4.15

Current passing through the 2 Ω resistor, I" =

0.96´2.71 = 0.55 A 4.71

Total current = I' + I" = 0.265 + 0.55 = 0.82 A 3-2.2

When the 12 V source is acting alone, 12 I= 4 =3A Current I when the 5 A source is acting alone 5×3 I= = 3.75 A 4

Actual current I flows in opposite direction to assumed direction. Now I = – 3.75A

Solution Manual (3) - 4

Now, current I = 3 – 3.75 = – 0.75A Now, current through the 4 Ω resistor = 4I = 4 × 0.75 =3A 3-2.3

Current flowing through the 20 Ω branch is I3 = 2.5 A The new voltage source I3 × ΔR = 2.5 × 6 = 15 V is inserted in the R3 branch to compensate the other branch currents.

Current I'3 =

15 = 0.5 A 30

This current is opposite to the current I3 in the branch R3.

Hence, the new current = 2.5 – 0.5 = 2 A The voltage across the other 20 Ω branch = 52 V The current in that branch = 2.6 A Total current = 2 + 2.6 = 4.6 A 3-2.4

Open-circuiting the 2 A current sources, 4 = 2I1 – 5(2I1) + 3I1 I1 = – 0.8 Þ Short-circuiting the 4 V voltage source

V V – 5V2 + =2 2 3 Þ

V 4V – =2 2 3

Þ

V = –2.4 V

Þ

I2 =

Þ

I2 = 3.2 A

∴

I1 = I2 – I1

2.4 – 5(–2.4 ) 3

= 3.2 – (– 0.8) = 4 A

Solution Manual (3) - 5

3-2.5

The given diagram can be drawn as given. (a) 10 = 2i'2 + i'2 + 2i'2

Þ i12 = 2 A

V V – 2i''2 + =3 2 1 V V +V + =3 Þ 2 1 6 V= 1.2 V Þ 5 V i"2 = – = – 0.6 A Þ 2 ' " ∴ i2 = i2 + i2 = 2 – 0.6 = 1.4 A (b)

(a) Current-source open circuit

3-3.1

(b) Short-circuit voltage source

Now, find the open-circuit voltage across 2 Ω

VOC – 50 = i1 40 VOC = 40i1 + 50 VOC = –20i1 + 10i1 VOC = –10i1 –10i1 = 40i1 + 50 i1 = –1 A VOC = 10 volts Now, find the short-circuit current. i1 =

50 40

10i1 20 50 1 æ –50 ö ÷÷ + çç iSC = 40 2 è 40 ø 5 iSC = A 8

and iSC = – i1 +

VTH = VOC = 10 V

VOC 10 = = 16 Ω I SC 5 / 8 10 i= A = 0.5 A 18

RTH =

3-3.2

Current through the 10 Ω resistor by Millman’s theorem,

By Millman’s theorem,

Solution Manual (3) - 6

V G +V G +V G V '= 1 1 2 2 3 3 G1 + G2 + G3 R'=

1 1 = = 0.55 Ω G1 + G2 + G3 1 + 0.5 + 0.33

G1 =

1 1 = =1 R1 1

G2 =

1 1 = = 0.5 R2 2

G3 =

1 1 = = 0.33 R3 3

(1 × 1) + (2 × 0.5) + (3 × 0.33) 1 + 0.5 + 0.33 3 = = 1.64 V 1.83

V '=

I10 =

1.69 10 + 0.55

I10 = 0.155 A

3-4.2

The circuit consists only of dependent sources. So apply the 1 A current source (independent source) across the terminals a–b so that we can find the Thevenin resistance across it, RTh =

Vab 1

(1 – i50 + 0.2i50) =i100 i100 + i50 = 0.2i50 + 1 (1 – 0.8i50) = i100 i50 × 50 = 20i100 + 100 × i100 i50 × 50 = 120i100 i100 =

5 i50 12

(1)

(2)

From (1) and (2), 1 – 0.8i50 = i100 =

5 i50 12

æ

1 = i50 ç ç0.8 +

è

5ö ÷÷ 12 ø

i50 = 2.106 A.

Solution Manual (3) - 7

Vab = i50 × 50 = 50 × RTh =

60 = 105.3 volts 73

Vab = 105.3 Ω 1

−i 100 5 So the equivalent circuit is

The short-circuit iSC =

3-7.1

The circuit can be replaced by Thevenin's equivalent circuit.

By opening RL and taking resistance seen into the terminals AB, RAB = [90 || 180 || 20] = 12 Ω VAB =

10´30 =6V 50

Thevenin's equivalent circuit is shown alongside.

Power delivered to the load P12 = 0.75 W Maximum current Imax =

6 = 0.5 A 12

Maximum voltage = 6 V 3-7.2

The circuit shown in the figure is replaced by Thevenin's circuit.

VAB = VA – VB =

100´20 100´10 – 30 30

= 33.33 V RAB = [(20 || 10) + (20 || 10)] = 13.34 Ω

Load RL can be connected to Thevenin's equivalent circuit. when RL = 13.34 Ω, the maximum power will be delivered to the load. ∴

33.33 = 1.25 A 26.68 2 PL = I L RL = 20.84 watts IL =

Solution Manual (3) - 8

CHAPTER 4 (PROBLEM SOLUTION)

1 1 = = 5 Hz T 0.2 1 (c) f = = 2 kHz 500μs

1 = 20 Hz 50 ms 1 (d) f = = 100 kHz 10μs

4-1.1

(a) f.

4-1.2

(a) T =

1 1 = 0.016 s = f 60

(b) T =

1 = 2 ms 500

(c) T =

1 = 1 ms 1000

(d) T =

1 =5s 200´103

(e) T =

1 = 0.2 ms 5´106

(b) f =

Vrms = 20 V; Vmax = 20 × 2 = 28.3 V The equation is v(t) = Vm sin ωt Instantaneous values (a) At 33º; v(t) = 28.3 sin 33º = 15.4 V (b) At 110º; v(t) = 28.3 sin 110º = 26.57 V (c) At 145º; v(t) = 28.3 sin 145º = 16.22 V (d) At 325º; v(t) = 28.3 sin 325º = –16.22 V 4-2.2 The peak value of the current IP = 100 mA i(t) = IP sin ωt (a) At 35; i(t) = 100 sin 35º = 57.36 mA (b) At 190º; i(t) = 100 sin 190º = –17.36 mA (c) At 200º; i(t) = 100 sin 200º = –34.2 mA (d) At 360º; i(t) = 100 sin 360º = 0 mA 4-3.1 4-2.1

A wave equation v(t) = 15 sin ωt B wave equation v(t) = 15 sin (ωt + 30º) Instantaneous value (a) At 30º; v(t) = 15 sin (ωt + 30) = 15 sin (30 + 30) = 12.99 volts (b) At 90º; v(t) = 15 sin (90 + 30) = 12.99 volts (c) At 45º; v(t) = 15 sin (45 + 30) = 14.4 volts (d) At 180º; v(t) = 15 sin (180 + 30) = –7.5 volts (e) At 300º; v(t) = 15 sin (300 + 30) = –7.5 volts 4-3.2 (i) V(t) = 25 cos ωt + 15 sin ωt = 25 sin (ωt + 90º) sin ωt 25 15 90º + 0º = 2 2 = 17.67 j + 10.60 = 20.60 59.04º Vrms = 20.60 (ii) i(t) = 100 sin ωt – 10 cos 2 ωt = 100 sin ωt + 10 sin (2 ωt π ) e

Solution Manual (4) - 1

2

⎛ 100 ⎞ ⎛ 10 ⎞ I rm = ⎜ + ⎝ 2 ⎠⎟ ⎝⎜ 2 ⎠⎟

2

= 71.06 A 4-3.3

As per the given data the waveform is

⎛ 200 – 0 ⎞ V (t ) = ⎜ t ⎝ 2 – 0 ⎟⎠ V(t) = 100t

0 < t < 2 time interval

2

∫ (100t ) dt 100 2

Vrms =

=

0

2

=

t dt 2 ∫ 2

0

3 2

100 t 2 3

2

100

∫ = 2 3 8 = 115.47 V 0

4-3.4

The waveform equation is (V – 0) =Vm – (– Vm) (t – T

⎡ 2Vm T ⎤ ⎢ I (t – 2 )⎥ dt ⎣ ⎦ 0

∫

Vrms =

=

)

2

T

=

2

T

2Vm T T 2Vm T T

=

T

⎛ 2 T2 ⎞ ⎜ t + 4 – tT ⎟ dt ⎝ ⎠ 0

∫

T

⎛ t3 T 2 t2 ⎞ t T + – ⎜3 4 2 ⎟⎠ ⎝ 0

∫

2Vm T T T T 2 3

Solution Manual (4) - 2

Vrms = 4-3.5

Vm

3

volts

I = 10 + 15 sin t

I rms =

⎛ Im ⎞ 2 I dc +⎜ ⎟

2

2

⎛ 15 ⎞ = 10 + ⎜ = 14.57 A ⎝ 2⎠ ⎝ 2 ⎟⎠ 2

4-3.6

The equation of the waveform in ⎡ –10 t , 0 < t < KT V (t ) = ⎢ KT ⎢ KT < t < T ⎣0, KT

100 t 3 ( KT )2 3 0 T

4-3.7

100 1 ( KT )3 ( KT )2 3 =4 T 12 K= = 0.12 100 The rms value of the waveform

52 + 4-3.8

=4

52 = 25 +12.5 = 6.12 A 2

V = 100 + 50 sin ωt Vrms =

1 2p ò( 100 + 50sinw t )2 d( w t ) 2p 0

=

1 2p ( 10000 + 2500sin 2 w t +10000sin w t )d w ò 2p 0

=

10000 +1250 = 106.06

4-3.9

Solution Manual (4) - 3

3p /4 p ù 1 éêp /4 14.14sin w td (w t ) + ò 14.14 d (w t ) + ò 14.14 sin w t d (w t ) ú ò ú p êë 0 p /4 3p /4 û

Vavg =

ù æp ö 1é ê14.14(1 – 0.707) +14.14 çç ÷÷ +14.14[1 – 0.707]ú úû p êë è2ø

=

= 9.71

ù æp ö p 1 éêp /4 (14.14)2 sin 2 w td (w t ) + (14.14) 2 çç ÷÷ + ò (14.14)2 sin 2 w td (w t ) ú ò ú p êë 0 è 2 ø 3p /4 û

Vrms =

1 1 + = 10.38 8p 2

= 14.14 4-3.10

v =40 (1 – e–100t) for 0 < t < 0.1 v = 40 e–50(t – 0.1) for 0.1 < t < 0.2 0.2 ù 1 éê0.1 (40) 2 (1– e –100t ) 2 dt +ò (40) 2 e –100(t – 0.1) dt ú ò ú 0.2 êë 0 0 û

Vrms =

0.1 éé ù –200t é e –10 2e –100t úù 1 úù ú êê e ê = 40 5 ê t + + + – ê –100 100 ú úú –100 úû êëê ë –200 ë ûû

5(0.095 + 4.5 ´ 10 –7 ) = 27.57

= 40 4-3.12

0 < t <1

⎡ 50 V =⎢ ⎣ –10

1< t < 2

1

2

1

2

∫ 50dt + ∫ (–10) dt 50 t ∫ –10 t ∫

Vav = 0 =

∫

2

∫

0

1

2 1

4-4.1

1

2

(50)2 dt + (–10)2 dt

2

∫

502 t + 102 t 2 Vrms =

0

50 (1– 0) –10 (2 –1) 50 –10 = = 20 2 2 1

Vrms =

=

1

2

0

∫ 1

2

V2rms = 13 Vrms = 36.05 volts v(t) 50 sin 500 t 1 1 = XC = w C 500C i(t) = 50 × 500 C sin 500 t = 25000 C sin 500 t

Solution Manual (4) - 4

CHAPTER 5 (SOLUTION MANUAL) 5-2.1

XL = JWL = j(2πf) L = j 1570.8 XC =

–j –j –j = = ω C (2 π f )C (314.15)C 250

I=

2 + j 1570.8 –

j 314.15 C

Magnitude of voltage across the coil, 280 = (5 + j1570.8)

280 =

(1250 )2 + (392700)2 1 ⎛ ⎞ 4 + ⎜1570.8 – ⎟ ⎝ 314.15C ⎠

2

C = 18.9 µF With this C, I = 0.178 –90°

5-3.1

f = 50 Hz ∴

ω = 2πf = 2 × π × 50 = 100 π

æ 1

Z = 150 – j ç ç

è w C1

+

1 ö÷ w C2 ÷ø

Solution Manual (5) - 1

250 –j ⎞ ⎛ ⎜⎝ 2 + j 1570.8 ⎟ 314.15C ⎠

æ ö 1 1 ÷ + ç100p ´100´10 –6 100p ´200´10 –6 ÷ è ø = 150 – j47.45 = 157.4 –17.65º 100 17.65 = 0.635 17.65º I= 157.4 = 150 – j ç

Total current Phase angle 5-3.2

5-3.3

θ = 17.65º lead

Z = 30 + j (15.7 + 31.41) = 30 + j(47.11) = 55.85 57.51º Phase angle θ = 57.51º lag At 100 Hz; ω = 2πf = 200π Z = R + jωL = 10 + j(200π × 0.01) = 11.8 32.13º I=

100 = 8.47 –32.13 11.8 32.13

At 3 kHz; ω = 2πf = 2π × 3 ×103 = 18840.5 Z = R + jωL = 10 + j(18849.5 × 0.04) = 88.75 86.96º I= 5-3.4

100 –86.96º = 0.53 –86.96º 188.75

The phasors of Va and Vb are,

⎛8⎞ 6

∴

⎛ – XC ⎞ ⎝ R ⎠⎟

tan–1 ⎜ ⎟ = 90 + tan–1 ⎜ ⎝ ⎠

XC = 0.75 R

(1)

The magnitude relation is V10 = 4 Va |I(6 + J8)| = |4[I(R – jXC)]| R2 + XC2 = 6.25 2

(2) 2

From Eq. (1), R + (0.75 R) = 6.25 R=2Ω ∴

XC = 0.75 × 2 = 1.5 Ω

1 = 1.5 ωc 5-3.5

C = 2.12 mF W = 2000 rad/s Vrms = irms = ∴

50 2 8 2

–25º

5º Z=

Vrms = 6.25 –30º irms

= 5.41 – j 3.125 The real part = R = 5.41 Ω

Solution Manual (5) - 2

The imaginary part = –j 3.125 =

–j ωC

C = 0.16 mF 5-3.6

500 rad/s Applied voltage, Vrms = 250 –45° Current, Irms = 17.91 (180 + 63.5)° ∴

total impedance, Z =

Vrms 250 –45° = I rms 17.91 243.5°

= 13.95 –288.5° = (4.42 + J13.23) Ω The real part is resistance, R = 4.42 Ω and L1 = 0.02 H is

X L1 = j L 1

= j(500)(0.02) = j 10 The total reactance is j13.23 Ω and X L = j 10. The remaining reactance = ( j13.23–j10) = j3.23.

1

The reactance of + j 3.23 gives inductance L2.

X L = j L 2 = J 3.23 ∴ 2

L2 = 6.46 mH. The elements are

5-3.7

Z = R + jωL |Z| =

R 2 + ( w L )2

|IT| =

15 15 = w L 200 L

= 0.075 L

|V| = |IZ| = 0.075 R 2 + ( w L )2

wL

= tan–1 R

Solution Manual (5) - 3

æ

–1 V = 0.075 L R 2 + ( w L )2 sin ç ç200t + tan

è

where

5-4.1

wLö ÷÷ Rø

ω = 200 rad/s

I1 =

50 0° = 10 +53.13° 3 – j4

I2 =

50 0° = 5 0° 10

IT = I1 + I2 = 10 +53.13º + 5 0º = 13.6 +36º

5-4.2

XC1 =

1 1 = 159.15 Ω = w C1 2p ´10´103 ´0.1´10 –6

XC2 =

1 1 = 318.31 Ω = 3 w C2 2p ´10´10 ´0.05´10 –6

XC3 =

1 1 = 53.05 Ω = w C3 2p ´10´103 ´0.3´10 –6

Writing nodal equations,

10 –V1 V1 V1 V = + + 1 154.15 –90º 318.31 –90º 53.05 –90º 700 ∴

V1 = 1.615 2.06º = 1.614 + j0.06

Solution Manual (5) - 4

V0.1µF = 10 – 1.614 – j0.06 = (8.386 – j0.06)V V0.05 µF = V0.3 µF = (1.614 + j0.06)V V700 = 1.615 2.06º

200 = 0.46 2.06º V 700 500 V500 = 1.615 2.06º × = 1.154 2.06º V 700 10 – V1 = 0.053 59.59º Total current = – j159.15 V 10 0º = = 188.67 –89.59º Total impedance Z = I 0.053 –89.59º V200 = 1.615 2.06º ×

5-4.3

Complex Impedance V = Vm cos ωt

Vm R Vm iL = wL ( R )( jw L ) Z= R + jw L iR =

|Z| =

Rw L 2

R + ( w L )2

V V = m R 2 + ( w L )2 Z Rw L V R iT = m R 2 + ( w L )2 cos (ωt + φ ) where φ = tan–1 wL Rw L

|iT| =

5-4.4

R = 5 Ω, L = 0.02 H, V = 100 sin (1000t + 50º) iT = ?, φ = ? |Z| =

( R )( w L ) R 2 + ( w L )2

=

5´1000´0.02 52 + ( 1000´0.02 )2

= 4.85

V 10.0 = = 20.61 A Z 4.85 R 1 φ = tan–1 = tan–1 1000´0.02 wL 5

|I| =

= 9075.96º = 14.04º I = 20.61 sin (1000t + 14.04º) V = 100 sin (1000t + 50º) The current lags the applied voltage by 80 – 14.04 = 35.96º

Solution Manual (5) - 5

iL = 5iC

5-4.5

V V =5 1 / wc wL 1 = ω2k 5 V 50 |Z| = = 25 = I 2 1 jw L jw L wL jw C = = 2 |Z| = 2 1 –w LC +1 –w LC +1 jw L + jw C 25´4 3000´ L = 25 Þ L = = 6.67 mH 1 3000 ´ 5 – +1 5 1 = 3.33 µF C= 5´6.6710 –3 ´30002 5-4.6

Z = 300 || 0.5 j ω ||

–j 10´10 –6 ´w

ω = 1000 Z = 300 || –125j = 115.38 –67.4° |Z| = 115.38 I=

5-5.1

V 200 = = 1.733 67.4° Z 115.38 –67.4

∴ i(t) = 1.733 sin (1000t + 67.4º) The circuit can be drawn as shown.

Total impedance ZT = {[[(2.5k + j1.57k)/3k] + 1k] || 1.5k} ZT = 944.2 3.67° Current

IT =

V 50 = = 0.053 –3.67° A ZT 944.2 3.67°

Phase angle θ = 3.67º lagging Current through the 1 Ω resistor =

0.053 –3.67º´1500 1500 +[[(2.5k + j1.57 k / 3k )] + 1k ]

Solution Manual (5) - 6

= 0.0198 –9.853º Current through the 3 kΩ resistor = 0.0198 –9.853º ×

2500 + j1570.1 5500 + j1570.1

= 0.0102 6.347º Voltage across the 3 kΩ resistor = 30.7 6.347º V

0.0198 –9853´3000 5500 + j1570.1 = 0.01038 –25.783º Voltage across the inductor = 0.01038 × 1570.1 90 – 25.783º

Current through the inductor =

= 16.3 64.217º V 5-5.2

XC1 =

1 1 = 15.915 Ω = w C1 2p ´100´10 –6

XL1 = 2π×100 × 0.1 = 62.83 Ω From the figure, the nodal equation can be written as

V – 50 V V + + =0 10 – j15.915 30 j 62.83 ∴ V = 38.93 32.43º 50 – 38.93 32.43º = 1.44 7.05º I= 10 – j15.915 V30 Ω = 38.93 = V0.1H V100µF = 1.44 7.05º × 15.915 90º = 22.9 –82.95º V10Ω = 14.4 V 5-5.3

100

Vrms =

2

0º , W = 5000 rad/s

= 70.7 0º XL = j L = j(5000)(0.002) = j10 Ω XC =

−j –j = = –j 10 Ω ωC ( 5000 )( 20 ×10 –6 )

i1 =

70.7 0° = 3.5 – j 3.5 = 4.95 – 45° A 10 + j10

i2 =

70.7 0° = 3.5 + j 3.5 = 4.95 45° 10 – j10

iT = i1 + i2 = 7 A

Solution Manual (5) - 7

CHAPTER 6 (Solution Manual) 6-2.1

–103 × i1 – 500i2

–V1 (i2 – i3) = 0 2

(1)

i2 – i1 = 10–3

(2)

2

–10 (i3 – i2) = 500ji3 = 0 J –i –4i3 = i2 Þ i3 = 2 4 æ i ö 500 –1000i1 – 500i2 – i2 + 2 ÷÷ = 0 i2 + 100j ç ç 4ø 2 è

(3)

–40i1 = (–30 + 5j)i2 and also i2 – i1 = 10–3 [from (2)] 40(i2 – 10–3) = (–30 + 5j)i2

Þ

i2 =

40´10 –3 = 5.7 × 10–4 4.08 70 – 5 j

2 R cos R = 0.081 mW P = Veff Ieff cos θ = I eff

6-2.2

Power delivered by the current source = ?

i1 = 1 0º A –j(i2 – i1) – (j2) (i2) – 0.25i2 = 0 i2 (–j + 2j – 0.25) + ji1 = 0 i2 =

–j = 0.97 165.96º A j – 0.25

V1 = j(i1 – i2) V1 = j(1 – 0.97 165.96 ) = j(1.94 – 0.235j) = 1.94j + 0.235 = 1.95 83.1º P = VI cos θ =

1.95 2

´

1 2

cos 83.1

= 0.114 Watts

Solution Manual (6) - 1

6-3.1

R=

100 = 20 Ω 5

150 = 30 Ω 5 125 XL = = 25 Ω 5 XC =

IXL = 125

Total impedance Z = R + j(XL – XC) ∴ Z = 20 – j5

R 20 = = 0.97 Z 20 – j 5

cos θ = 6-3.2

P =100 W;

i(t) = 8 sin (50t + 45º)

8

I=

2

45º = 5.66 45º ∴ V = 24.99 90º

VI cos θ = 100 I2R = 100 R=

100 (5.66) 2

= 3.12 Ω

R = 0.0707; Z XL = 3.12 Ω = 3.12 + j3.12

cos θ =

L= 6-3.3

∴ Z =

3.12 = 9.93 mH 2p ´50

3.12 = 4.413 Ω 0.707

[milli Henry]

v = 100 sin (377 t + 10º) = 70.71 10º i = im sin (377 t + 10 – 30º) = im sin (377 t – 20º) = I rms – 20º P = Vrm Irm cos 400 = 70.71 × Irms cos 30º Irms = 6.53 A Complex power S = Vrms Irms = 70.71 × 6.53 = 461.88 VA Total impedance Z = {[j4 || – j10] + 3 + j4} = j6.67 + 3 + j4 = 11.10 74.3º Power factor cos θ = 0.27 Input power

6-3.4

I=

200 2 = 12.73 A 11.10 2

Power dissipated = 12.73 3 = 486.5 watts

(

6-3.5

)

Total impedance Z = (5 + j8) || (3 + j4) = (27.17 + j38.53) Ω V = 100 rms V 100 I= = = 2.12A Z 47.15 Power dissipated I2R = (2.12)2 × 27.17 = 122.11 watts Power factor

cos

ோ

= = 0.576 ௓

Solution Manual (6) - 2

6-3.7

PT = 1100 W Data for inductions reactance –j4 will be j4

Z eq =

Z1 Z 2 10(3 + 4 j ) 10(5 53.13º 50 53.13° = = = 13 + 4 j 13.60 17.10° Z1 + Z 2 10 + (3 + 4 j )

Z eq = 3.67 36.03º = (2.96 + j 2.15) W PT = I2 Req

Req

PT = I2 2.96 = 1100

I=

V Z eq

I = 19.27 –36.03º A

Z2 10 19.27 –36.03º I= Z1 + Z 2 10 + (3 + 4 j ) 10 19.27 – 36.03°=14.13 – 53.13º = 13.60 17.10º Z1 3+4 j 5 53.13º 19.27 –36.03º = 19.27 –36° I2 = I= 10 + (3 + 4 j ) 13.60 17.10º Z1 + Z 2

I1 =

I 2 = 7.08 0º PZ1 = I12 R1 = 600 W PZ2 = I22 R2 = 500 W 6-3.8

200 30º = 66.42 + j5.81 30 25º 200 30º Z2 = = 3.76 + j1.368 50 10º

Z1 =

True power in the branch 1 = VI1 = cos = 200 × 30 cos 5° = 5977.2 watts Apparent power in the branch 1 = VI1 = 6000 VA Reactive power in the branch 1 = VI1 sin θ = 6000 × 0.087 = 522.93 VAR True power in the branch 2 = VI2 cos θ = 9396.9 watts Apparent power in the branch 2 = VI2 = 10000 VA Reactive power in the branch 2 = VI2 sin θ = 3420.2 VAR Total current I = I1 + I2 = 30 25º + 50 10º = 80.08 15.48º The angle between voltage and current = 30º – 15.48º θ = 14.52º Power factor = cos θ = 0.97 Total active power in the circuit = VI cos θ = 15.396 kW Total reactive power = VI sin θ = 3944 VAR

Solution Manual (6) - 3

Total apparent power = 15.87 kVA 6-3.9

ZT = ( j4) || 3 || (5 + j5) = 1.796 38.93° = 1.397 + j1.128 I2R = 500 W 500 I= =18.92 A 1.397 V = IZ = 18.92 × 1.796 = 33.98 volts Hence, the current lags the voltage by 38.93°. θ = 38.93°; pf = cos θ = 0.778 Preactive = VI sin θ = 33.98 × 18.92 × sin 38.93° Prea = 403.98 VAR 6-3.10

⎡ V − 60 25° ⎤ V − 60 25° V V +j + = 0.5 × 12 × ⎢ ⎥ 8 j + 12 20 20 ⎣ 12 + 8 j ⎦ ⎡ 1 j 1 0.5 × 12 ⎤ 60 25° 60 25° × 0.5 × 12 V⎢ + + − − ⎥= 12 + 8 j ⎣ 8 j + 12 20 20 12 + 8 ⎦ 8 j + 12 V = – 49 – 36.6 j V − 60 25° −5.37 + 11.25 j = I= 12 + 8 j 12 + 8 j = 0.86 81.66 V = 60 25° = 81.86 – 25 = 56.86° Power dissipated by the voltage source, P = IV cos 60 0.86 = × × cos 56.86 2 2 = 14.1 W XL = L = 2 fL = 2 × 50 × 19.1 × 10–3 = 6Ω

6-5.1

Z1 = 10Ω , Z2 = R + j × 2 = 8 + 6j

I1 =

100 0º = 10 0º A 10

Solution Manual (6) - 4

I2 =

100 0º 100 0º = = 10 –36.86º A 8 + 6 j 10 36.86º

I s = I1 + I 2 = 10 0°+ 10 –36.86º = (10 + j 0) + (8 – 6 j ) = 18 – 6 j = 18.97 -18.43° A Real power, Total PT = V I Cos = 100 × 18.97 cos 18.43º = 1800W Reactive power, Total QT = VI sin = 100 × 18.97 sin 18.43º = 600 VAR Real power, PZ1 = VI1 Cos 1 = 100 × 10 × cos 0º = 1000 W Reactive power, QZ1 = VI1 sin °1 = 100 × 10 × sin 0° = 0 VAR Real power PZ2 = V I2 cos 2 = 100 × 10 cos 36.86º = 800 W Reactive power QZ2 = VI2, sin 2 = 100 × 10 sin 36.86º = 600 VAR

I1 =

6-5.2

230 0º 230 0º = 4 + 5 j 6.40 51.34º

= 35.93 –51.34ºA I2 =

230 0º 230 0º = 8 + j10 12.8 51.34º = 17.96 –51.34º A

(i)

(ii)

IS = I1 + I2 I s = 35.93 –51.34º + 17.96 –51.34º = (22.44 – j 28.05) + (11.21 – 14.02j) = 33.65 – j 42.07 = 53.87 –51.34º Total admittance (Y) Z=

Vs 230 0º = = 4.26 51.34º W ls 53.87 –51.34º

Y=

1 1 = = 0.234 –51.34º Z 4.26 51.34° IS = 53.87A, Power factor = cos = cos 51.34 = 0.624 kg give

(iii) IC = IT Sin 51.34 for unity pt IC = 53.89 Sin 51.34 VS = 42.08 XC VS 230 = = 5.46W 42.08 42.08 1 XC = wc 1 1 C= = = 5.83´10 –4 F X CW 5.46 (314) XC =

Solution Manual (6) - 5

6-5.3

v(t) = 100 sin 500 t 1 1 XC = = 2 × 500 × 25 × 10 −6 c = 12.75 Ω XL = L = 2 × 500 × 0.05 = 157.1 Ω Z = 10 + j157.1 – j12.75 = (10 + j144.35) Ω 100 V= 0° = 70.71 0° 2 V 70.71 I= = = 0.49 −86.04° Z 144.69 86.04° Power factor = cos = 0.069 Power delivered by the source = I2 R = (0.49)2 × 10 = 2.4 watts Reactive power supplied by the source = I2xz = (0.49)2 ×144.35 = 34.66 VAR Reactive power supplied by the capacitor = (0.49)2 × XC = (0.49)2 × 12.75 = 3.06 VAR Reactive power by the inductor = (0.49)2 XL = (0.49)2 × (157.15) = 37.72 VAR

6-5.4

The circuit can be redrawn as shown.

ω = 50 rad/s XL1 = 0.05 × 50 = 2.5 Ω XL2 = 0.01 × 50 = 0.5 Ω XC1 =

1 = 400 Ω 50´50´10 –6

XC2 =

1 = 333.3 Ω 50´60´10 –6

Total impedance Z = 5 + j2.535 = 5.61 26.89º V = 99.98 0º ;

6-5.5

I=

99.98 0º = 17.82 –26.89º 5.61 26.89º

Power factor = cos θ = 0.891 Active power = VI cos θ = 1587.7 watts Reactive power = VI sin θ = 806.2 VAR Apparent power = VI = 1781.9 VA ZT = (2 + j) || (5 + j3) = 1.61 27.78º = 1.42 + j0.752

cos θ =

R = 0.88 Z

P = 1000; I2R = 1000 W. I = 26.53 A

Solution Manual (6) - 6

V = IZ = 26.53 × 1.61 27.78 = 42.83 27.78º volts Papp = VI = 42.83 × 26.53 = 1136.36 VA Preactive = VI sin θ = 529.6 VAR 6-5.6

100 – 20i1 + 3j (i1 – i2) = 0 3j(i2 – i1) – 2ji3 = 0 i3 – i2 = 0.1 Vx = 0.1 (–3j) (i1 – i2) i3 = i2 (1 + 0.3j) + 0.3ji1 3j(i2 – i1) – 2j [i2 (1 + 0.3j) – 0.3ji1] = 0 i2 ( j + 0.6) + i1 [–3j – 0.6] = 0 i (3 j + 0.6) 3.06 78.69 i2 = 1 = i1 0.6 + j 1.166 59.04

= i1 2.624 19.65° = [2.47 + 0.88 j ] i1 100 – 20i1 + 3j [i1 – 2.47i1 – 0.88ji1] = 0 i1 [–20 + 3j – 7.41j + 2.64] + 100 = 0 i1 [–17.36 – 4.91j] + 100 = 0 100 100 i1 = = = 5.88 −14.25° 17.36 + 4.41 j 17.19 14.25 Power factor = cos (–14.25) = 0.969 100 × 5.58 Pavg = VI cos = × 0.969 2 100 × 5.587 Preactive = VI sin = × sin 14.25° 2 = 68.66 = 68.67 VAR 100 × 5.58 Preactive = VI = = 279 VAR 2

Solution Manual (6) - 7

CHAPTER 7 (Solution Manual) 7-1.1

The node equation can be taken as

V – 50 0º V V =0 + + 5 + j6 – j5 9 + j5 é 1 1 1 ù 50 0º ú= Vê + + êë5 + j 6 – j 5 9 + j 5 úû 5 + j 6 V = 35.5 –68.25º

∴

Ix = 7-1.2

V = 3.4 –97.3º 9 + j5

Given network consists of three meshes Apply KVL for the mesh 1

5I1 + j5(I1 – I2) – 30 0° = 0 (5 + j5) I1 – j5I2 = 30 0°

(1)

Apply KVL for the mesh 2.

(2 + j3) I2 + 6(I2 – I3) + j5(I2 – I1) = 0 –j5I1 + (8 + j8)I2 – 6I3 = 0

(2)

Apply KVL for the mesh 3.

4I3 + 20 0° + 6(I3 – I2) = 0 –6I2 + 10I3 = –20 0°

(3)

⎡5 + j 5 – j 5 0 ⎤ ⎡ I1 ⎤ ⎡ 30 0° ⎤ ⎢ – j 5 8 + j8 –6 ⎥ ⎢ I ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ 0 –6 10 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ –20 0°⎥⎦

I2 =

5 + j5

30 0°

0

– j5

0

–6

0 –20 0° 10 5 + j5 – j5 0 – j 5 8 + j8 –6 0

–6

+10

Solution Manual (7) - Page 1

I2 =

5 + j 5(–120 0°) – 30 0°(– j 50) 5 + j 5[(8 + j8)10 – 36] + j 5(– j 50)]

I2 =

–( 5 + j 5)120 + 30( j 50 ) 5 + j 5[80 + j80 – 36 ] + 250

=

–600 – j 600 + j1500 400 + j 400 – 180 + j 400 – 400 – 180 j + 250

=

–600 + j 900 70 + j 620

= 1.325 + j1.11 = 1.734 40° 7-1.3

V – 50Vx V –100 V =0 + + 2 1+ 3 j – j4 Vx =

V ×1 1+ 3 j

50V V –100 V 1+ 3 j + + =0 2 1+ 3 j –4 j é1 1 3 j j 15 5 ù + – Vê + – – j ú = 50 ê ë2 10 10 4 4 4 úû 50 50 V= = –3.15 –1.3 j 3.4077 –157.57 V–

7-1.4

Given data Vrms = 10 V,

XL = 20 Ω,

XC1 = 10 Ω

Thevenin's theorem: To find current 'I' in the '5 Ω ' resistor.

Step 1: Calculate Rth (or) Zth. Step 2: Calculate VOC. Calculation of Equivalent Impedance: To calculate equivalent impedance between A and B Short ckt the voltage source. = 20 Ω || j20 Ω then 20 × j 20 j 400 = 20 + j 20 28.28 45°

=

400 90° = 14.14 45° Ω.

28.28 45°

–j10 Ω || –j10 Ω then

Solution Manual (7) - Page 2

XC2 = 10 Ω

– j10 × – j10 – 100 100 5 = = = = 5 –90° Ω – j 20 – j 20 j 20 j Now, 14.14 45° in series with 5 –90° 14.14 45° + 5 –90° = (9.99 + j9.99) + (0 – j5) ZAB = (9.99 + j4.99) Ω = Zth Calculation of Vth: I1 =

10 20 + j 20

=

10 28.28 45°

I1 = 0.35 –45° A 10 10 = j0.5 = 0.5 90° A = – j10 – j10 – j 20

I2 =

(j20)0.35 –45° + Vth – (–j10)0.5 90° = 0 Vth = (–j10)(0.5 90° ) – (j20)(0.35 –45° ). = –j10(0.5j) – j20(0.247 – j0.247) = 5 – j4.94 – 4.94 = (0.06 – j 4.94) V I = =

I=

V Z th + R

0.06 – j 4.94 9.99 + j 4.99 + 5

0.06 – j 4.94 4.94 –89.3° = 14.99 + j 4.99 15.79 18.41°

I = 0.25 –107.71° A. 7-2.1

(7 – j8)I1 – (4 –j8)I2 = 100 0º – (4 – j8)I1 + (4 – j4)I2 = –50 90º Δ=

T – j8 –(4 – j8) = 44.18 5.19º –(4 – j8) (4 – j 4)

Solution Manual (7) - Page 3

100 0º –(4 – j8) = –j600 –50 90° (4 – j 4) D 600 –90º I1 = 1 = = –(1.23 + j13.52) A D 44.18 5.19

Δ1 =

Δ2 =

7 – j8 100 = –j1150 –4 – j8 –50 j

I2 =

D2 = – (2.3 + j25.9) A D

Current I' due to the 5 0º source = 5 0º Current I" due to the 3 20º source = 0 Current I"' due to the 10 0º source = 0 VAB = 5 0º = (2 – i) + 10 0º VAB = 20 – j5 ZAB = Z – j IN =

V AB 20 – j 5 = 9.2 12.52º = Z AB 2– j

IN = (8.99 + j2) A 7-2.2

Nodal equation can be written as

V V V = 10 0º + + 3 – j10 2 + j 5 V = 24 10.34º P10 = V I = 240 watts (from the mesh equation) Δ3 = Δ=

j10 0 = 300 j 3 – j10 30 j10

2 – j5

3 – j10

j10

V = I3Z = 24 126.18º

I3 = 2.38 – j3.8

= –(56 – j35) ∴

P = 240 W

Solution Manual (7) - Page 4

Δ2 =

0

2 – j5

30 5º

j10

= +161.5 111.8º

7-2.3

X1µF = –j100 X50mH = –jωL = 500j 1k Ω || 1µF

(103 )(– j100) 1000 – j100 (1000)(– j100) = = 9.995 –84.29º 1004.987 – 5.71

=

i1 =

1 ( 2)1000

A

–9.995 –84.29º (i2 – i1) –500 × 103 × i2 – 0.5 × 500 × 103 × i2 – 500ji2 = 0 i2[9.995[ –84.29º + 500 × 103 + 0.5 × 500 × 103 + 500j]] = 9.995 –84.29º i1 I2 = 9.43 × 10–9 90º A. ∴

V1 = (500 × 103)i2 4.715 × 10–3 90º volts.

7-3.1

(a) When V1 = 50 90° volts, the source is acting alone.

j5 and (3 + j5) Ω are parallel, =

( 3 + j 5) j 5 ( 3 + j 5) + j 5

(–25 + J15) = (0.688 + j2.7) Ω 3 + J 10

Solution Manual (7) - Page 5

Total Z = 5 + (0.688 + j2.7) = (5.688 + j2.7) Ω

∴

IX =

50 90° j 50 = (3.4 + j7.174)A = 5.688 + j 2.7 5.688 + j 2.7

By current-division rule, I1 =

(3.4 + j 7.174 )( j 5) –35.87 + j17 = (0.572 + j3.758) A = ( 3 + j 5) + ( j 5 ) 3 + j10

(b) When V2 = 50 0° volts, the source is acting alone.

5 Ω and (3 + j5) Ω are parallel,

5(3 + j 5) = 2.753 + j1.4 5 + 3 + j5

Total impedance = (j5) + (2.753 + j1.4) = (2.753 + j6.4) Ω Iy =

50 0° = (2.836 – j6.6) A ( 2.753 + j 6.4 )

I2 =

–(2.836 – j 6.6 )(5) –14.18 + j 33 = – (0.572 + j 3.758) 5 + ( 3 + j 5) 8 + j5

By the superposition theorem, I = I1 + I2 = (0.572 + j3.758) – (0.572 + j3.758) = 0 7-4.1

By converting star-to-delta and delta-to-star transformation, we have

Solution Manual (7) - Page 6

I=

100 0º 2.66 + j 5.4 +1.32 – j 2 + 4 – j 5

I = 12.3 11.34º VAB = I(5.99 – j3.3) = 84.132 –17.51º ZAB = {(1.99 + j1.7) || (5.99 – j3.3)} + 1.99 + j1.7 ∴

ZAB = 4.75 32.62º

∴

IAB =

84.132 17.51º 19.17 + j 2.56

IAB = 4.38 9.84º 7-4.2

(a) Zth = ZN:

⎛ 10 × 6 ⎞ ( j15)(– j8) Zth = Zn = ⎜ + ⎝ 10 + 6 ⎠⎟ j15 + (– j8)

= (3.758 – j17.14) Ω (b) Vth = VOC:

∴

I1 =

100 0° = 6.25 0° A 10 + 6

I2 =

100 0° = –j14.28 A ( j15 – j8)

Vth = (j15)I2 – (10I1) = (j15)(–j14.28) – (10(6.25)) = 214.2 – 62.5 = 151.7 volts.

(c) IN = ISC:

Solution Manual (7) - Page 7

KVL to the loop 1: –100 + 10(I1 – I2) + 6(I1 – I3) = 0 16I1 – 10I2 – 6I3 = 100

(1)

KVL to the loop 2: j15 I2 + 10(I2 – I1) = 0 –10I1 + (10 + j15) I2 = 0

(2)

KVL to loop 3: (–j8) I3 + 6(I3 – I1) = 0 –6I1 + (6 – j8) I3 = 0

(3)

From equations (1), (2) and (3),

16 100 –6 –10 0 0 0 (6 – j 8) –6 = (2.11 – j4.628) A I2 = 16 –10 –6 0 –10 10 + j15 0 6 – j8 –6 16 –10 100 –10 10 + j15 0 –6 0 0 = (3.96 + j3.82) A I3 = 16 –10 –6 –10 10 + j15 0 0 6 – j8 –6

IN = ISC = I3 – I2 = (1.85 + j8.448) A

By Thevenin's theorem,

∴

I =

Vth 151.7 0° = Z th + Z L (3.75 – j17.14 ) + ( 5 + j 4 )

= (5.326 + j8) A

Solution Manual (7) - Page 8

By Norton's theorem, I=

I N .Z ZN + ZL

(1.85 + j8.448)(3.75 – 17.14 ) = (5.32 + j8) A (3.75 – j17.14 ) + ( 5 + j 4 ) The result is verified.

=

7-4.3

–Vab 90º (100 – j30) 166.7 –Vab 90º 104.4 16.699º = 166.7

VCC =

= –Vab 0.626 73.3º = Vab [0.179 – j0.6]

–Vab 90º 166.7 V R = OC = 104.4 –16.7º I SC

ISC =

7-4.4

= (100 – j30) Ω As i = 0, consider the open-circuit voltage across ab Then Vab = 1 0º × 1j VOC = Vab = 1 90º VOC = 1 90º = VTh And short the terminals ab to calculate circuit current.

I = –1 J(I –1 0º ) 4

Solution Manual (7) - Page 9

æ1

ö + j ÷÷ = + 1.J è4 ø

Iç ç

I (1 + 4 j ) = 1.J 4 4j I= q 1+ 4 j ISC = I =

4 17

θ = tan–1(1/4) ZTh =

VOC 1 90º = 4 I SC q 17

= ZTh = 7-5.1

17 90 – q 4 17 tan –1 (4) 4

VA = I1(j2) VB = I2(–j) I1 =

10 0º 10 0º = 5 + j 2 5.38 21.8º

I1 = 1.86 –21.8º I2 =

5 0º = 1.21 14º 4– j

VAB = VA – VB = 3.72 68.2º – 1.21 –76º = 4.8 76.8º ZAB = RAB = (5 || j2) + (4 || – j) = (0.93 + j0.78) Ω IAB =

4.8 76.8 = 4 36.81º = (3.2 + j2.4) A 1.2 39.99º

Thevenin’s equivalent circuit: 4.8 76.8º = (1.1 j4.7) in series with (0.93 + j0.75) Ω 7-5.2

Norton’s equivalent circuit: (3.2 + j2.4)A current source parallel with (0.93 + j0.75) Ω From the circuit, VAB = I × (5 + j5) where

I = 25 0º ×

– j3 = 7.29 –119.05º 9 + j5

VAB = 7.29 –119.05º × (5 + j5) = 51.54 –74.05º ZAB = 4|| (5 + j5) = 2.75 15.95º = (2.64 + j 0.75) Ω 15.95º = (2.64 + j 0.75) Thevenin's theorem VAB in series with ZAB

Solution Manual (7) - Page 10

Norton's theorem IAB in parallel with ZAB *

7-6.1

ZAB = (2 || – j3) + (4 || j5) = 3.964 15° 50 0° × 3 −90° = 41.67 −33.69° 2 − j3 50 0° × 4 VB = = 31.25 −51.34° 4 + j5 VA =

VAB = VA – VB = 15.15 + j12.9 = 15.2 4.87° 15.2 = 1.989 A 3.82 + 3.82 Pmax = I2R = (1.989)2 × 3.82 = 15.11 watts I=

7-6.2

Loop equations are (10 –j3)I1 – (–j3)I2 = 50 0º –(–j3)I1 + (8 + j1)I2 = 0 Δ=

10 – j 3

j3

j3

8 + j1

= 93.06 –3.65º ;

Solution Manual (7) - Page 11

Δ=

I1 =

5 0º

j3

0

8 + j1

= 4.33 7.12º

D1 433 7.12º = 4.65 15.77º = D 93.06 –8.65º

P = 50 0º × 4.5 15.77º = 232.5 15.77º Power P = 232.5 watts 7-6.3

To obtain Thevenin's equivalent, calculate Thevenin's impedance (Zth), remove voltage source with short circuit.

10(3 – j 4 ) + 8 = Zth 10 + 3 – j 4 30 – j 40 + 8 = 2.97 –j2.16 + 8 13 – j 4

Zth = (10.97 – j2.16) Ω Calculation of Thevnin's Voltage (Vth) Apply KCL, V V – 20 0° =0 + 3 – j4 10

10 V + (3 – j4)(V – 20 0° ) = 0 10 V + 3V – 60 0° – j4V + (j4)20 0° = 0 13 V – 60 0° – j4V + 4 90° 20 0° = 0 (13 – j4)V = 60 0° – 80 90° (13 – j4)V = 60 – j80

Solution Manual (7) - Page 12

V=

60 – j80 = (5.945 – j4.324)V 13 – j 4

Apply KVL. –10 45° + Vth – (5.945 – j4.324) = 0 Vth = (5.945 – j4.324) – 10 45° Vth = 5.945 – j4.324 – 7.071 + j7.071 Vth = –1.126 – 11.395 j = 11.39 264.4° Volts = 11.39 264.4° Volts voltage source is in series with the impedance of (10.97 – j2.16) Ω 7-6.4

In the circuit shown, the resistance R2 is fixed. Here, the maximum power transfer theorem does not apply. Maximum current flows in the circuit when Rg is minimum. For the maximum current Rg = 2 Ω. ∴ Z = Rg + j5 + R2 = 2 + j5 + 10 = (12 + j5) Ω I=

100 0° = 7.1 – j2.95 12 + j 5

= 7.688 –22.56° Maximum power = I2R2 = (7.688)2 10 = 593 watts

Solution Manual (7) - Page 13

CHAPTER 8 (Solution Manual)

8-3.1

v(t) = 50 sin ωt

BW = 500 rad/s VC = 400 V

Z = R = 100 Ω

Vmax = 50;

Vrms =

50 =35.35 volts 2

VC = 400 V Q=

VC 400 = = 11.3 V 35.35

ω2 = ω1 = 500 rad/s Q=

∴

wr w 2 – w1

= 11

ωr = 5500 rad/s fr = 875.35 Hz

R ; L

ω2 – ω1 = fr =

ω=

8-3.2

1 2p LC

∴L=

100 = 0.2 H 500

∴C=

1 = 0.165 µF (2p f r ) 2 L

f1 = fr –

R = 836.28 Hz 4p L

f2 = fr +

R = 914.42 Hz 4p L

1 1 = LC 0.5´20´10 –6

ω = 316.22 rad/s f = 50.3 Hz I=

20 1 1 = 0.4 A; XC = =158 Ω = 50 w C 36.22´ 20´10 –6 VC = IXC = 63.2 V Q=

8-3.3

w L 316.22´0.5 R

=

50

= 3(approx.)

Q = 10; ω = 200 rad/s I = 0.5 A,

V = 10 V

ZT = 40 Ω

R=

V 10 = = 20 Ω I 0.5

Solution Manual (8) – Page No. 1

Q=

wL R

=

200´ L = 10 20

L=1H Total impedance Z =

R 2 + ( X L – X C )2 (20) 2 + ( X L – X C ) 2

40 =

XL – XC = 34.64

∴

XC = 200 – 34.64 = 165.36 Ω

∴

C = 30 µF Q = 10;

ω = 100 rad/s

Imax = 10 A;

V = 100 V

8-3.4

R=

100 = 10 Ω 10 R 2 + ( X L – X C )2

Z = 14.14 Ω = Q=

∴

w L 100 L =

R

10

= 10

L=1H

(XL – XC)2 = (14.14)2 – (10)2 XL – XC =

(14.14) 2 – (10) 2 = 9.99 Ω

XC = 90 8-3.5

∴ C = 1.11× 10–4 F

v(t) = 15 sin 1800t Vrms =

15 = 10.6 V 2

ωr = –

1 LC

= 316 rad/s

fr = 50.3 Hz Q=

wL R

=

316´5 = 105 15

f1 = fr –

15 R = 50.3 – = 50.06 Hz 4p ´5 4p L

f2 = fr +

R = 50.54 Hz 4p L

Bandwidth = f2 – f1 = 0.48 Hz

Solution Manual (8) – Page No. 2

8-3.6

Imax when L = 0.5 H

Im

when L = 0.2 H

2 ω=

1 LC

ωL –

1 =R wc

∴ Q= 8-3.7

∴

C=

1 2

w L

=

1 = 50 µF (200) 2 ´0.5

R = 60 Ω

1 1 = = 1.66 ≈ 1 w cR 200´50´10 –6 ´60

The frequency for which the voltage across the inductor is maximum, fL =

1

1

2 π LC

1–

R 2C 2L

1

= fr

1–

1 2Q 2 1

= 1000

1–

= 1002.5 Hz

1 2 × 102

The maximum voltage across the inductor, (VL) max = IXL =

V ⎛ 1 ⎞ R 2 + ⎜ wL L – wL c ⎟⎠ ⎝

wL2 .V .

=

2

⎛ ⎞ 1 [(2 π)2 (1002.5)2 ] (10) ⎜ 2 3 2⎟ ⎝ (2 π) (10 ) ⎠ 1 ⎛ (2 π)2 (1002.5)2 ⎞ +⎜ – 1⎟ 102 ⎝ (2 π)2 (103 )2 ⎠

8-3.8

wL c

R c wL2 + (w 2 LC – 1)2 2 2

1 wr2

⎞ 1 ⎛ 2 1 + ⎜ wL . 2 – 1 ⎟ 2 Q ⎝ wr ⎠

=

wL LV .

wL L =

2

= 100.37 V

The frequency at which the maximum voltage across the capacitor, fC =

=

1 2 π LC

1–

R2 .LC 2L

Solution Manual (8) – Page No. 3

1 1 R2 – 2 π LC 2 L

=

R C

1

2 π LC

R2c

–

1 2

⎡R C ⎤ ⎡ 1 L⎤ 1 1 = ⎢ .

– ⎥ .⎢ ⎥ 2 2 ⎣ L ⎦ ⎣ 2π C L ⎦ R c

fr Q

=

Q = 25 = Q= fr =

LC =

L 2

–

1 1 L and fr = R C 2 π LC

where Q =

8-3.9

L R 2C

wL R

VC = 25 V 1 2p LC

∴

VC = 25 × 5 = 125 V

= 200 × 103 Hz

1 200´103 ´ 2p

L = 3.17 mH

wL R

= 25

Imax = 8-3.10

∴

R = 159 Ω

5 = 0.031 A 159

(a) VL at resonance = Q V = 10 × 10 = 100 V (b) The frequency for which the voltage across the inductor is maximum, fL =

1

1

2 π LC

1

= fr

1– ⎡ 1 1 L⎤ and Q = ⎢ where fr = ⎥ R C⎦ 2 π LC ⎣

1

= 1000

1–

1 2 × 102

= 1002.5 Hz 8-5.1

Equivalent parallel impedance

Solution Manual (8) – Page No. 4

1–

1 2Q 2

R2 c 2L

= (5 + j3) || (10 – j30) = 5.97 20.35º = 5.59 + j2.07

∴ 8-5.2

X3 = 2.07 Ω

ωr =

=

1 LC

1 2´3´10–6

= 408.25 rad/s

Voltage applied

10 × 12 = 84.85 V 2

V = IR =

IL =

V

wr L

=

Q= ω2 – ω1 =

84.58 = 0.104 A 408.25´2

I L 0.104´ 2 = 0.015 = 10 I 1 1 = RC 12´3´10 –6

f2 – f1 = 4420.96 Hz ω1 =

1 + 2 RC

æ 1 ö2 1 ç ÷÷ + = 5.99 rad/s ç LC è 2 RC ø

f1 = 0.95 Hz ω2 = 13888.9 + 13894.8 = 27783.7 rad/s f2 = 4421.9 Hz 8-5.3

Zeq = (3 + j12) || (RC – j12.5) =

(3 + j12)( RC – j12.5) 3 + RC – j 0.5

Imaginary part must be zero.

–112.5 + 36 RC – 37.5 RC + 12 RC2 + 1.5 RC + 75 = 0 12 RC2 – 37.5 = 0 ∴

RC = 1.77 Ω

8-5.6 (a) Resonance frequency, fr =

1 2π

=

R2 1 – 2L LC L

1 1 22 – 2 π (10 −3 )(10 × 10 –6 ) (10 –3 )2

Solution Manual (8) – Page No. 5

= 1559.4 Hz

(b) Dynamic resonance =

L CRL

=

(c) Q-factor =

10 –3 (10 × 10 –6 )(2)

= 50 Ω

wr L (2 πfr )L 2 π(1559.4 )10 –3 = = RL RL 2 = 4.99

Bandwidth =

fr 1559.4 = 318.3 Hz = Q 4.99

Solution Manual (8) – Page No. 6

CHAPTER 9 (Solution Manual) 9-4.1

Taking VRN as reference, IR =

254 0º = 25.4 0º A 10 0º

IY =

254 –120º = 25.4 –146.8º A 10 26.8º

IB =

254 120º = 25.4 146.8º A 10 –26.8º

Neutral current IN = –(IR + IY + IB) = 17.1 0º 2

2

2

Load power = I R × 10 + IY × 8.925 + I B × 8.925 = 17967.7 W 9-4.2

Assume RYB phase sequence VRN =

440 3

= 254 V

Taking VRN as reference, VRN = 254 0º V VRN = VYN = 254 –120º ; and

VBN = 254 –240º V

The three line currents are (i)

(ii)

IR =

254 0º = 34.89 –29.74º A 7 + j4

IY =

254 –120º = 70.44 –153.69º A 3 + j2

IB =

254 –240º = 27.55 –252.52º A 9 + j2

The neutral current IN = – (IR + IY + IB) = – (34.89 –29.74º + 70.44 –153.69º + 27.55 –252.52º ) = 46.758 28.41º A

9-4.3

VL = 440 V

Z1 = 2 + 3j = 3.6 56.3

VP = 440 V

Z2 = 1 – 2j = 2.2 –63.43 Z3 = 3 – 4j = 5 –53.13

Solution Manual 9 – Page No. 1

9-6.1

Taking R-phase voltage as reference VRN = 254 0º ;

VYN = 254 –120º ;

VBN = 254 120º

The three line voltages VRY, VYB and VBR are 30º ahead of their respective phase voltages.

IR =

254 0º = 50.8 0º A 5

IY =

254 –120º = 25.4 –120º A 10

IB =

254 120º = 16.936 120º A 15 2

PR = I R × 5 = 12903.2 W 2

PY = IY × 10 = 6451.6 W 2

PB = I B × 15 = 4302.4 W 9-6.2

Phase currents

Line currents

IRY =

440 0º = 22 –30º 20 30º

IYB =

440 –120º = 22 –120º 20 0º

IBR =

440 120º = 22 150º 20 –30º

IR = IRY – IBR = 38.1 – j22 = 44 –30º A IY = IYB – IRY = –30.05 – j8.05 = 31.1 194º A IB = IBR – IYB = –8.05 – j30.05 = 31.1 104º A

9-6.3

Taking VRY as reference, VRN = 231 –30º ; VYN = 231 –150º ; VBN = 231 90º

231 –30º 231 –150º 231 90º + + 25 10 15 = 53.25 183.6º VO'O = 1 1 1 + + 25 10 15 Voltage across the load phases VRO' = VRO – VO'O = 231 30º – 58.25 183.6º = 281.32 –23.42º VYO' = VYO – VO'O = 231 –150º – 58.25 183.6º

Solution Manual 9 – Page No. 2

= 180.6 218.25º VBO' = VBO – VO'O = 231 –270º – 58.25 183.6º = 241.8 76º Phase currents IR =

9-6.4

281.32 –23.42° = 11.25 –23.42º 25

IY =

180.6 –218.25° 218.25° = 18.06 218.25º 10

IB =

241.8 76º = 16.12 76º 15

1 10 –90

1 , 10 0

Y2 =

1 , 5 53.14

Y1 = 0.1 0 ,

Y2 =

1 , Y3 = 0.1 90 5 -53.14

Y1 =

Von =

=

=

V1 Y1 + V2 Y2 + V3 Y3

Y3 =

100 = 57.73V Y1 Y2 Y3 3 57.73 –30 ´ .1 0 + 57.73 – 150 ´q .2 – 53.14 + 57.73 90 ´ .1 9 V=

(.1 + .2 –53.14 + .1 j ) 5.77 – 30 + 11.54 – 203.14 + 5.77 180

(0.1 + 0.2 – 53.14 + 0.1 j )

4.99 - 2.88 j + (– 10.6 + 4.5 j ) + (-5.77 + 0 j ) (.319 – .16 j ) -11.38 + 1.62 j 11.49 171.89 = = .35 – 26.6 .35 – 26.6

=

VON = 87.43 -142.98º Volts 9-7.1 From Additional Solved Problem 9.3, it may be seen that when connected in delta, the power drawn would be three times as that for star connection. ∴

9-7.3

P4 = 1000 W;

PΔ = 3 × PY = 3000 W

Converting star-connected impedances into delta, the equivalent impedance/phase of the system is given by

10 30º × 15 35º = 6 32º 10 30º + 15 35º The phase currents

IR =

173 0º = 28.83 –32º A 6 32º

IY = 28.83 –152º and The line currents are

I1 = 50 –62º ;

and

I3 = 50 58º

IB = 28.83 88º A

I2 = 50 182º

Solution Manual 9 – Page No. 3

Power =

9-7.4

3 × 173 ×50 cos 32º = 12705 watts.

VL = 440V 50 HZ

XC =

1

1 2 p fc 2p ´50´100 m F

=

= 31.83Ω For Delta Connection,

VL = VP and I L = 3I P V P = 440 V,

IP =

VP 440 = Z P 31.83

= 13.83 IL =

3´13.83A

= 23.94. A For star connection IL = IP = 23.94 A VP = 254 V ZP =

VP 254 = I P 23.94

= 10.61Ω ZP = XC

1 2pfc 1 1 C= = = 173m f 2p fxc 2p ´50´10.61 XC =

9-7.5

ZR ZY + ZY ZB + ZB ZR = (3 + j2)j9 + j9(3) + 3(3 + j2) = 60.67 98.53º Equivalent impedances in delta connection ZR =

60.67 93.53° = 20.22 98.53º 3

ZY =

60.67 98.53 = 16.82 64.83º 3 + j2

ZB =

60.67 98.53° = 6.74 8.55º j9

Solution Manual 9 – Page No. 4

9-8.1

Source phase voltage = VRN = 230 60º Line voltage =

3 × phase voltage and 30º ahead of the respective phase voltage

∴

VRY =

3 × 230 90º

Total load power = 15000 watts 3VPh IPh cos φ = 15000 3VRh IR cos φ = 15000 ∴

IR = 15.688 A

IR lags VRY by 36.86º (∵ cos p = 0.8) IR = 15.688 53.14º All phase currents are equal in magnitude and displaced by 120º from one another.

Line current in the line 1, I1 =

3 × 15.688 = 27.172 A

lag IR by 30º I1 = 27.172 23.4º

∴

All the line currents are equal in magnitude and displaced by 120º from one another.

9-8.2

ZR = 7 10º

VL = 400

400 = 231V 3

VP =

ZY = 8 30º ZB = 8 50º

VR = 231 0º ,

IR =

231 0 , 7 10

IR =33 –10 , 9-8.4

VY = 231 –120 ,

IY =

231 –120º , 8 30

I Y = 28.87 –150 ,

VB = 231 –240

IB =

231 –240 8 50

I B = 28.87 –29

VL = 440 V

VP =

440 = 254 V 3

IP = 20A, IL = 20A f = 40º

ZP =

VP 254 = = 12.7 40 20 IP

XL = 8.16 8.16 = wL 8.16 = 2 fL

L=

ZP = 9.72 + j8.16

8.16 8.16 = = 25.9mH 2p f 100p

P = 3. VL I L cos f = 3. 440.20 ´ cos 40°

Solution Manual 9 – Page No. 5

9-9.1

P = 23.35kW W1 = VL IL cos (30 + ) = 440 ´ 20 ´ Cos (30 + 40) W1 = 3009.7 W W2 = VL IL cos (30 – ) = 440 × 20 × cos (30 – 40) W2 = 8666.3 W Total power = 3500 – 300 = 3200 W tan θ =

⎛ 3500 + 300 ⎞ = 2.05 3⎜ ⎝ 3500 – 300 ⎟⎠

Power factor = cos θ = 0.437 9-11.1

(i)

Input = 1000 + 250 = 1250 W

(ii)

tan θ =

3

ω 2 – ω1 = 1.04 ω 2 + ω1

PF = cos θ = 0.693 (iii)

Input power = ∴

(iv)

IL =

3 VL IL cos θ 1250 3 × 440 × 0.693

= 2.36 A

Motor output = 0.8 × 1250 × = 1000 watts

Solution Manual 9 – Page No. 6

CHAPTER 10 (Solution Manual) 10-2.1

(a)

M=K

L1 L2 40 × 150 × 10–3 = 54.22 mH

= 0.7 (b)

M=

L1 L2 = 77.46 mH

L1 + L2 + 2M = 0.8

10-2.2

L1 + L2 – 2M = 0.5 Solving the above equations, 4M = 0.3 M = 0.075 H 10-2.3

Let i be the current in the circuit. v = 13

∴

di dt

L = 13 H

10-2.4

Mesh equations are

di2 d (i1 – i2) – 2 dt dt

(1)

di2 di2 di di di d d + (i2 – i1 ) + 2 2 – 2 (i1 – i2 ) + 2 + 3 2 + dt dt dt dt dt dt dt

(2)

v=2

0=

2

10-2.5

di1 di –4 2 =v dt dt

–4

di1 di + 12 2 = 0 dt dt

∴

3v = 2

Hence,

L=

v = v1 + v2

di1 dt

∴

v=

2 di1 3 dt

2 H 3

(1)

and

v1 = 1.

di di2 di – 0.5 3 + 0.5 2 dt dt dt

v2 = 1.

di di2 di – 0.5 3 + 0.5 2 dt dt dt

i1 = i2 + i3

Solution Manual (10) – Page No. 1

(2)

v = 1.

di3 di di di – 0.5 2 – 0.5 2 = Leq . 1 dt dt dt dt

Since v = v1 + v2 di3 di di di2 di di – = 1.5 2 – 0.5 3 + 1.5 2 – 0.5 3 dt dt dt dt dt dt

2

di3 di =4 2 dt dt di3 di =2 2 dt dt

From Eq. (2),

(3) di di1 di = 2 + 3 dt dt dt

(4)

Substituting Eq. (3) in Eq. (4), di1 di di = 2 +2 2 dt dt dt di2 1 di1 = 3 dt dt di3 1 di1 2 di1 =2× = 3 dt 3 dt dt

∴

Leq.

di3 di1 di =3 2 – dt dt dt

Leq.

di1 1 di1 2 di1 1 di1 =3× – = 3 dt 3 dt 3 dt dt

The equivalent inductance between the terminals ab is Leq. = 10-3.1

v1 = L1

di1 di +M 2 dt dt

v2 = L2

di2 di +M 1 dt dt

1 H 3

L1, L2 are self-inductances of the coils and M is the mutual inductance between coils.

10-5.1

v1 = L1

di1 di –M 2 dt dt

v2 = L2

di2 di –M 1 dt dt

di1 = 100 cos (40t –30º) dt di2 = 80 cos (40t – 30º) dt v1= 181.44 cos (40t – 30º) V

Solution Manual (10) – Page No. 2

v2= 202.88 cos (40t – 30º) V 10-5.2

For the first mesh, j2i1 + j3i2 + j4i3 + j5i3 + j4i1 + j3i2 = 10º 0º j6i1 + j6i2 + j9i3 = 10 0º

(1)

For the second mesh, j6i2 + j3i1 + j3i3 – j5i3 – j4i1 + j3i2 = 0 – ji1 + j3i2 – j2i3 = 0 10-5.3

(2)

For the first mesh, (5 + j2)i1 + j4(i1 – i2) + j3i2 = V1 (5 + j6)i1 – ji2 = V1

(1)

For the second mesh, (j1 – j4 + j3)i2 + (i2 – i1)j4 + j2i2 + j2i2 = j3i2 – j3(i2 – i1) = 0 –ji1 + j2i2 = 0

10-6.1

10-7.1

v1 = 2

di di1 di +5 2 –2 3 dt dt dt

v2 = 4

di2 di +5 1 dt dt

v3 = 6

di3 di –2 1 dt dt

(2)

Inductance matrix ⎡ 2 5 –2 ⎤ ⎢5 4 0⎥ ⎢ ⎥ ⎣⎢ –2 0 6 ⎥⎦

The inductance matrix

⎡ L11 ⎢ L = ⎢ L21 ⎢⎣ L31

L12 L22 L32

L13 ⎤ ⎡ 8 –2 1 ⎤ ⎥ L23 ⎥ = ⎢⎢ –2 4 –6 ⎥⎥ L33 ⎥⎦ ⎢⎣ 1 –6 6 ⎥⎦

The diagonal elements (8, 4, 6) in the matrix represent the self-inductances of three coils 1, 2, and 3 respectively. The second element in the first row (–2) is the mutual inductance between coils 1 and 2, the negative sign indicates that the current in the first coil enters the dotted terminal, and the current in the second coil enters at the undotted terminal. Similarly, the remaining elements are fixed.

Solution Manual (10) – Page No. 3

VS = –j4i1 + j4(i1 – i2) + j2i2

10-8.1

i2 = 0.5 A VS = (–j4 + j4)i1 – j4i2 + j2i2 VS = –j2i2 = 1 –90º V

10-9.4

Let us consider i1 and i2 mesh currents in the primary and secondary windings.

As the currents i1 and i2 are entering the dotted terminals, the sign of the mutual inductance is to be positive.

Using KVL for the first mesh, V1 = i1(R1 + j L1) + i2 j m 10 = i1 (1 + 2j) + i2 j(4)

(1)

Similarly, for the second mesh, i2(R2 + j L 2) + i1 j m = 0 i2(2 + j8) + i1 j(4) = 0

(2)

(1 + 2j) i1 + j4i2 = 10 j4 i1 + (2 + j8)i2 = 0 10

j4

0 2 + j8 i1 = j4 1+ 2 j j4

=

(20 + j80 ) = 6.756 – j0.54 (1 + 2 j)(2 + j8) + 16

2 + j8

1 + 2 j 10 j4 0 – j 40 = i2 = (1 + 2 j)(2 + j8) + 16 2 + j12 = (–3.24 – j0.54) A V2 = i2(2) = 2(–3.24 – j0.54)

V2 (−6.48 − j 1.08) = = –0.648 – j0.108 V1 10 = 0.657 –170.53° 10-9.5

Applying KVL for the first and second meshes, V = L1

di1 di +M 2 dt dt

(1)

O = L2

di2 di +M 1 dt dt

(2)

From Eq. (2), Substituting

di2 –M di1 = L2 dt dt

di2 value in Eq. (1), dt

Solution Manual (10) – Page No. 4

V = L1

Leq .

∴

⎛ –M di1 ⎞ di1 +M ⎜ dt ⎝ L2 dt ⎟⎠

⎛ di1 M 2 ⎞ di = ⎜ L1 – L2 ⎟⎠ dt dt ⎝ ⎛ M2 ⎞ L – Leq. = ⎜ 1 L2 ⎟⎠ ⎝

10-10.1 Length of the flux path = πD = π × 25 = 78.54 cm = 0.7854 m Area of flux path = 4 × 10–4 m2 mmf = flux × reluctance; µr = 800

l

Reluctance of the iron =

μo μr A

=

0.7854 4π ×10 × 800 × 4 ×10 –4 –7

= 2.4868 × 1011 × 10–5 × 0.7854 = 0.1953 × 107 Reluctance of the air gap =

lg 1.5 ×10 –3 = μo A 4π ×10 –7 × 4 ×10 –4

=

15 × 107 = 0.2984 × 107 16 π

Total reluctance = 0.1953 × 107 + 0.2984 × 107 Total mmf = flux × reluctance = 0.48 × 10–3 × 0.4937 × 107 = 359520 AT mmf = NI I=

∴

mmf 359520 = 299.6 A = N 1200

the current required, I = 299.6 A

10-11.1 Length of the flux path = πD = π × 25 cm =

π × 25 m = 0.7854 m 100

Cross-sectional area = πr2 = π =

d2 4

π × 32 cm2 4

= 7.0686 cm2 = 7.0686 × 10–4 m2 Given N = 700; Assume µr = 1000

Solution Manual (10) – Page No. 5

Reluctance of the steel =

l

μ o μr A

0.7854

=

4π ×10

–7

×1000 × 7.0686 ×10 –4

= 8.842 × 105 Reluctance of the air gap =

lg

μo A

=

1.5 ×10 –3 4π ×10 –7 × 7.0686 ×10 –4

= 0.0168868 × 108 = 168.868 × 104 Total reluctance = 8.842 × 105 + 16.887 × 105 = 25.729 × 105 Magnetomotive force = NI = 700 × 2 = 1400 AT Flux =

mmf 1400 = = 78.4 × 10–5 Wb reluctance 25.729 × 105 = 0.784 mWb

Flux density =

φ A

=

0.784 ×10 –3 7.0686 ×10

–4

= 1.12 Wb/m2

Solution Manual (10) – Page No. 6

CHAPTER 11 (Solution Manual) 11-2.1

d 2 i(t) dt

2

+ 10

di(t) + 25i(t) = 0 dt

i(0+) = 2;

di(0 + ) =0 dt

Taking transforms on both sides, [S2I(S) – Si(0+) – i(0+)] + 10[SI(S) – i(i–)] + 25I(S) = 0 I(S) [S2 + 10S + 25] = 2S + 20 2 S + 20

I(S) =

S + 10 S + 25 2

=

10 (S + 5)

2

+

2 S+5

i(t) = 10te–5t + 2e–5t i(t) = e–5t (10t + 2) 11-2.2

The steady-state current through the circuit before closing the switch at t = 0 is i(0–) =

V A R1 + R2

After closing the switch at t = 0, the circuit consists of the resistance R2, inductor L, and the voltage V. The current through the circuit for t ≥ 0 is i(t).

Applying KVL for the circuit, V = R2 i + L L

di dt

di + R2 i = V dt

R di V + 2 i= dt L L

(1)

We compare the differential equation (1) with a non-homogeneous differential equation dx + Px = k whose solution is x = e–Pt∫k ePt dt + c e–Pt dt

where 'c' is an orbitary constant.

Rt

Here, i = c e – R2t / L + c e – R2t / L ∫ i = c. e

–

R2 t L +

V L2 .e dt L

V R2

We know that at t = 0, i =

V R1 + R2

V V = c. 1 + R1 + R2 R2

Solution Manual (11) – Page No. 1

c=

⎡ R – R1 – R2 ⎤ –VR1 V V = v⎢ 2 – ⎥= R1 + R2 R2 R (R + R ) R (R ⎣ 2 1 2 ⎦ 2 1 + R2 ) Rt

∴ i(t) =

– 2 –VR1 V e L + R2 (R1 + R2 ) R2

(2)

When R1 = R2 = 100 ohms, V = 10 V,

L=1H

At time t = 5 ms, i(t) =

–10 × 100 e 100(100 + 100 )

–100 t 1 +

10 100

=

–10 –100 × 5 ×10 –3 1 + e 200 10

=

–1 –0.5 1 = 0.0696 A + e 20 10

= 69.6 mA Voltage drop across R2 = i(t) R2 = 0.0696 × 100 = 6.96 volts Voltage across L = 10 – 6.96 = 3.04 volts 11-2.3

When the switch k is closed at time t = 0, the inductor will not allow sudden change in current, i.e., the current flows in the loop consisting of the capacitor and resistor.

At

t = 0;

and c at

V(t) = 0

dv –10 = – 40 = dt 1 / 4

t = 0,

dv = – 40 dt

Applying KVL at node A, c.

d v 1 (v – 10 ) + + ∫ v dt = 0 dt R L

1.

dv + 4v + 3∫ v dt = 0 dt

Differentiating the above equation, d2v dt

2

+4

dv + 3v = 0 dt

D1 = –1, D2 = –3 The complementary function is v(t) = c1 e–t + c2 e–3t Substituting initial voltage V(0) = 10 at t = 0 10 = c1 + c2

(1)

Solution Manual (11) – Page No. 2

dv = –c1 e–t – 3c2 e–3t dt

At

dv = – 40 dt

t = 0;

–40 = –c1 – 3c2 c1 + 3c2 = 40

(2)

Solving equations (1) and (2), we get c1 = –5 and

c2 = 15

∴ the voltage across the inductor

v(t) = –5e–t + 15e–3t volts 11-2.4

In the steady state, i(0–) = i(0+) = 2 A After t > 0, Applying Kirchhoff's voltage law to the circuit, 5i(t) + 2

di(t) + 5i(t) = 0 dt

10i(t) + 2

di(t) =0 dt

(10 + 2D) i(t) = 0 D = –5 i(t) = C1 + C2 e–5t As

t → ∞ i(∞) = C1 = 0

At

t=0 i(0) = C1 + C2 = 2 C2 = 2 i(t) = 2e–5t

11-2.5

When the switch is closed, i2(0) =

100 2 × 10

3

=

1 20

Capacitor C1 is charged to 100 V, i1(0) =

3 20

(i1 – i2)103 = 100 i1 – i2 = 1/10 I1(s) – I2(s) =

1 10 S

Solution Manual (11) – Page No. 3

3 × 103i2(t) – 103i1(t) +

1 10 –6

∫ i2 dt = 0

By KVL to Loop 2: 3 × 103

di2 (t) di t – 103 1 + 106 i2(t) = 0 dt dt

3 × 103 [SI2(S) – i2(0)] – 103 [SI1(S) – i1(0)] + 106 I2(S) = 0 I2(S) =

5 5 + 50000t

i2(t) = 5 e –5 ×10 t 4

and i1(t) = 9.99 – 8.49 e –5 ×10 t 4

11-2.6

When the switch K is closed at t = 0, application of KVL to the closed path gives 100 = Ri + +

1 i dt C∫

Differentiating the above equation, we get R

di 1 + i =0 dt C

1 di + i =0 dt RC

(1)

The solution for the above equation is i(t) = K e–t/RC The initial value of current i(0+) = 0.01 =

V at t = 0 R

0.01 = k (1) ∴

i(t) = 0.01 e–t/RC

Since the capacitor never allows sudden changes in voltage, at t = 0 it will act as a short circuit.

V = 0.01 R

i(0+) = R=

100 = 10 kΩ 0.01

The current disappears after 0.1 second.

0 = 0.01 e–0.1/RC e–0.1/RC = 0 Usually, after 4 times the time constant, the transient current disappears.

0.1 = 4 × τ = 4 RC 4C= C=

0 .1 0 .1 = 10–5 = R 104

1 × 10–5 = 2.5 µF. 4

Solution Manual (11) – Page No. 4

11-2.7

At switch K is position 1, i(t) =

V (t) 100 0° = Z 2 + j(2 π × 50 × 1m)

i(t) = 48.796 – 7.665 i = 49.39 –8.93 A At

t = 0;

I(0) = 49.39 A

When the switch is in the position '2', the closed path consists of inductance and capacitances only and apply KVL around that closed path.

L

di 1 + i dt = 0 dt C ∫

di 1 + i dt = 0 dt LC ∫ d 2i dt

2

+

1 i =0 LC

1 ⎞ ⎛ 2 D + ⎟i = 0 ⎝⎜ LC ⎠

(D2 + (100)2)i = 0

D = ±100j

The solution of the above equation is i(t) = (c1 cos 100t + c2 sin 100t) A

(1)

At t = 0 i(t) = 49.39 A Then from Eq. (1), 49.39 = c1 cos 100(0) c1 = 49.39 Differentiate Eq. (1) w.r.t. to 't'. di(t) = –100c1 sin 100t + c2 (100 cos 100t) (2) dt

–100c1 sin cos + 100 c2 cos 100t = We know that at t = 0; L

di(t) dt

di 1 + i dt = 0 dt C ∫

Solution Manual (11) – Page No. 5

Since the capacitor does not allow sudden change in voltage, L

di =0 dt

di =0 dt

From Eq. (2),

0 = –100 c1 sin 100t + c2 100 cos 100t

At

c2 = 0

t = 0;

Therefore,

i(t) = 5 cos 100t

11-2.8

50 = 10 (i1 + i2) +

1 i dt 2∫1

(1)

di2 +10i2 dt

(2)

50 = 10 (i1 + i2) + 2 Taking transforms,

1 ⎡I 200 ⎤ 50 = 10I1(S) + 10I2(S) + ⎢ 1 + 2 ⎣S S ⎥⎦ S ⎡ ⎣

I1(S) ⎢10 +

1 ⎤ –50 = – 10I2(S) S 2 S ⎥⎦ I1(S) =

–100 – 20SI 2 (S) 20S + 1

50 = 10I1(S) + 20I2(S) + 25I2(S) – 4 S

(3)

(4)

Substituting (3) into (4), we get I2(S) =

I2(S) =

80 S 2 + 2004 S + 50 40 S (S 2 + 5.05S + 0.5)

101.21 30.9 52.11 + – S S + 0.1 S + 4.94

i2(t) = 101.21 + 30.9 e–0.1t – 52.11e–4.94t 11-2.9

Solution Manual (11) – Page No. 6

10(i1 + i2) + 106

∫ i1 dt = 100 I1 (S) 100 = S S

10I1(S) + 10I2(S) + 106 ∴

I1(S) =

(1)

100 10 + 10 S 6

–

10 S 10 + 10 S

100 = 10(i1 + i2) + 0.1

6

I 2 (S)

(2)

di2 + 10i2 dt

10I1(S) + 20I2(S) + 0.1SI2(S) =

(3)

100 S

(4)

Substituting (2) into (4), I2(S) =

90 S + 10 7 S(S 2 + 10019S + 20 × 10 5 ) 3

i2(t) = 108 [5 + 4.99e–200.2t + 0.005e–99.899 × 10 t] V1 = 0.1

di2 dt 3

= 0.1 × 108 [– 4.99 × 200.2e–200.2t – 0.005 × 99.899 × 103e–99.9 × 10 t] 3

= –107 [998.99e–200.2t + 499.495e–99.9 × 10 t] V2 = Ri2(t) 3

= 109 [5 + 4.99e–200.2t + 0.005e–99.9 × 10 t] 11-2.10

d

∫ i1 dt + 3 × 10 dt (i – i ) = 100

(1)

I1 (S) 100 + 3 × 10–3 S[I(S) – I2(S)] = S S

(2)

10i1(t) + 106

–3

1

2

Taking transforms, 10I1(S) + 106

d (i2 – i1) = 0 dt

(3)

10I2(S) + 3 × 10–3 [SI2(S) – SI1(S)] = 0

(4)

10i2(t) + 3 × 10–3

From Eq. (4), I2(S) =

SI1 (S) S + 3.33 × 103

Solution Manual (11) – Page No. 7

Substituting in Eq. (2), I1(S) =

5(S + 3333.33) (S + 3419.92)(S + 48246.73)

i1(t) = 5e–3419.92t – 9.5e–48246.73t i2(t) = 0.92e–48246.73t – 0.07e–3419.92t 11-2.11

Change on the capacitor = 200 × 50 × 10–6 = 0.01 C 100i +

di

1 50 × 10

–6

∫ i dt + 0.1 dt = 0

Taking transforms, 100I(S) +

I(S) =

1 50 × 10

⎡ I(S) 0.01 ⎤ – + 0.1 SI(S) = 0 S ⎥⎦ ⎣ S

–6 ⎢

2000

S + 1000 S + 20000 2

I(S) =

2000 (S + 276.4)(S + 723.6)

I(S) =

A B + S + 276.4 S + 723.6

i(t) = 4.47 [e–276.4t – e–723.6t] 11-2.12 Let iL(t) be the current through the inductor and VC(t) be the voltage across the capacitor.

Then

iL(0–) = iL(0+) = 4 A VC(0–) = VC(0+) = 8 volts 12 = 2i(t) + VC(t)

or

Solution Manual (11) – Page No. 8

12 = 2i(t) + 2iL(t) + V1(t)

di (t) dVC (t) = iL(t), V1(t) = 10 L dt dt

i(t) = C

⎛ ⎝

12 = 2 ⎜10

–4 dVC (t)

dt

VC(t) = 2iL(t) + 10

⎛

12 = 2 ⎜10

–4

⎝

12 = 410–4

2

di (t) ⎞ + iL (t) ⎟ + 2iL(t) + 10 L ⎠ dt

diL (t) dt

diL (t) d 2 i (t) ⎞ di (t) + 10 L ⎟ + iL(t) + 2iL(t) + 10 L dt dt dt ⎠

d 2 iL (t) diL (t) di (t) + 2 × 10–3 + 4iL(t) + 10 L 2 dt dt dt

12 = (2 × 10–3 D2 + (10 + 4 × 10–4)D + 4) i(t) D1 = –0.4, D2 = –5000 i(t) = C1 + C2 e–0.4t + C3e–5000t i(∞) = C1 = 3 i(0) = C1 + C2 + C3 = 4 C2 + C3 = 1 Now consider, VC(0–) = VC(0+) = 8 V So,

VC(t) = 2iL(t) + 10

diL (t) dt

VC(0+) = 8 = 2iL(0+) + 10

diL (t) / t( 0 + ) dt

8 = 2 × 4 + 10 = (C2 (–0.4)e–0.4t + C3(–8000) e–5000t)t = 0+ 0 = (C2 (– 0.4) + C3(–5000)) C2 = 1.00008 ≈ 1 C3 = –8 × 10–5 or

i(t) = 3 + 1e–0.4t – 8 × 10–5e–500t

Voltage across the inductor V1(t) = 10

di(t) = 10[(–0.4)e–10.4t – 8 × 10–5 (–5000)e–5000t] dt

V1(t) = 10(–0.4 e–0.4t + 0.4 e–5000t)

Solution Manual (11) – Page No. 9

V1(t) = 4e–5000t – 4e–0.4t 11-3.1

Applying KCL, i2 = i3 + i1 I2(S) = I3(S) + I1(S)

(1)

Applying KVL,

6i2 = 4i1 + 2i2 + 0.1

4i2 = 4i1 + 0.1

di2 dt

di2 dt

4I2(S) = 4I1(S) + 0.1[SI2(S) – 5]

I1(S) =

4 I 2 (S) + 0.5 – 0.1SI 2 (S) 4

(2)

Applying KVL,

10i3 + 2i2 + 0.1

di2 =0 dt

10I3(S) + 2I2(S) + 0.1[SI2(S) – 5] = 0

I3(S) =

0.5 – I 2 (S) [ 2 + 0.15] 10

(3)

From (1), (2), and (3),

I2(S) =

0.5 – [I 2 (S)(2 + 0.15)] 4 I 2 (S) + 0.5 – 0.15I 2 (S) + 10 4

Solution Manual (11) – Page No. 10

20I2(S) = 1 – 4I2(S) (2 + 0.15) + 20I2(S) + 2.5 – 0.5SI2(S)

⇒

I2(S) =

3.5 = 4 + 0.75

5 40 S+ 7

Applying inverse Laplace transform, i2(t) = 5e–5.71t 11-3.4

1 ⎞⎛ 1 ⎞ ⎛ ⎜⎝10 + ⎟⎠ ⎜⎝10 + ⎟ 0.52 0.25S ⎠ Z(S) = 20 + 1 1 + 20 + 0.55 0.25S

Z(S) =

500 S 2 + 130 S + 8 S(20 S + 5)

I(S) =

100 S(20 S + 6 ) × S 500 S 2 + 180 S + 8

=

1 ⎡ 19.37 0.62 ⎤ + ⎢ 5 ⎣ S + 0.05 S – 0.308 ⎥⎦

i(t) = 3.87e–0.05t + 0.12e–0.308t 11-3.5

250i(t) +

1 0.5 × 10 –6

∫ i(t) dt = 100 sin (200t + 30º)

Taking transforms,

⎡ I(S) 25 × 10 –6 ⎤ 100 ⎡ S ⎤ – – 100 3 ⎥ 250I(S) + 2 × 10 ⎢ ⎥= ⎢ S ⎦ ⎣ S ⎦ 5–2 ⎣2 6

Solution Manual (11) – Page No. 11

⎡ 47.8 2.195S – 239.02 50 ⎤ – + ⎥ 2 2 S + 8000 ⎦ S + 200 ⎣ S + 8000

I(S) = 0.004 ⎢ ∴

i(t) = – 0.0088e–8000t + 0.0087 cos 200t – 0.0047 sin 200t = – 0.0088e–8000t + 0.0098 sin (200t – 61.62º)

11-3.6

At initial condition, i.e., t < 0, Capacitor is open

⎛ V – 50 ⎞ VA + – 0.1 V5 = 0 Applying KCL at the node 'A', ⎜ A ⎝ 2 ⎟⎠ 5 VA – 50 VA + – 0.1(VA ) = 0 2 5 ⎛1 1 ⎞ VA ⎜ + – 0.1⎟ = 25 ⎝2 5 ⎠

V5 = VA = I5Ω =

25 = 41.66 volts 0 .6

VA 41.66 = 8.33 A = 5 5

V0.2F = V5 + (4 * 0.1 V5) = 58.32 volts ∴ voltage drop across capacitor V0.2F = 51.5 volts; for t < 0

At

t > 0,

switch is opened.

Apply KCL at the node A, VA d – 0.1 V5 + 0.2 (VA – 58.5) = 0 dt 9

0.2

dVA V – 0.1VA + A = 0 dt 9

Solution Manual (11) – Page No. 12

dVA + 0.056 VA = 0 dt

(D + 0.056)VA = 0 VA = c e–0.056t

At

(1)

VA = 58.32 volt

t = 0,

Substituting VA in Eq. (1), we get c = 58.32 VA = 58.32 e–0.056t volts V5 = (58.32 e–0.056t) volts 11-3.8

When the switch k is opened, the current I divides into 3 elements.

Applying KCL at node A I=

V 1 dv + ∫ v dt + c R L dt

5=

V 1 dv + v dt + 1. ∫ 1/ 4 1/ 4 dt dv + 4 v + 4 ∫ v dt = 5 dt

Differentiating the above equation, d2v dt

2

+4

dv + 4v = 0 dt

(D2 + 4D + 4)V = 0 (D + 2)2v = 0 The complementary current V(t) = (c1 + c2t) e–2t At

(1)

t = 0, V = 0 0 = (c1 + 0)(1) c1 = 0

At

t=0 c.

dv =5A dt dv = 5 v/S dt

From Eq. (1),

dv = (c1 + c2t) e–2t (–2) + e–2t c2 dt

5 = 0. + c2 c2 = 5 ∴

–2t

v(t) = 5t e

Solution Manual (11) – Page No. 13

CHAPTER 12 (Solution Manual)

12-1.1

T = 2π 2π

a0 =

1 –50 (ωt – 2π).dωt .

2 π ∫0 π

=

1 ⎛ –50 ⎞ .⎜ ⎟ (ωt – 2 π).d ωt 2 π ⎝ π ⎠ ∫0

2π

−25 π

[(ωt ) – 2

2

2

=

π .ωt ]20 π =

−25 π

[4π2 – 4π2 + 0 – 0] = 0 2

y – y1 =

y2 – y1 . (x – x1) x2 – x1 (2π, 0), (0, 100)

a0 = 0 2π

an =

1 . V(t) cos. nωt dωt π ∫0 2π

1 –50 (ωt – 2π). cos nωt dωt = .∫ π 0 π

=

25 × 2 π2

x1 y1

x2 y2

V(t) =

100 (ωt – 2π) –2 π

=

–50 (ωt – 2π) π

2π

∫ (2 π – ωt ).cos nωt d ωt 0

2π

=

⎤ 25 × 2 ⎡ –1 ⎛ sin nωt ⎞ . ⎢(2 π – ωt ). ⎜ + – (–1). 2 . cos nωt ⎥ ⎟ 2 ⎝ n ⎠ π n ⎣ ⎦0

=

25 × 2 ⎡ –1 ⎤ (1–1)⎥ 2 ⎢ 2 π ⎣n ⎦

[ ∴ cos 2x = 1 for all values of integers]

an = 0 2π

bn =

1 . V(t) sin nωt dωt π ∫0 2π

=

1 50 .∫ (2 π – ωt ).sin nωt . d ωt π 0 π

=

25 × 2 ⎡ ⎛ cos nωt ⎞ ⎧ –1 ⎫⎤ (2 π – ωt ). ⎜ – (–1). ⎨ 2 .sin nωt ⎬⎥ ⎟ 2 ⎢ ⎝ –n ⎠ π ⎣ ⎩n ⎭⎦ 0

=

25 × 2 ⎡ 2 π 1 ⎤ +50 × 2 0+ – 2 ( 0 – 0 )⎥ = = bn 2 ⎢ n nπ n π ⎣ ⎦

2π

a0 = 0, an = 0, bn =

+50 × 2 nπ

Using these term coefficients and the average term, the series is

Solution Manual (12) – Page No. 1

∞

V(t) = a0 + ∑

(an . cos nωt + bn. .sin nωt)

n =1 ∞

= ∑

n =1

+

50 × 2 .sin nωt nπ

V(t) =

–200 ⎡ 1 1 ⎤ sin ω t + .sin 2ω t + .sin 3ω t + ......⎥ 2 3 π ⎢⎣ ⎦

a0 =

1 . V sin ωt . d ωt 2 π ∫π

12-1.2

π

2

=

V –V V π . [ – cos ωt ] π = [–1– 0] = 2π 2π 2π 2

π

an =

1 . V sin ωt .cos nωt d ωt 2 π ∫π 2 π

=

V 1 . . [sin (n + 1)ωt + sin(1– n)ωt ]. d ωt π 2 ∫π 2 π

V ⎡ – cos(n + 1)ωt cos(1– n)ωt ⎤ = .

– ⎥ 2 π ⎢⎣ n +1 1– n ⎦ π/2

=

–V ⎡ 1 ⎡ π⎤ 1 ⎡ π ⎤⎤ .⎢ cos( n + 1)π – cos(n + 1). ⎥ + cos(1– n)π – cos(1– n) ⎥ ⎥ ⎢ ⎢ 2π ⎣ n +1 ⎣ 2 ⎦ 1– n ⎣ 2 ⎦⎦ π

bn =

1 . V sin ωt .sin nωt . d (ωt ) π ∫π 2 π

=

V . 2 sin ωt .sin nωt . d (ωt ) 2 π ∫π 2 π

=

V . [cos(ωt – nωt ) – cos(ωt + nωt )]d ωt 2 π ∫π 2

Solution Manual (12) – Page No. 2

π

=

π

V V . [cos (1– n)ωt d ωt – ∫ cos (1 + n)ωt d ωt 2 π ∫π 2 π π/2 2 π /2

π

V V . cos (1– n)ωt d ωt – ∫ cos(1 + n)ωt dωt 2 π ∫π 2 π π/2

bn =

2 π

=

π

V ⎡ sin(1– n)ωt ⎤ V ⎡ sin(1 + n)ωt ⎤ – ⎢ ⎥ ⎢ ⎥ 2 π ⎣ 1– n ⎦ π/2 2 π ⎣ 1 + n ⎦ π/2

π⎤ ⎡ sin(1 + n) ⎥ V ⎡ sin(1– n)π – sin(1– n)π / 2 ⎤ V ⎢ sin(1 + n)π 2 – = ⎥ ⎥ – 2π ⎢ 1 + n + n 2π ⎢⎣ 1– n 1 ⎦ ⎢ ⎥ ⎣⎢ ⎦⎥ =

V π⎤ V π⎤ ⎡ ⎡ sin(1– n)π – sin(1– n) ⎥ – sin(1 + n)π – sin(1 + n) ⎥ ⎢ ⎢ 2 π(1– n) ⎣ 2 ⎦ 2 π(1 + n) ⎣ 2⎦

∴ V(t) =

∞ V –V ⎡⎛ 1 π⎞ 1 π ⎤ (cos(1 + n)π – cos(1 + n) ⎟ + (cos(1– n)π – cos(1– n) ) ⎥ cos nω t +∑ ⎢⎜ 2π n =1 2π ⎣⎝ 1 + n 2 ⎠ 1– n 2 ⎦

∞

V ⎡ 1 ⎛ π⎞ 1 π⎤ sin(1– n)π – sin(1– n) ⎟ – (sin(1 + n)π – sin(1 + n) ⎥ sin nω t ⎜ ⎢ 2 ⎠ 1+ n 2⎦ n −1 2π ⎣ (1 − n) ⎝

+ ∑ 12-1.3

π /2

a0 =

=

1 . 10. d ωt 2 π ∫0

10 10 ⎛ π ⎞ 10 5 [ωt ]0π/2 = = ⎜ – 0⎟ = 2π 2π ⎝ 2 ⎠ 4 2 π /2

an =

2 . 10.cos nωt . d ωt 2 π ∫0 π/2

= =

10 ⎡ sin nωt ⎤ . π ⎢⎣ n ⎥⎦ 0

=

10 ⎡ ⎛ n π ⎞ ⎤ sin ⎜ ⎟ – 0 2 πn ⎢⎣ ⎝ 2 ⎠ ⎥⎦

10 nπ .sin nπ 2 π /2

bn =

2 . 10.sin nωt . d ωt 2 π ∫0

Solution Manual (12) – Page No. 3

π/2

=

10 ⎡ – cos nωt ⎤ . ⎥ 2 π ⎣⎢ n ⎦0

=

10 ⎛ nπ ⎞ ⎜1– cos ⎟⎠ nπ ⎝ 2

=

–5 ⎡ ⎛ nπ ⎞ ⎤ cos ⎜ ⎟ – 1⎥ ⎝ 2 ⎠ ⎦ n π ⎢⎣

∞ ⎡⎛ 10 ⎤ nπ ⎞ nπ ⎞ 5 10 ⎛ + ∑ ⎢⎜ .sin . cos nω t + 1– cos .sin nω t ⎥ ∴ V(t) = ⎟ ⎜ ⎟ nπ ⎝ 12 12 ⎠ 12 ⎠ ⎝ nπ ⎦ n =1 ⎣ 12-1.4

The equation for the waveform shown in the figure is given by v(t) =

–10 ωt + 10 π

v(t) = 0

for 0 ≤ ωt ≤ π for π ≤ ωt ≤ 2π

The coefficients of the exponential series cn =

=

π ⎤ 1 ⎡ ⎛ –10 ⎞ ωt + 10 ⎟ e– jn(ωt ) d (ωt )⎥ ⎢∫ ⎜ ⎠ 2 π ⎣⎢ 0 ⎝ π ⎦⎥ π ⎤ 1 ⎡ ⎛ –10 ⎞ ωt e – jn(ωt ) d (ωt ) ⎟ + 10e – jn(ωt ) d (ωt )⎥ ⎢∫ ⎜ ⎠ 2 π ⎢⎣ 0 ⎝ π ⎥⎦

π π⎤ ⎡ 10 ⎢ –1 e – jn(ωt ) 1 – jn(ωt ) ⎥ = (– jnω t –1) – e ⎥ 2π ⎢ π (– jn)2 jn 0⎦ 0 ⎣

=

⎤ ⎡ 1 – Jn( π) 1 0 ⎤ ⎤ 10 ⎡ –1 ⎡ e – jnπ e –0 – e ⎥⎥ ⎢ ⎢ 2 2 (– jnπ –1) – 2 2 (0 –1)⎥ – ⎢ e 2 π ⎢⎣ π ⎣ j n jn ⎦ ⎥⎦ j n ⎦ ⎣ jn

=

10 ⎡ –1 ⎡ ⎡ e – jnπ e – jnπ ⎤ 1 ⎤ ⎡ e – jnπ 1 ⎤ ⎤ – ⎥⎥ ⎢ ⎢ ⎢ 2 2 × – jnπ – 2 2 ⎥ + 2 2 ⎥ – ⎢ 2 π ⎣⎢ π ⎣⎢ ⎣ j n jn ⎦ ⎦⎥ j n ⎦ j n ⎦⎥ ⎣ jn

=

10 ⎡ ⎡ −e – jnπ e – jnπ 1 ⎤ e – jnπ 1 ⎤ + ⎥ ⎢ ⎢ 2 2 × – jnπ + 2 2 – 2 2 ⎥ – 2 π ⎣⎢ ⎣ j n π jn jn ⎦⎥ j n π j n π⎦

=

10 ⎡ e – jnπ e – jnπ 1 e – jnπ 1⎤ ⎢ – 2 + 2 – + ⎥ jn ⎥⎦ 2 π ⎢⎣ jn jn n π n π

Solution Manual (12) – Page No. 4

=

10 ⎡ – e – jnπ 1 1 ⎤ 10 ⎡ – e – jnπ + 1 1 ⎤ + ⎥ ⎢ 2 + 2 + ⎥= ⎢ 2π ⎣ n π jn ⎦ n π jn ⎦ 2 π ⎣ n2 π

∴ the exponential Fourier series is,

∝ v(t) =

∑ Cn . e jnωt n=– ∝ ∝

=

10 ⎡ – e – jnπ + 1 1 ⎤ jnωt ∑ + ⎥e ⎢ 2 π ⎣ n2 π jn ⎦ n=– ∝ ⎡ ⎤ 1 ⎤ – j 3ωt 1 – j 2ωt ⎛ 1 1 ⎞ ⎡ 1 +j + ⎜ 2 + j ⎟ e – jωt ⎥ e ⎢… + ⎢ 2 + j 6 π ⎥ e ⎝π 4π 2π ⎠ ⎣ 9π ⎦ ⎥ ⎢ 1 ⎛ 1 ⎥ 1 ⎞ jωt 1 j 2ωt ⎛ 1 1 ⎞ j 3ωt + ⎜ 2 – j ⎟e + …⎥ e ⎢ + + ⎜⎝ 2 – j ⎟⎠ e – j ⎝ ⎠ π π π 4 2 4 6 π π 9 ⎣ ⎦

= 10 ⎢

Spectrum 12-2.1

⎧ –25 . ωt 0 < ωt < π ⎪⎪ V(t) = ⎨ π ⎪ 50 (ωt – π) – 25 π < ωt < 2 π ⎪⎩ π

(0, 0) (π, –25) m1 =

–25 π

a0 = 0.

( π, – 25) (2 π, 25) x1 y1 x2 y2

Since the area under the curve of V(t) between 0 and π is

V1(t) + 25 =

25 – (–25) . (ωt – π) 2π – π

V1(t) + 25 =

50 . (ωt – π) π

equal to the area under the curve of V(t) between π and 2π, T

an =

2 . V (t ) . cos nωt . dωt T ∫0

Solution Manual (12) – Page No. 5

V1(t) = –75 +

50 ωt π

=

π 2π ⎤ 2 ⎡ ⎛ –25 50 ⎞ ⎛ ⎞ . ωt ⎟ cos nωt . d ωt – ∫ ⎜ –25 + (ωt – π) ⎟ cos nωt . d ωt ⎥ ⎢∫ ⎜ ⎠ ⎝ ⎠ 2 π ⎢⎣ 0 ⎝ π π ⎥⎦ π

π 2π 2 ⎡ –25 ⎛ sin nωt 1 sin nωt ⎧⎪ 50 ⎛ sin ωt 1 ⎞ ⎤ ⎡ ⎞ ⎫⎪⎤ ⎢ + 2 cos nωt ⎟ ⎥ + –(25 + 50) + ⎨ ⎜ ωt . + 2 .cos nωt ⎟ ⎬⎥ ⎢ ⎜ ωt . ⎠0 ⎥ ⎢ ⎝ ⎠ π ⎪⎥ π n n n 2 π ⎢⎣ π ⎝ n n ⎪ ⎦ ⎣ ⎩ ⎭⎦

=

2 ⎡ –25 ⎛ 1 1 ⎞⎤ ⎡ 50 1 1 ⎤ 0 + 2 cos nπ – 0 + 2 ⎟ ⎥ + ⎢ –(25) {0 – 0} + (2π (0 )) + 2 (1) – 0 – 2 . cos nπ ⎥ 2π ⎣⎢ π ⎝⎜ π n n ⎠⎦ ⎣ n n ⎦

an =

–25 × 2 2π n

2 2

[1 + (–1)n ] +

50 n π 2

[1– (–1)n ]

T

bn =

=

2 V (t ) . sin nωt.dω T ∫0 π 2π ⎤ 2 ⎡ ⎛ –25 50 ⎞ ⎧ ⎫ . ωt ⎟ .sin nωt . d ωt + ∫ ⎨ –25 + (ωt – π)⎬ .sin nωt. d ωt ⎥ ⎢∫⎜ ⎠ 2 π ⎣⎢ 0 ⎝ π π ⎭ π ⎩ ⎦⎥ π

2π

⎞⎤ ⎡ ⎞⎤ 2 ⎡ –25 ⎛ ⎛ – cos nωt ⎞ 1 ⎛ – cos nωt ⎞ 50 ⎛ ⎛ – cos nωt ⎞ 1 = ωt. ⎜ ⎢ ⎟⎠ + 2 .sin nωt ⎟ ⎥ + ⎢(–75) ⎜⎝ ⎟⎠ + ⎜ ωt ⎜⎝ ⎟⎠ + 2 .sin nωt ⎟ ⎥ ⎜ ⎝ ⎠⎦0 ⎣ ⎠⎦π n n π ⎝ n 2π ⎣ π ⎝ n n

=

−25 × 2 ⎡ – π ⎛ –1 ⎞ 50 ⎡ –2 π ⎤ ⎛ –1 ⎞ 50 ⎛ – π ⎞ ⎤ (–1)n + 0 + 0 − 0⎥ + (–75) ⎜ ⎟ + ⎢ + 0⎥ − (–75) ⎜ .(–1)n ⎟ – ⎜ .(–1)n + 0 ⎟ 2 ⎢ n ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ n π n n π n 2π ⎣ ⎦ ⎣ ⎦ = =

+25 × 2 75 100 75 50 (–1)n + – – (–1)n + (–1)n 2 πn n n n n

25 × 2 25 25 (–1)n – – .(–1)n 2 πn n n

⎤ 50 ⎡ ⎛ 1 ⎞ n bn = ⎜⎝ –1⎟⎠ (–1) –1⎥ ⎢ n ⎣ 2π ⎦

From a0, an and bn, the series is ∞ V(t) = a0 + ∑ [an. cos nωt + bn. sin nωt]

a1 = a2 =

–25 2π

2

–25 4π

2

(0) +

50 100 (2) = π π

+0=

–25 4π2

⎡⎛ 1 ⎤ –25 ⎞ b1 = 25 ⎢ ⎜ + 1 ⎟ – 1 ⎥ = ⎝ ⎠ π 2 ⎣ ⎦ 2π ⎤ 25 25 ⎡ ⎛ 1 ⎞ b2 = − 25 ⎜ ⎟ – 1 − 1⎥ = 2 ⎢⎣ ⎝ 2 π ⎠ ⎦ 4π

n =1

⎫ 25 ⎛ 100 ⎞ ⎧⎛ –25 ⎞ ⎛ 25 ⎞ = 0+ ⎜ . cos nω t – .sin nω t ⎟ + ⎨⎜ 2 ⎟ . cos 2ω t + ⎜ – 25 ⎟ .sin 2ω t ⎬ + ......

2π ⎝ π ⎠ ⎩⎝ 4π ⎠ ⎝ 4π ⎠ ⎭

Solution Manual (12) – Page No. 6

12-2.2

π/2

1 . 100 cos nω t .d ω t 2 π –∫π

a0 =

2

=

100 100 100 π/2 . [sin ωt ]– π/2 = [1– (–1)] = π 2π 2π π /2

an =

2 . ∫ 100 cos ωt .cos nωt d ωt 2 π – π/2

cos A • cos B =

1 [cos(A+ B) + cos(A – B)] 2

cos A • sin B =

1 [sin (A + B) – sin (A – B)] 2

π/2

=

2 × 100 1 . . [cos(n + 1)ωt + cos(n –1)ωt ]. d ωt 2 π 2 – ∫π2

=

2 × 100 ⎡ sin(n + 1)ωt sin(n –1)ωt ⎤ . + 4 π ⎢⎣ n +1 n –1 ⎥⎦ – π/2

=

200 ⎡ 2 –π 2 π⎤ .⎢ sin(n + 1) + .sin(n –1) ⎥ 4π ⎣ n + 1 2 n −1 2⎦

π/2

π /2

bn =

2 . ∫ 100 cos ωt. sin nωt . dωt.

2 π – π/2 π/2

=

100 1 . . [sin( n + 1)ωt – sin( n –1)ωt ]. d ωt 2 π 2 – ∫π2

=

100 ⎡ – cos(n + 1)ωt cos( n –1)ωt ⎤ . + ⎥ 2 π ⎢⎣ n +1 n –1 ⎦ – π/2

=

π π 100 ⎡ –1 ⎛ 1 ⎡ ⎛ –π ⎞ ⎞ ⎛ – π ⎞⎤⎤ =0 cos( n + 1) – cos( n + 1) ⎜ . ⎢ cos( n –1) – cos( n –1) ⎜ ⎢ ⎟⎠ ⎟ + ⎜ ⎝ ⎝ 2 ⎟⎠ ⎥⎦ ⎦⎥ 2π ⎣ n + 1 ⎝ 2 2 ⎠ n −1 ⎣ 2

π/2

an =

200 ⎡ 2 π 2 (n + 1)π ⎤ sin( n + 1) + .sin ⎢ 4π ⎣ n +1 2 n –1 2 ⎥⎦

⎡ ⎤ ⎢ sin (n –1) . π ⎥ 200 ⎢ ⎥ 2 . 0+ a1 = ⎥ 4π ⎢ ⎛ n –1 ⎞ ⎢ ⎥ ⎜⎝ ⎟⎠ 2 ⎥ n =1 ⎦ ⎣⎢

Solution Manual (12) – Page No. 7

⎡ ⎤ ⎢ sin (n –1) . π ⎥ 100 ⎢ 2 ⎥ . = ⎥ 4 ⎢ ⎛ n –1 ⎞ ⎢ ⎜ ⎥ ⎟⎠ π ⎝ 2 ⎢⎣ n =1 ⎥ ⎦

x→0

sin X =1 X

=

100 ⎡ Lt sin x ⎤ = 25 4 ⎢⎣ x → 0 x ⎥⎦

a2 =

200 ⎡ 2 3π π⎤ .sin + 2.sin ⎥ ⎢ 4π ⎣ 3 2 2⎦

= (n –1) π = 2x

100 ⎡ –2 ⎤ 100 4 100 + 2⎥ = × = 4 π ⎢⎣ 3 ⎦ 4 π 3 3π

=n=

=

V(t) =

12-2.3

Lt

100

π

∞

+∑

an .cos nωt

n =1

=

100 100 ⎡ ⎤ + 2 ⎢25.cos ωt + cos 2ωt + ......⎥ π 3π ⎣ ⎦

=

100 ⎡ π 2 ⎤ 1 + cos ωt + cos 2ωt + ......⎥ ⎢ 3 π ⎣ 2 ⎦

The equation for the waveform shown in the figure is given by v(t) = 10

for

–π π ≤ ωt ≤ 2 2

= –10

for

π 3π ≤ ωt ≤ 2 2

The coefficients of the exponential series π /2 3 π /2 ⎤ 1 ⎡ – jn( ωt ) e d t 10 ( ) –10 e – jn(ωt ) d (ωt )⎥ ω + ⎢ cn = ∫ ∫ 2 π ⎣⎢ – π /2 π /2 ⎦⎥

=

π /2 3π /2 ⎤ 10 ⎡ –1 – jn ω t 1 – jn ωt ⎢ e ⎥ – e 2π ⎢ jn – jn –π /2 π /2 ⎥ ⎣ ⎦

=

3π /2 ⎤ –10 ⎡ – jn ωt π /2 e – e – jn ωt ⎢ – π /2 π /2 ⎥ 2π jn ⎣ ⎦

Solution Manual (12) – Page No. 8

⎛ n –1 ⎞ ⎜⎝ ⎟ π =x 2 ⎠

2 x+1 π

=

–10 – jnπ /2 ⎡e – e jnπ /2 – e – jn3 π /2 + e – jnπ /2 ⎤⎦ 2 πjn ⎣

=

–10 ⎡2e – jnπ /2 – e jnπ /2 – e – j 3nπ /2 ⎤ ⎦ 2 πjn ⎣

∴ the exponential Fourier series is,

∝ ∝ –10 ⎡ – jnπ /2 jnωt ∑ = ∑ .

2e – e jnπ /2 − e– j 3nπ /2 ⎤⎦ e jnωt c e v(t) = n 2 πjn ⎣ n=– ∝ n= – ∝ =

20 ⎡ 1 1 1 1 ⎤ … + e− j 5ωt – e – j 3ωt + e− jωt + e jωt – e j 3ωt + e j 5ωt …⎥ 5 3 3 5 π ⎢⎣ ⎦

Frequency spectrum

12-3.1

The equation for the waveform shown in the figure is given by v(t) = 50 cos ωt for

–π π ≤ ωt ≤ 2 2

The coefficients of the exponential series π /2

cn =

1 50 cos ωt . e–jnωt d (ωt) π – π∫/2 π /2

=

1 [e jωt + e – jωt ] – jnωt 50 e d (ωt ) π – π∫/2 2

=

25 ⎡ e jωt – jnωt + e – jωt – jnωt ⎤ d (ωt ) ⎦ π – π∫/2 ⎣

π /2

=

π /2 π /2 ⎤ 25 ⎡ ωtj (1– n ) d (ωt ) + ∫ e – jωt (1+ n ) d (ωt )⎥ ⎢ ∫ e π ⎣⎢ – π /2 – π /2 ⎦⎥

Solution Manual (12) – Page No. 9

=

π /2 π /2 ⎤ 25 ⎡⎢ e jωt (1– n ) e – jωt (1+ n ) ⎥ + – j(1 + n) π ⎢ j(1– n) ⎥ – π /2 – π /2 ⎣ ⎦

π π π ⎡ j π (1– n ) – j (1– n ) – j (1 + n ) j (1 + n ) ⎤ 25 ⎢ e 2 e 2 e 2 e 2 ⎥ = – – + ⎢ j (1– n) j (1 + n) j(1 + n) ⎥ π j(1– n) ⎢ ⎥ ⎣ ⎦ π π ⎡ j π (1– n ) – j π (1– n) – j (1 + n ) j (1+ n ) ⎤ 2 2 2 25 ⎢ e –e e –e 2 ⎥ = – ⎢ ⎥ j(1– n) j (1 + n) π ⎢ ⎥ ⎣ ⎦

=

25 ⎡ 2 sin((1 – n)π / 2) 2 sin((1 + n)π / 2) ⎤ + ⎥ π ⎢⎣ (1– n) (1 + n) ⎦

For n = 1, π /2

c1 =

=

1 50 cos ωt e–jωt d(ωt) ∫ π – π /2 π /2 ⎛ e j ωt + e – j ωt ⎞ – j ωt 1 50 d (ωt ) ⎜ ⎟e π – π∫/2 ⎝ 2 ⎠ π /2

=

50 (1 + e –2 jωt ) d (ωt ) 2 π – π∫/2

=

50 ⎡ e –2 jωt ⎤ ⎢ωt – ⎥ 2π ⎣ 2j ⎦

π /2

– π /2

π π ⎡ –2 j +2 j ⎤ 50 ⎢ π e 2 π e 2⎥ = + – – 2π ⎢ 2 2j ⎥ 2j 2 ⎢ ⎥ ⎣ ⎦

=

50 ⎡ 1 j π − j π ⎤ (e – e ) ⎥ 2 π ⎢⎣ 2 j ⎦

=

50 (0) = 0 2π

Solution Manual (12) – Page No. 10

12-3.2

The equation for the waveform shown in the figure is given by v(t) =

–V ωt + V π

v(t) = 0

for

0 ≤ ωt ≤ π

for

π ≤ ωt ≤ 2π

The coefficients of the exponential series. cn =

π ⎤ 1 ⎡ ⎛ –V ⎞ . ωt + V ⎟ e – jn(ωt ) d (ωt )⎥ ⎢∫ ⎜ ⎠ 2 π ⎢⎣ 0 ⎝ π ⎥⎦

=

π 1 ⎡ ⎛ –V ⎞⎤ . ωt e – jn(ωt ) d (ωt ) + 10 e – jn(ωt ) d (ωt ) ⎟ ⎥ ⎢∫ ⎜ ⎠⎥ 2 π ⎣⎢ 0 ⎝ π ⎦

=

π π⎤ ⎡ V ⎢ ⎛ –1 ⎞ e – jn(ωt ) 1 – jn(ωt ) ⎥ (– jn t –1) – e ω ⎥ 2π ⎢ ⎜⎝ π ⎟⎠ (– jn)2 jn 0⎦ 0 ⎣

=

⎤ ⎡1 V ⎡ –1 ⎡ e – jnπ e –0 1 ⎤⎤ . ⎢ ⎢ 2 2 (– jnπ –1) – 2 2 (0 –1)⎥ – ⎢ e – jnπ – e0 ⎥ ⎥ jn ⎦ ⎦⎥ 2 π ⎣⎢ π ⎣ j n j n ⎦ ⎣ jn

=

V ⎡ –1 ⎡ ⎡ e – jnπ e – jnπ ⎤ 1 ⎤ ⎡ e – jnπ 1 ⎤⎤ . ⎢ ⎢ ⎢ 2 2 × (– jnπ ) – 2 2 ⎥ + 2 2 ⎥ – ⎢ – ⎥⎥ jn ⎥⎦ ⎥ 2π ⎢ π ⎢⎣ ⎢⎣ j n j n ⎥⎦ j n ⎥⎦ ⎢⎣ jn ⎣ ⎦

=

V ⎡ ⎡ e – jnπ e – jnπ 1 ⎤ e – jnπ 1⎤ . ⎢ ⎢ 2 2 × (+ jnπ ) + 2 2 – 2 2 ⎥ – – ⎥ 2π ⎣⎢ ⎣⎢ j n π jn jn ⎦⎥ j nπ j n ⎦⎥

=

V ⎡ e – jnπ e – jnπ 1 e – jnπ 1⎤ .⎢ – 2 + 2 – + ⎥ 2 π ⎣ jn jn jn ⎦ n π n π

=

V ⎡ – e – jnπ 1 1⎤ .⎢ 2 + 2 + ⎥ 2π ⎣ n π n π jn ⎦

=

V ⎡ – e – jnπ + 1 1 ⎤ + ⎥ ⎢ 2 π ⎣ n2 π jn ⎦

∴ the exponential Fourier series is

Solution Manual (12) – Page No. 11

∝ v(t) =

=

∑ cn . ejnωt n=– ∝ ∝ V ⎡ e – jnπ + 1 1 ⎤ jnωt ∑ + ⎥e ⎢ 2 π ⎣ n2 π jn ⎦ n=– ∝

⎡ 1 ⎞ – j 3 ωt 1 – j 2 ωt ⎛ 1 1 ⎞ – j ωt 1 ⎤ ⎛ 1 + j +⎜ 2 + j + ⎥ e ⎟e ⎢… + ⎜⎝ 2 + j 6 π ⎟⎠ e ⎝ π π⎠ 4 2 4⎥ π 9π ⎢ =V ⎢ ⎛ 1 ⎥ 1 ⎞ j ωt 1 j 2 ωt ⎛ 1 1 ⎞ + ⎜ 2 – j ⎟ e j 3 ωt + … e ⎢ + ⎜⎝ 2 – j ⎥ ⎟⎠ e – j ⎝ ⎠ 2π 4π 6π 9π ⎣ π ⎦

12-3.3

The equation for the given waveform is v(t) =

V V (ωt + π) = ωt + V π π

v(t) =

V ωt – V for π

for

–π ≤ ωt ≤ 0

0 ≤ ωt ≤ π

The Fourier coefficients are cn =

0 π ⎤ 1 ⎡ ⎛V ⎞ – jn(ωt ) ⎛V ⎞ d (ωt ) + ∫ ⎜ ωt – V ⎟ e – jn(ωt ) d (ωt )⎥ ⎢ ∫ ⎜ ωt + V ⎟ e ⎠ ⎝π ⎠ 2 π ⎣⎢ – π ⎝ π 0 ⎦⎥

0 0 π π ⎤ V ⎡ 1 1 – jn( ωt ) – jn( ωt ) – jn( ωt ) t e d t e d t t e d t e – jn(ωt ) d (ωt )⎥ ( ) ( ) ( ) – ω ω + ω + ω ω = ⎢∫ ∫ ∫ ∫ 2 π ⎢⎣ – π π π ⎥⎦ –π 0 0

Solution Manual (12) – Page No. 12

=

0 0 π π ⎤ V ⎡⎢ 1 ⎡ e – jnωt e – jnωt 1 e – jn ( ωt ) e – jnωt ⎤⎥ (– jn t –1) (– jn t –1) – ω + + ω ⎢ ⎥ π (– jn)2 2 π ⎢ π ⎣ (– jn)2 (– jn) – jn ⎥ ⎦–π –π 0 0⎦ ⎣

=

V ⎡ e –0 e jnπ 1 e jnπ e – jnπ 1 ⎤ + + 2 2 (– jnπ –1) + 2 2 ⎥ ⎢ 2 2 (–1) – 2 2 ( jnπ –1) – 2π ⎣ j n π jn jn j n π j n π o n π⎦

=

V ⎡ 1 e jnπ 1 e jnπ e – jnπ 1 ⎤ – 2 (– jnπ –1) – 2 ⎥ + ⎢ 2 + 2 ( jnπ –1) – 2π ⎣ n π n π jn jn n π n π⎦

=

V ⎡ e jnπ e jnπ 1 e jnπ e – jnπ e – jnπ ⎤ × + + × + π π jn – – jn ⎢ ⎥ 2π ⎣⎢ n 2π jn jn n 2π n 2π n 2π ⎦⎥

=

V ⎡ – e jnπ e jnπ 1 e jnπ e – jnπ e – jnπ ⎤ – 2 – – + + 2 ⎥ ⎢ 2 π ⎣ jn jn jn n π jn n π⎦

=

V ⎡ – e jnπ + e jnπ ⎡ e – jnπ + 1⎤ ⎤ –⎢ ⎢ ⎥⎥ 2 π ⎣⎢ n2 π ⎣ jn ⎦ ⎦⎥

The Fourier series is

∝

v(t) =

V ⎡ – e jnπ + e – jnπ ⎡ e – jnπ + 1 ⎤ ⎤ jnωt ∑ –⎢ ⎢ ⎥⎥ e 2 π ⎢⎣ n2 π ⎣ jn ⎦ ⎥⎦ n=– ∝

=

∝ V ⎡ –2 j sin nπ ⎡ e – jnπ + 1 ⎤ ⎤ jnωt ∑ –⎢ ⎢ ⎥⎥ e 2 π ⎣⎢ n2 π jn ⎦ ⎦⎥ ⎣ n=– ∝

Solution Manual (12) – Page No. 13

12-3.4

Fourier series

∝ V(t) = a0 + ∑ (an cos nωt + bn sin nωt) n =1

∝ = a0 + ∑ cn cos (nωt – ∝ n) n =1 ⎛b ⎞ ∝n = tan–1 ⎜ n ⎟ ; cn = ⎝ an ⎠

an2 + bn2 a0 =

T

1 A A . d ( ωt ) = . T = A T ∫0 T

T

an =

2 A .cos nωt d (ωt ) T ∫0 T

=

2 ⎡ sin nωt ⎤ 2A A. ⎢ ⎥ = Tn [sin nT ] T n ⎣ ⎦0 T

bn =

2 A .sin nωt d (ωt ) T ∫0 T

=

2 A ⎡ – cos nωt ⎤ ⎥ T ⎢⎣ n ⎦0

=

2 A ⎡ – cos nt 1 ⎤ + ⎥ T ⎢⎣ n n⎦

æ wT ö ÷÷ Sin çç 2A ⎡2A ⎤ è w ø [sin nT ] + [– cos nT + 1]⎥ = 2T ∴ v(t) = A + ∑ ⎢ wT Tn ⎣ Tn ⎦ n =1 2

∝

Solution Manual (12) – Page No. 14

12-3.5

x(t) cos 10t x(t) = 2 cos 10t 2

x(ω) = ∫ 2 cos10t e – jωt dt –2

2 ⎡ e j10 t + e – j10 t ⎤ – jωt = 2∫ ⎢ dt ⎥e 2 ⎦ –2 ⎣

2 (10 j + jω )t ⎡ 2 e(10 j – ωj )t ⎤ e dt + ∫ dt ⎥ = 2⎢∫ 2 2 –2 ⎣⎢ –2 ⎦⎥ 2

2

⎡ e(10 j – jω )t ⎤ ⎡ e –(10 j + jω )t ⎤ = ⎢ ⎥ +⎢ ⎥ ⎣ 10 j – j ω ⎦ –2 ⎣ –(10 j + j ω ) ⎦ –2 ⎡ e(10 j – jω )2 e –(10 j – jω )2 e –(10 j + jω )2 e(10 j + jω )2 ⎤ = ⎢ + – – ⎥ 10 j – jω –(10 j + jω ) –(10 j + jω ) ⎦ ⎣ 10 j – j ω

= 2 sin c[2(ω –10)] + 2 sin c[2(ω + 10)]

12-5.1

Solution Manual (12) – Page No. 15

∞

V(ω) = ∫ V (t ) e – jωt dt –∞

–1

0

1

2

–2

–1

0

1

= ∫ 2. e – jωt dt + ∫ 4. e – jωt dt + ∫ 4. e – jωt dt + ∫ 2. e – jωt dt –1

0

1

2

⎛ e– jωt ⎞ ⎛ e– jωt ⎞ ⎛ e – jωt ⎞ ⎛ e– jωt ⎞ + 4 + 4 + 2 = 2⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ – jω ⎠ –2 ⎝ – jω ⎠ –1 ⎝ – jω ⎠ 0 ⎝ – jω ⎠1

⎡ e jω ⎡ e − jω ⎡ e –2 jω ⎡ 1 e2 jω ⎤ 1 jω ⎤ 1 ⎤ 1 – jω ⎤ = 2⎢ e ⎥ + 4⎢ e ⎥ + + + + ⎥ + 4⎢ ⎥ +2⎢ jω ⎦ jω ⎣ – jω jω ⎦ ⎣ – jω ⎣ – jω jω ⎦ ⎣ – jω ⎦

=

2 jω 2 2 jω 2 – jω 2 –2 jω – – e + e e e jω jω jω jω

=

2 jω − jω 2 2 jω –2 jω [e – e ] + [e – e ] jω jω

= 4 sin ω + 4 sin 2ω 12-5.2

The equation for the wave is 40 (ωt ) T = 20

v(t) =

for

0 ≤ ωt ≤ T/2

for

T/2 < ωt ≤ T

T /2

= V( ) =

T

40 – j ωt – jωt ∫ T (ωt ) e dt (ωt ) + ∫ 20. e d(ωt ) 0 T /2

=

20 20 [1– e– jT /2 ] + [e – jT /2 – e– jT ] j j

=

20 20 – jT /2 20 – jT /2 20 – jT – – e + e e j j j j

V( ) =

20 [1– e– jT ] j ∞

12-5.3

(i) V(ω) = ∫ V (t ) e – jωt dt –∞ ∞

= ∫ e – t u(t ) e – jωt dt –∞

Solution Manual (12) – Page No. 16

∞

= ∫ e – t (1+ jω ) dt –∞

∞

0

= ∫ e – t (1+ jω ) dt + ∫ e – t (1+ jω ) dt –∞

0

∞

0

⎡ e – t (1+ jω ) ⎤ ⎡ e – t (1+ jω ) ⎤ = ⎢ ⎥ +⎢ ⎥ ⎣ – (1 + jω ) ⎦ – ∞ ⎣ – (1 + jω ) ⎦ 0

=

1 1 2 + = 1 + jω 1 + jω (1 + jω) ∞

(ii) V(ω) = ∫ e– t u(−t ) e – jωt dt –∞

∞

0

= ∫ e – t [ −u(t )] e – jωt + ∫ e – t u(t ) e – jωt dt –∞

0

∞

– t (1 + jω ) ⎤ 0

⎡+e ⎡ e – t (1+ jω ) ⎤ = ⎢ ⎥ +⎢ ⎥ ⎣ + (1 + jω) ⎦ – ∞ ⎣ –(1 + jω) ⎦ 0

= 12-5.4

(i)

1 1 2 + = 1 + jω 1 + j ω 1 + jω

F(eat cos bt) ∞

e jbt + e – jbt – jωt e dt 2 –∞

F(cos bt) = ∫ =

jω ( jω)2 + b2

If (F(t) = F(jw) F(e–at t(I)) = F(jω + a) jω + a

So F[e–at cos bt] = (ii)

( j ω + a )2 + b 2

F(sin(ωt + θ)) ∞

e jωt – e – jωt – jωt e dt 2j –∞

F(sin ωt) = ∫ =

ω ( j ω )2 + ω 2

F(f(t)) = F(jω) F(f(t + a)) = e–ajω F(jω) ω

∴ F(sin (ωt+ θ)) = e–θj

ω ( j ω )2 + ω 2

Solution Manual (12) – Page No. 17

12-6.1

By frequency shifting,

X(ω) =

A (ω + 2a) 2a

for

–2a < ω < 0

=

–A ω+A 2a

for

0 < ω < 2a

By inverse Fourier transform, ∝

x(t) =

1 x ( ω ) e j ωt d ω 2 π –∫∝

=

0 2a ⎤ 1 ⎡ A ⎛ –A ⎞ (ω + 2a) e jωt dω + ∫ ⎜ ω + A ⎟ e jωt dω ⎥ ⎢∫ ⎝ 2a ⎠ 2 π ⎢⎣ –2 a 2a ⎥⎦ 0

=

A sin at πt

By frequency shifting properties, if x(t) ↔ x(ω) then x(t) e jω 0 t ↔ x(ω – ω0).

As two triangles are shifted to –w0 and ω0, x(t) =

2 A sin at cos ω 0 t A sin at jω 0t (e + e – jω 0t ) = πt πt

Solution Manual (12) – Page No. 18

CHAPTER 13 (Solution Manual)

13-2.1

The sine wave originates at t = 0 and ends at t = 4 s The wave equation is f(t) = sin ωt for 0 < t < 2.

In terms of unit step functions, the equation is f(t) = sin ωt [(u(t) – u(t – 4)] Here, time period T = 4 s

Frequency, f

1 T

1 4

w = 2p f = 2p f(t) = sin

1 p = 4 2

p t[u(t ) - u(t - 4)] 2

13-2.2

•

13-2.3

(a) As Ú f (t ) d (t - a)dt = f (a) -•

3

3

I = Ú (t 3 + 2)d (t )dt + Ú (t 3 + 2)8d (t –1) dt -1

-1

Solution Manual (13) – Page No. 1

= ÈÎ0 2 (1) + 2(1)˘˚ + ÈÎ(12 + 2)8˘˚ = 26 2

2

2

(b) I = Ú t 2d (t ) dt + Ú t 2d (t - (-1.5) dt + Ú t 2d (t - 3) dt -1

-1

-1

=0+0+0=0 13-2.4

Area under the function is Œ

A = Ú f (t ) dt -Œ Œ

Œ /p dt Œ2 + t 2 -Œ

= Ú

Œ

=

Œ 1 dt 2 Ú p Œ Œ +t 2

=

ŒÈ1 Ê t ˆ˘ tan -1 Á ˜ ˙ Í Ë Œ¯ ˚ -Œ p ÎŒ

Œ

[As ή 0]

1 È tan -1 (μ) - tan -1 (- μ)˘˚ pÎ 1 Èp Ê p ˆ ˘ = Í -Á- ˜˙ =1 p Î2 Ë 2¯˚ =

As ή 0 , the above function becomes an impulse function.

13-2.5

f (t ) = t (0.5)t U (t ) = t (e -0.7 )t Ut = t e -0.7 t U (t ) F ( s) =

1 –e–s = (s + 0.7)2 (–0.5 e –5 )(log(0.5 e – s ))2

13-2.6

Solution Manual (13) – Page No. 2

Straight line (a) gives the – (20t +400) u(t + 20)line.

Straight line (b) gives the + (40t + 400) u(t + 10)line.

Straight line (c) gives the + (400 – 40t) u(t – 10) line.

Straight line (d) gives the + (20t – 400) u(t – 20) line.

13-2.7

(a)

1 (s + a )

(b)

w s + w2 2

(c) sin(w t + q ) = sin w t ◊ cos q + cos w t ◊ sin q

w s a (sin(w t + q ) = cos q ◊ 2 + sin q ◊ 2 2 s +w s +w2 (d)

s s = 2 2 s -1 s -1 2

(e) cos h(t + q ) = cos h t ◊ cos h q + sin ht ◊ sin hq

s 1 a (cos h(t + q )) = cos h q ◊ 2 + sin hq ◊ 2 s -1 s -1 13-2.8

(a) a (t 2 e - at ) =

2!

2 = 2 +1 (s + a) ( s + a )3

(b) f (t ) = e - at sin h b t

Êd ˆ a Á f (t )˜ = sf (s) - f (0) Ë dt ¯

F ( s) =

B ( s + a )2 – B 2

f(0) = θ0, sin h0 = 0

s.B Èd ˘ a Í (e - at sin h b t )˙ = 2 2 Î dt ˚ (s + a ) - b (c) a [q t ◊ f (t )] =

-d F ( s) ds

f(t) = cos t

F ( s) =

s s + w2 2

a [t ◊ cos w t ] =

=

-d È s ˘ ds ÍÎ s 2 + w 2 ˙˚

s2 - w 2 ( s 2 + w 2 )2

Solution Manual (13) – Page No. 3

13-2.9

(a)

Èd ˘ As a Í (d1 (t ))˙ = sF1 (s) - f1 (0) Î dt ˚

F1 (s) =

w ( s + a )2 + w 2

F ( s) = s ◊

(b)

w ( s + a )2 + w 2

È ˘ F (s ) As a Í Ú f (t )dt ˙ = 1 s Î0 ˚

F1 (s) =

(s + a ) ( s + a )2 + w 2

F ( s) =

(s + a) s[(s + a)2 + w 2 ] a

d Èd ˘ (c) (i) By definition, a Í (e - at sin w t )˙ = Ú (e - at sin w t )e - st dt Î dt ˚ 0 dt a

Integrating by parts, we get we get = e - st (e - at sin w t )a0 + Ú se - st (e - at sin w t ) dt 0

a

= 0 + Ú s e- ( s + a ) t sin w t dt 0

a

È e-( s+ a )t ˘ = s◊Í [ - s sin w t - w cos w t ]˙ 2 2 Î (s + a) + w ˚0 = s◊

w +w2 (s + a )2 - sx

Èt ˘ aÊ ˆ (ii) By definition, a Í Ú e - ax cos w t dx ˙ = Ú Á Ú e - ax cos w t dx ˜ e – sx dx ¯e Î0 ˚ 0Ë0 a

a Ê e - sx x - ax ˆ 1 - sx - ax Integrating by parts, we get = Á [ e cos t dx ] w + Ú Ú e e cos x dx ˜ Ë -s 0 ¯0 s 0

(

)

a

= 0+

1 -(s+a ) x cos w x dx e s Ú0

=

a 1 - ( s + a ) x Ê e jw x + e - jw x ˆ e ÁË ˜¯ dx s Ú0 2

=

a a ˘ 1 È - ( s + a - jw ) x e dx e - ( s + a + jw ) x dx ˙ + ÍÚ Ú 2s Î 0 0 ˚ a

a

1 È e - ( s + a - jw ) x ˘ 1 È e - ( s + a - jw ) dx ˘ = Í˙ + Í ˙ 2s Î (s + a) - jw ˚ 0 2s Î (s + a + jw ) ˚ 0

Solution Manual (13) – Page No. 4

=

(s + a) s((s + a)2 + w 2 )

(a) and (b) are proved (or) verified. a

13-2.10 (a)

By definition, μ [ f (at )] = Ú f (at ) e - st dt 0

1 dt = dt a

Let at = a

s - t

μ [ f (at )] = Ú f (t ) e a 0

1 dt a

a

(b)

=

s - t 1 a ( ) f t e dt a Ú0

=

1 Ê sˆ F a ÁË a ˜¯

By definition, a

F (s) =μ ( f (t )) = Ú f (t ) e - st dt 0

Integrating both sides from ut, a

a a È ˘ - st F ( s ) ds = Ús Ús Í Ú0 f (t ) e dt ˙ ds Î ˚ By changing the order of integration, we get

μ μ Èμ ˘ Èμ ˘ = Ú f (t ) Í Ú e - ut du˙ dt = Ú F (t ) Í Ú e – st ds ˙ dt 0 0 Îs ˚ Îs ˚ μ

μ

= Ú f (t ) 0

e - ut e – st dt = Ú f (t ) dt t t 0

μ

μ

Ê f (t ) ˆ - ut Ê f (t ) ˆ È f (t ) ˘ – st È f (t ) ˘ = ÚÁ = ÚÍ e dt ◊ = μ Á e dt = μ Í ˜ ˜ ˙ ˙ Ë ¯ Ë ¯ t t t ˚ Î t ˚ 0 0 Î 13-2.11

s+2 s 2 + 5s + 6

s 3 + 7s 2 + 18s + 20 s 3 + 5s 2 + 65 – –– – – – – – – – – 2 s 2 + 12s + 20 2 s 2 + 10s + 12 ––––––––––– 2s + 8

2s + 8 Y ( s) = s + 2 + 2 s + 5s + 6 = s+2+

2(s + 4) (s + 2)(s + 3)

Solution Manual (13) – Page No. 5

2s + 8 A B = + (s + 2)(s + 3) s + 2 s + 3 A = (s + 2) ◊

2s + 8 4 = =4 (s + 2)(s + 3) s = -2 1

B = (s + 3) ◊

Y( s ) = s + 2 +

2s + 8 2 = = -2 (s + 2)(s + 3) s =-3 -1

4 2 s+2 s+3

Applying inverse Laplace transform,

y (t ) =

13-2.12 F (s) =

d d (t ) + 2d (t ) – 2 ◊ e -3t U (t ) + 4 ◊ e -2 t U (t ) dt

10(s 2 + 119) A Bs + C = + 2 2 (s + 5)(s + 10s + 169) s + 5 s + 10s + 169

10(s 2 + 119) = A(s 2 + 10s + 169) + Bs + C (s + 5) Comparing equal terms, 10 = A + B

–

(1)

0 = 10A + 5B +C

–

(2)

–

(3)

1190 = 169A + 5C Solving equations (1), (2) and (3), A = 10; B = 0; C = –100

F ( s) =

10 100 s + 5 (s + 5)2 - (12)2

Applying inverse Laplace transform,

100 ˘ È 10 ˘ -1 È L-1 [ F (s)] = L-1 Í -L Í 2 2 ˙ ˙ Îs + 5˚ Î (s + 5) - (12) ˚ f (t ) = 10e -5t f (t ) = 10e -5t 13-2.13 (a)

F ( s) =

100 -5t ◊ e ◊ sin12t 12

25 -5t ◊ e ◊ sin12t 3

18s 2 + 66s + 54 (s + 1)(s + 2)(s + 3)

Applying partial fractions to F(s),

18s 2 + 66s + 54 A B C = + + (s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3

Solution Manual (13) – Page No. 6

18s2 + 166s + 54 = A(s + 2) (s + 3) + B(s + 1) (s + 3) + C(s + 1) (s + 2) 18 = A + B + C

–

(1)

66 = 5A + 4B + 3C

–

(2)

54 = 6A + 3B + 2C

–

(3)

Solving equations (1), (2), and (3), A = 3, B = 6, C = 9

F ( s) =

3 6 9 + + s +1 s + 2 s + 3

Applying inverse Laplace transforms,

6 9 ˆ Ê 3 L-1 [ F (s)] = f (t ) = L-1 Á + + Ë s + 1 s + 2 s + 3 ¯˜ f(t) = 3e–t + 6e–2t + 9e–3t (b)

F ( s) =

11s 2 + 172s + 700 (s + 2)(s 2 + 12 s + 100)

Applying partial fractions,

F ( s) =

A Bs + C + s + 2 s 2 + 12s + 100

11s2 + 172s + 700 = A(s2 + 125 + 100) + Bs + C(s + 2) Comparing equal terms, 11 = A + B

–

(1)

172 = 12A + 2B + C

–

(2)

700 = 100A + 2C

–

(3)

Solving equations (1), (2), and (3), A = 5, B = 6, C = 100

F ( s) = =

=

5 6s 100 + 2 + 2 s + 2 s + 125 + 100 s + 125 + 100 5 6s 100 + + s + 2 (s + 6)2 + 82 (s + 6)2 + 82

5 6(s + 6) - 36 + 100 5 6(s + 6) 64 + = + + 2 2 2 2 s+2 s + 2 (s + 6) + 8 (s + 6)2 + 82 (s + 6) + 8

Applying inverse Laplace transform,

f (t ) = 5 ◊ e -2 t + 6 e -6 t ◊ cos 8t + 8 e -6 t ◊ sin 8t (c)

F ( s) =

56 s 2 + 112s + 5000 s(s 2 + 14 s + 625)

Solution Manual (13) – Page No. 7

F ( s) =

A Bs + C + 2 s s + 14s + 625

56s2 + 112s + 5000 = A(s2 + 14s + 625) + (Bs + C)s Comparing equal terms, 56 = A + B

–

(1)

112 = 14A + C

–

(2)

5000 = 625A

–

(3)

From equations (1), (2), and (3), A = 8; B = 48; C = 0

F ( s) =

8 48s + 2 s s + 14s + 625

Applying inverse Laplace transform,

48s Ê 8ˆ Ê ˆ L-1 [ f (s)] = L-1 Á ˜ + L-1 Á 2 ˜ Ë s¯ Ë s + 14s + 625 ¯ È 48s ˘ Ê 8ˆ = L-1 Á ˜ + L-1 Í 2 2 ˙ Ë s¯ ( s + 7) + 24 Î ˚ È 48(s + 7) ˘ -1 È 48 ¥ 7 ˘ Ê 8ˆ = L-1 Á ˜ + L-1 Í -L Í 2 2 ˙ 2 2 ˙ Ë s¯ Î (s + 7) + 24 ˚ Î (s + 7) + 24 ˚ f(t) = 8 + 48e–7t cos 24t – 14e–7t × sin 24t 13-2.14

(a) F (s) =

F ( s) =

=

40 (s 2 + 4 s + 5)2 40 ( s + 2 - j )2 (s + 2 + j )2

K1 K2 K1* K 2* + + + ( s + 2 - j )2 ( s + 2 - j ) ( s + 2 + j )2 ( s + 2 + j )

K1* and K 2* are conjugate values of K1 and K2 K1 =

40 = -10 (s + 2 + j )2 s =-2 + j

K2 =

˘ d È 40 = 80– - 90∞ Í ˙ ds Î (s + 2 + j )2 ˚ s = -2 + j

K1* = -10, K 2* = 80–90∞ F ( s) =

-10 80– - 90∞ (-10) 80–90∞ + + + 2 2 (s + 2 - j ) (s + 2 + j ) (s + 2 + j ) (s + 2 - j )

Solution Manual (13) – Page No. 8

By inverse Laplace transform,

f (t ) = -10t e( -2 + j ) t + 80– - 90∞ e( -2 + j ) t - 10t e - (2 + j ) t + 80–90∞ e- (2 + j ) t = –20 e–2t (tcos t – sin t ) (b)

F ( s) =

5s 2 + 29s + 32 (s + 2)(s + 4)

s 2 + 6s + 8

5 2

5s + 29s + 32 5s 2 + 30 s + 40 –

–

–

-s - 8 F ( s) = 5 -

s+8 (s + 2)(s + 4)

s+8 A B = + (s + 2)(s + 4) s + 2 s + 4 A = (S + 2)

6 =3 2

A=

B = (s + 4) ◊

=

s+8 (s + 2)(s + 4) s =−2

s+8 (s + 2)(s + 4) s = - 4

4 = -2 -2

F ( s) = 5 -

3 2 + s+2 s+4

Applying inverse Laplace transform,

f (t ) = 5 d (t ) - 3 e -2 t + 2 e -4 t (c)

F ( s) =

2 s3 + 8s 2 + 2 s - 4 s 2 + 5s + 4

s 2 + 5s + 4

2s - 2 3

2 s + 8s 2 + 2 s - 4 2s3 + 10 s 2 + 8s –

–

–

2

-2 s - 6 s - 4 -2s 2 - 10s - 8 +

+

+

4s + 4

Solution Manual (13) – Page No. 9

F ( s) = 2 s - 2 +

4(s + 1) (s + 5s + 4)

F ( s) = 2 s - 2 +

4(s + 1) (s + 1)(s + 4)

2

F (S ) = 2 S - 2 +

4 S+4

Applying inverse Laplace transform,

f (t ) =

2d d (t ) - 2d (t ) + 4 ◊ e - 4 t dt

13-3.1

x (t ) = [u(t ) - u(t - T )] + ( -2)[u(t - T ) - u(t - 2T )] +2[u(t - 2T ) - u(t - 4T )] + ( -2)[u(t - 4T ) - u(t - 5T )]

By Laplace transform,

x (s ) = =

1 e - Ts 2e - Ts 2e -2Ts 2e -2Ts 2e -4TS 2e 4TS 2e 5TS + + + s s s s s s s s 1 [1 - 3eTS + 4e -2 TS - 4e -4TS + Ze 5TS ] s

13-3.2

The time period T =

2p w

i(t ) = 0 for 0 to = 2 sin w t for

p w

p 2p <t< w w

Solution Manual (13) – Page No. 10

T

Hence, L(i(t)) =

1 f (t ) e - st dt 1 - e - sT Ú0 2 p /w Èp / w ˘ - st + o e dt ( ) 2 sin w t e - st dt ˙ Ú Ê 2p ˆ Í Ú - sÁ ˜ p /w ˚ Ëw¯ Î 0

1

=

1- e

=

=

2 p /w

È e - st ˘ (- s sin w t - w cos w t )˙ 2 -2 p s Í 2 s w + ˚ p /w 1- e w Î 2

-p s È -2wp s ˘ 2 w w e e Í ˙ -2 p s È ˘ 2 Î ˚ 2 w e s w + 1 [ ] Í ˙ Î ˚

13-4.1 As s = j ,

1 4 + s St p ( jw ) 2p J Ú 9 + s 1 4+s = p e St d ( jw ) 2p j Ú 9 + s 1 = F (s) e St ds 2p j Ú (4 + s) p F(s) 9+s 9 ˘ È 4 f ( s) = p Í +1s + 9 ˙˚ Îs + 9 By inverse Laplace transform, f (t ) = p (4e -9 t + d (t ) - 9 e -9 t ) Given f(t) =

= p (d (t ) - 5 e -9t ) 13-4.2

(a)

1 s2 + 9 1 1 3 = . 2 s + 9 3 s + 32 2

3 ˘ 1 È1 Applying inverse Laplace transform, L-1 Í ◊ 2 = sin 3t 2 ˙ Î3 s + 3 ˚ 3 (b)

2p s+p Applying inverse Laplace transform,

È 2p ˘ -p t L-1 Í ˙ = 2p ◊ e Î s - ( -p ) ˚ (c)

8 (s + 3) (s + 5)

Solution Manual (13) – Page No. 11

8 A B = + (s + 3) (s + 5) s + 3 s + 5 A = (3 + 3) ◊

8 8 = =4 (s + 3)(s + 5) s =-3 2

B = (s + 5) ◊

8 =-4 (s + 3)(s + 5) s =-5

=

4 4 s+3 s+5

Applying inverse Laplace transform,

4 e -3t - 4 ◊ e -5t 5 s +9

(d)

2

5 5 3 = ◊ 2 s + 9 3 s + 32 2

Applying inverse Laplace transform,

3 ˘ 5 È5 L-1 Í ◊ 2 = sin 3t 2 ˙ Î3 s + 3 ˚ 3 K1 K 2 K 3 + 2 + 3 s s s

(e)

Applying inverse Laplace transform,

K K ˘ K + K2 t + K3t 2 ÈK L-1 Í 1 + 22 + 33 ˙ = 1 2 s s ˚ Î s 13-4.3

F ( s) =

1 A B = + 2 s + 2 (s + 2)2 (s + 2)

1 = A(s + 2) B. Comparing equal terms, 0 = A; 1 = 2A + B

F ( s) =

B=1

1 (s + 2)2

È 1 ˘ = t e –2 t U (t ) f (t ) = L-1 Í 2 ˙ ( + 2) s Î ˚ 13-5.1

(a) F (s) =

18s 2 + 66s + 54 (s + 1)(s + 2)(s + 3)

Initial value theorem: Lt f (t ) = Lt sF (s) tÆ0

sƕ

Solution Manual (13) – Page No. 12

= Lt s ◊ sÆ•

18s 2 + 66s + 54 (s + 1)(s + 2)(s + 3) 66 s

s 3 18 Lt

s

s3 1

1 s

1

54 s2 2 s

1

3 s

f (0) = 18 Final value theorem: Lt f (t ) = Lt sF (s) t→∞

= Lt s ◊ tÆ•

f (•) =

(b)

F ( s) =

t→∞

18s 2 + 66 s + 54 (s + 1)(s + 2)(s + 3)

0 =0 6

11s 2 + 172s + 700 (s + 2)(s 2 + 12 s + 100)

Initial value theorem: Lt f (t ) = Lt s ◊ F (s) = Lt s ◊ tÆ0

sƕ

sƕ

11s 2 + 172 s + 700 (s + 2)(s 2 + 12s + 100)

172 700 ˆ Ê ÁË 11 + s + 2 ˜¯ s = Lt s3 ◊ sÆ• 2 12 100 ˆ Ê ˆ Ê s3 Á 1 + ˜ Á 1 + + 2 ˜ Ë ¯ Ë s s s ¯ f (0) =

11 = 11 1

Final value theorem: Lt f (t ) = Lt sF (s ) = Lt s ◊ tÆ0

sÆμ

s Æq

11s 2 + 172s + 700 (s + 2)(s 2 + 12s + 100)

f (•) = 0 (c)

F ( s) =

56 s 2 + 112s + 5000 s(s 2 + 14 s + 625)

Initial value theorem : Lt f (t ) = Lt s ◊ tÆ0

sƕ

=

56s 2 + 112s + 5000 s (s 2 + 14s + 625)

56 = 56 = f (0) 1

Final value theormem : Lt f (t ) = Lt s ◊ tÆ•

sÆ0

56s 2 + 112s + 5000 s (s 2 + 14 s + 625)

Solution Manual (13) – Page No. 13

f (•) =

13-5.2

(a)

F ( s) =

5000 =8 625

7s 2 + 63s + 134 (s + 3)(s + 4)(s + 5)

Initial value theorem: Lt f (t ) = Lt sF (s) tÆ0

sƕ

7s 2 + 63s + 134 = Lt s ◊ 3 sÆ• s (s + 3)(s + 4 )( s + 5) 63 134 ˆ Ê s3 Á 7 + + 2 ˜ Ë 7s + 63s + 134 s s ¯ = Lt s ◊ = Lt sÆ• Ê 3ˆ Ê 4 ˆ Ê 5ˆ s Æ • 3 Ê 3ˆ Ê 4 ˆ Ê 5ˆ s3 Á 1 + ˜ Á 1 + ˜ Á 1 + ˜ s Á1 + ˜ Á1 + ˜ Á1 + ˜ Ë s¯ Ë Ë s¯ Ë s¯ Ë s¯ s¯ Ë s¯ 2

f(0) = 7 Final value theorem Lt f (t ) = Lt sF (s) tƕ

F (S ) =

sÆ0

7S 2 + 63S + 134 (S + 3)(S + 4)(S + 5) 7s 3 + 63s 2 + 134s =0 s Æ 0 (s + 3)(s + 4)(s + 5)

Lt sF (s) = Lt

sÆ0

(b)

F ( s) =

4 s 2 + 7s + 1 s(s + 1)2

Initial value theorem:

Lt f (t ) = Lt sF (s)

tÆ0

sƕ

7 ˆ Ê s 2 Á 4 + + 1˜ Ë 4 s + 7s + 1 s ¯ Lt s ◊ = Lt sÆ• sÆ• 2 1ˆ s (s + 1)2 Ê s2 Á 1 + + 2 ˜ Ë s s ¯ 2

f (0) = 4 Final value theorem: Lt f (t ) = Lt s ◊ F (s) tÆ•

Lt s ◊

sÆ0

sÆ0

4 s 2 + 7s + 1 4 s 3 + 7s 2 + s = Lt s Æ 0 s3 + 2 s 2 + s s(s + 1)2

4 s 2 + 7s + 1 =1 s Æ 0 s2 + 2s + 1 Lt

f (•) = 1 (c)

F ( s) =

40 ( s + 4 s + s )2 2

Solution Manual (13) – Page No. 14

40

Initial value theorem: Lt f (t ) = Lt sF (s) = Lt s ◊ tÆ0

sƕ

sƕ

Ê 4 5ˆ s4 Á1 + + 2 ˜ Ë s s ¯

f (0) = 0 Final value theorem: Lt f (t ) = Lt s ◊ F (s) = Lt s ◊ tÆ•

=0 13-5.2

sÆ0

sÆ0

40 ( s + 4 s + s) 2 2

f (•) = 0

Lt g (t ) = Lt sg (s)

tÆ0

sƕ

Ê 1ˆ s Á1 + ˜ Ë s¯ s +1 Lt s ◊ = Lt =1 sÆ• s Æ • Ê 2ˆ s (s + 2) s Á1 + ˜ Ë s¯ Verification:

g ( s) =

s +1 s(s + 2)

=

s 1 + s(s + 2) s(s + 2)

=

1 È 0.5 0.5 ˘ + s + 2 ÍÎ s s + 2 ˙˚

y(t) = e–2t + 0.5 – 0.5 e–2t = 0.5 e–2t + 0.5 Initial value Lt ( y(t )) = 0.5(1) + 0.5 = 1 tÆ0

Solution Manual (13) – Page No. 15

2

CHAPTER 14 (Solution Manual) 14-2.1

The circuit representation in the S-domain of the above problem is given below

500 Ω, 16 × 10–3 S and ∴

1 25 × 10 –9 S

1 1 1 1 = + + –3 Y1 500 16 × 10 S 1 / 25 × 10 –9 S

=

1 1000 + + 25 × 10 –9 S 500 16S

=

16S + 5 × 10 +5 + 200 × 10 –6 S 2 8000 S Y1 =

8000 S 16 S + 5 × 105 + 200 × 10 –6 S 2

Overall impedance, Z = 2000 +

=

= 14-2.2

are in parallel,

8000 S 16 S + 5 × 105 + 200 × 10 –6 S 2

0.4 S 2 + 4000 S + 109 (200 × 10 –6 S 2 + 16 S + 500000) 2000(S + 50000)2 (S 2 + 80000 S + 25 × 108 )

The circuit representation in the S-domain is given below

⎛ 1 ⎞ × 10 –9 S ⎟ Z = (1000 + 0.5 S) || ⎜ ⎝ 0.4 ⎠

⎛ 1 ⎞ × 10 –9 S ⎟ (1000 + 0.5 S ) ⎜ ⎝ 0.4 ⎠ = 1 1000 + 0.5 S + 0.4 × 10 –9 S

Solution Manual (14) – Page No. 1

=

= 14-2.3

1000 + 0.5 S 0.4 × 10

–6

S + 0.2 × 10 –9 S 2 + 1

2.5 × 109 (S + 2000) (S 2 + 2000 S + 5 × 109 )

The S-domain equivalent circuit is given below

1 160 I(S) = 0.25S S

(a) By KVL: 4.8 I(S) + 4 SI(S) + 160

I(S) =

(b) I(S) =

1 ⎞ ⎛ S ⎜ 4.8 + 4 S + ⎟ ⎝ 0.25S ⎠

40 S + 1.2S + 1 2

=

40 S + 1.2 S + 1 2

40

=

(S + 0.6) + 0.8 2

2

=

50 × 0.8 (S + 0.6)2 + 0.82

By inverse Laplace transform, i(t) = 50 e–0.6t sin (0.8t) A (c) V(S) = I(S) . (4S) =

(d) V(S) = = =

16 s s + 1.2s + 1 2

160 S S + 1.2 S + 1 2

160(S + 0.6) (S + 0.6) + 0.8 2

2

160(S + 0.6) (S + 0.6) + 0.8 2

2

–

–

160 × 0.6 (S + 0.6)2 + 0.82 120 × 0.8 (S + 0.6)2 + (0.8)2

By inverse Laplace transform, v(t) = 160 e–0.6t cos (0.8t) – 120 e–0.6t sin (0.8t) = 200 e–0.6t cos(0.8t + 36.87º)

Solution Manual (14) – Page No. 2

14-2.4

(a) The S-domain equivalent circuit is given below

By KCL at the node 1, SV1 +

V1 – V2 160 = S S

V2 160 1⎞ ⎛ = ⎜⎝ S + ⎟⎠ V1 – S S S

(1)

By KCL at the node 2,

V2 – V1 V2 + + S 3 ⇒

15 S =0 15

V2 –

–V1 ⎛ 1 1 1 ⎞ 1 + ⎜ + + ⎟ V2 = S ⎝ S 3 15 ⎠ S

(2)

From equations (1) and (2),

V1(S) =

160 S

⎛ –1 ⎞ ⎜⎝ ⎟⎠ S

1 S

⎛1 6 ⎞ ⎜⎝ + ⎟⎠ S 15

=

1⎞ ⎛ ⎜⎝ S + ⎟⎠ S

1 – S

1 S

⎛1 6 ⎞ ⎜⎝ + ⎟⎠ S 15

–

960 S + 2415 S (6 S 2 + 15S + 6)

1 ⎞ 160 ⎛ ⎜⎝ S + ⎟⎠ S S 1 1 – 15S 2 + 2415 S S V2(S) = = 1 1 S(8S 2 + 15S + 6) – S+ S S 1 ⎛1 6 ⎞ – ⎜ + ⎟ S ⎝ S 15 ⎠

(b)

V1(S) =

960 S + 2415 S (6 S 2 + 15S + 6)

=

160 S + 402.5 S(S + 2)(S + 0.5)

Solution Manual (14) – Page No. 3

160S 402.5 + (S + 2)(S + 0.5) S(S + 2)(S + 0.5)

=

⎡ –106.6 106.6 ⎤ ⎡ 402.5 134 536.66 ⎤ + + + – = ⎢ S + 0.5 ⎥⎦ ⎢⎣ S S + 2 S + 0.5 ⎥⎦ ⎣ S+2 =

402.5 27.4 430 + – S S + 2 S + 0.5

By inverse Laplace transform, V1(t) = 402.5 + 27.4 e–2t – 430 e–0.5t V2(S) = =

15S 2 + 2415 S (6 S 2 + 15S + 6)

=

2.5S 2 + 402.5 S(S + 2)(S + 0.5)

2.5S 402.5 + (S + 2)(S + 0.5) S(S + 2)(S + 0.5)

⎡ 3.33 0.833 ⎤ ⎡ 402.5 134 536.66 ⎤ – + – = ⎢ ⎥+⎢ S + 2 S + 0.5 ⎥⎦ ⎣ S + 2 S + 0.5 ⎦ ⎣ S =

402.5 137.33 537.5 + – S S+2 S + 0.5

By inverse Laplace transform, V2(t) = 402.5 + 137.5 33e–2t – 537.5 e–0.5t (c) V1(0+) = 402.5 + 27.4(1) – 430(1) = 0 V2(0 +) = 402.5 + 137.33(1) – 537.5 = 0 (d) Steady-state values of V1 = 402.5 volts V2 = 402.5 volts 14-2.5

The S-domain equivalent circuit is given below

(a) Owing to the voltage source

Solution Manual (14) – Page No. 4

20 V S + V + =0 2 1.25S 1 / 0.05S

V–

By KCL at the node 1:

1 ⎛1 ⎞ 10 + 0.05S ⎟ = V⎜ + ⎝ 2 1.25S ⎠ S

V= =

25 0.125S + 1.25S + 2 2

=

(200) (S + 2)(S + 8)

100 100 – 3(S + 2) 3(S + 8)

By inverse Laplace transform, V(t) =

100 –2t –8t (e – e ) volts 3

(b) Owing to the current source

By KCL at the node 1: =

V V V 5 + + = 2 1.25S 1 / 0.05S S

1 ⎛1 ⎞ 5 V⎜ + + 0.05S ⎟ = ⎝ 2 1.25S ⎠ S

V =

100 100 – 6(S + 2) 6(S + 8)

By inverse Laplace transform, V(t) =

100 –2t –8t (e – e ) volts 6

(c) By superposition, ⎡ 100 –2 t –8t ⎤ ⎡100 –2 t –8t ⎤ V= ⎢ (e – e ) ⎥ + ⎢ (e – e )⎥ ⎣ 3 ⎦ ⎣ 6 ⎦

=

300 –2t –8t (e – e ) volts 6

= 50 (e–2t – e–8t) volts

Solution Manual (14) – Page No. 5

14-2.6

The S-domain circuit is shown below.

Nodes 1 and 3 combined forms a supernode.

By KCL to the supernode,

V –V V V1 V –V + 1 2 + 3 2 + 3 =0 1 / 0.02 S 25 S 25 25

⎡ ⎡ 1 1 ⎤ 1⎤ ⎡1 1⎤ ⎢0.02 S + 25 S ⎥ V1 – ⎢ 25 S + 25 ⎥ V2 + ⎢ 25 + 25 ⎥ V3 = 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ and But

(1)

V3 – V1 = 20 IS(S) IS(S) =

V1 0.02 S

⎛ V1 ⎞ V3 – V1 = 20 ⎜ ⎝ 0.02 S ⎟⎠

∴

⎛ 1000 ⎞ – ⎜1 + ⎟ V1 + V3 = 0 ⎝ S ⎠

(2)

By KCL at the node 2,

V2 – V1 V2 – V3 100 + = S 25 S 25 –

V ⎡ 1 1 1⎤ 100 V1 + ⎢ + ⎥ V2 – 3 = 25 S 25 S ⎣ 25 S 25 ⎦

(3)

From equations (1), (2) and (3),

Solution Manual (14) – Page No. 6

0

⎛ 1 1 ⎞ + ⎟ –⎜ S 25 25 ⎝ ⎠ 0

100 S

⎛ 1 1 ⎞ + ⎟ ⎜ ⎝ 25 S 25 ⎠

0

V1(S) =

=

1 25

⎛ 1000 ⎞ – ⎜1 + S ⎟⎠ ⎝

0

1

1 25 S

⎛ 1 1 ⎞ + ⎟ ⎜ S 25 25 ⎝ ⎠

1 25 S

–

–4 4 – S2 5 = –(1.6 + 1.6016 S + 0.0024S 2 + 0.0008 S 3 ) S2

1 25

4S + 4 0.0008S + 0.0024 S 2 + 1.6016 S + 1.6 3

5000 S + 5000 (S + 1)(S + 45.75)(S – 43.75) ⎛ 1 ⎞ ⎜⎝ 0.02 S + 25 S ⎟⎠

0

2 25

⎛ 100 ⎞ – ⎜1 + ⎟ ⎝ S ⎠

0

1

1 25 S

100 S

– V2(S) =

–

2 25

–

=

1

⎛ 1 1 ⎞ –⎜ + ⎟ ⎝ 25 S 25 ⎠

0.025S +

=

2 25

–

1 25

1 0.02 S + 25 S

⎛ 1 1⎞ –⎜ + ⎟ ⎝ 25 S 25 ⎠

2 25

⎛ 100 ⎞ – ⎜1 + ⎟ ⎝ S ⎠

0

1

–1 25S

⎛ 1 1⎞ ⎜⎝ 25 S + 25 ⎟⎠

–1 25

=

⎛ 2S 2 + 85 + 804 ⎞ –⎜ ⎟ S2 ⎝ ⎠ –(1.6 + 1.6016 S + 0.0024 S 2 + 0.0008 S 3 ) S2

2500 S 2 + 10000 S + 1005000 (S + 1)(S + 45.75)(S – 43.75)

(a) Io(t) =

V1 (S ) V (S ) – V2 (S ) + 1 1 / 0.02S 25S

⎛ 1 ⎞ 1 2 ⎜⎝ 0.02S + 25 S ⎟⎠ [5000 S + 5000] – 25 S [2500 S + 10000 S + 1005000] = (S + 1)(S + 45.75)(S – 43.75) =

100 S 3 – 200 S – 40000 S(S + 1)(S + 45.75)(S – 43.75)

=

20 20 53 0.2 + + – S S + 1 (S + 45.75) (S – 43.75)

Solution Manual (14) – Page No. 7

(b) By initial value theorem, i(0+) = Lt SI(S) S →∝

100 S 3 – 200 S – 40000 S →∝ (S + 1)(S + 45.75)(S – 43.75)

= Lt

= 100 A By final value theorem, i(∝) = Lt SI(S) S→ 0

100 S 3 – 200 S – 40000 s → 0 ( S + 1)( S + 45.75)( S – 43.75) Lt

= 20 A (c) In Case (b), agreeing to the results, for t = 0 +, the inductor acts as a short circuit and the capacitor acts open circuit and For t = ∝, the inductor acts as open circuit and the capacitor acts as short circuit.

20 20 53 0.2 + + – (d) I0(S) = S S + 1 S + 45.75 S – 43.75 By inverse Laplace transform, I0(t) = (20 – 20e–t + 53e–45.75t + 0.2 e43.75t) A 14-2.8

The S-domain circuit is given below

(i)

When

60 voltage source is acting alone S

By KCL,

60 V0 S + V (0.5 × 10 –6 S ) + =0 01 2000 (S + 4000)

V01 –

⎛ 1 ⎞ 1 60 + 0.5 ×10 –6 S + V01 ⎜ ⎟= S + 4000 ⎠ 2000 S ⎝ 2000

Solution Manual (14) – Page No. 8

V01 =

(ii)

60(S + 4000) S (10 S + 5S + 6000) –3 2

When the

=

60000(S + 4000) S(S 2 + 5000 S + 6 × 106 )

12 current source is acting alone S

By KCL at the node 1:

V02 V02 V –V + + 02 x = 0 2000 1 / 0.5 × 10 –6 S S

V 1⎞ ⎛ 1 + 0.5 × 10 –6 S + ⎟ V02 – x = 0 ⎜⎝ ⎠ S S 2000

By KCL at the node 2: –

Vx – V02 V 12 + x = S 4000 S

V02 ⎛ 1 1 ⎞ 12 Vx = +⎜ + ⎟ S ⎝ S 4000 ⎠ S

From equations, V02 =

=

= (a)

(1)

(2)

0

–1 / S

12 S

1 ⎞ ⎛1 ⎜⎝ + ⎟ S 4000 ⎠

10 –3 S 2 + S + 2000 2000 S

–

–1 S

1 ⎞ ⎛1 ⎜⎝ + ⎟ S 4000 ⎠

1 S

12 × 8 × 106 S(10 –3 S 2 + 5S + 6000) 96 × 109 S(S 2 + 5000 S + 6 × 106 )

By superposition, V0 = V01 + V02 =

6000(S + 4000) + 96 × 109 S(S 2 + 5000 S + 6 × 106 )

Solution Manual (14) – Page No. 9

=

12 / S 2 10 –3 S 2 + 5S + 6000 (8 × 106 S )

(b)

=

6000 S + 96.024 × 109 S(S + 2000)(S + 3000)

V0 =

6000 S + (96.024 × 109 ) S(S + 2000)(S + 3000)

=

16004 48006 32002 + – S S + 2000 S + 3000

∴ by inverse Laplace transform,

V0(t) = (16004 – 48006 e–2000t + 32002 e–3000t) volts 14-2.8

The S-domain circuit is given below

(a) The unit step response is Ig =

1 S

V0 (S ) V0 (S ) 1 + = 1 2+S S 0.1S

By KCL,

V0(S) =

10 S + 20 S (S 2 + 2 S + 10)

=

A B B* + + S (S + 1– J 3) (S + 1 + J 3)

=

2 (–1– J1.33) (–1 + J1.33) + + S (S + 1 – J 3) (S + 1 + J 3)

By inverse Laplace transform, V0(t) = [2 + 3.33 e–t cos (3t – 126.87°)] volts (b) The unit impulse response: is Ig = 1

By KCL,

V0 (S ) V0 (S ) + =1 2 + S 1 / 0.1S

V0(S) =

10S + 20 S + 2 S + 10 2

10( S + 1) 10 = (S + 1)2 + 32 + (S + 1)2 + 32

By inverse Laplace transform,

Solution Manual (14) – Page No. 10

V0(t) = 10 e–t cos 3t + 14-2.9

10 –t e sin 3t = 10.54 e–t cos (3t – 126.87º) volts.

3

The S-domain circuit is given below

⎡ ⎤ 750 ×10 –3 –3 V0(S) = ⎢ ⎥ 20 ×10 S –3 –3 ⎢⎣ (100 + 5 ×10 S + 20 ×10 S ) ⎥⎦ =

⎞ 3 ⎛ S + 4000 3⎛ S 4000 ⎞ = ⎜ – ⎜ ⎟ 5 ⎝ S + 4000 ⎠ 5 ⎝ S + 4000 S + 4000 ⎠⎟

By inverse Laplace transform VO(t) =

3 [S(t) – 4000 e–4000t] volts.

5

14-2.10 Vi is not given, assume Vi = 160 volts.

When the switch is in the position a for long time, both the capacitors of 0.5 µF and 2 µF are charged to 80 V and the capacitor of 1.5 µF is charged to 0 V. (a) v1(0) = 80 V (b) v1(0 –) = 80V (c) v3(0 –) = 0 (d) When the switch is in the position b, the equivalent circuit is given below

I(S) =

160 / S 1 1 1 + + –6 –6 0.5 × 10 S 2 × 10 S 1.5 × 10 –6 S

= 50.5 × 10–6 i(t) = 50.5 δ(t) µA (e) v1(0 +) = 80 V

Solution Manual (14) – Page No. 11

(f) v2(0 +) = 80 V (g) v3(0 +) = 0 V 14-2.11

I R1

0 1 1 0

5 0.5

R2 L1

2 0 (1 / 8) 1 2 1 PRINT DC V(2) PROBE END

14-2.12 f =

4 = 0.63 2π

V 1 0 sin(0 2 0.63 0) R 1 2 6 L 2 3 1 1 C 3 0 IC = 2 9 TRAN 0.1 PROBE PLOT TRAN I(C) END

14-3.1

The S-domain circuit is given below:

By KCL,

V0 (S ) V0 (S ) + = I 0 (S ) 2 + S 1 / 0.1S

V0 (S ) = I 0 (S )

14-3.2

1 1 + 0.1S 2+S

=

10(2 + S ) (S 2 + 2 S + 10)

(a) By Laplace transform, V0(S) = L[10000 e–70t (cos 240t . cos θ – sin θ sin 240t)]

Solution Manual (14) – Page No. 12

= L [10000 e–70t [0.96 cos 240t – 0.28 sin 240t]] = 9600

(S + 70) (S + 70)2 + 2402

–

2800(240) (S + 70)2 + 2402

=

9600 S (S + 70)2 + 2402

For unit impulse, input × (S) = 1 ∴ transfer function, H(S) =

For unit step, input × (S) =

9600 S (S + 70)2 + 2402

1 S

∴ output, VO = H(S) × (S)

=

9600 (S + 70)2 + 2402

∴ the response, VO(t) = 40 e–70t sin 240t

14-4.1

The S-domain circuit is

V0 =

Vi S +1

When Vi is the unit impulse function (t), V0 = e–t v(t) By convolution integral, t

For

t

0 < t < 1: V0 = ∫ e – τ dt = – e – τ = [1 – e – t ] volts.

0

0

t

For 1 < t < ∝: V0 =

∫e

–τ

dt = (e – 1)e–t volts.

t –1

14-4.2

The S-domain circuit is given below

Solution Manual (14) – Page No. 13

V0 =

S Vi 1+ S

When Vi is the unit impulse function δ(t), V0 = [δ(t) – e–t u(t)] By convolution integral, t

For 0 < t < 1: V0 = ∫ [δ(t ) – e – τ u(t )] dt 0

= (1 + e–τ)ot = (e–t – 1). t

for 1 < t < ∝: VO =

∫ [δ(τ) – e u(τ)] d τ –τ

t –1

= e–t (1 – e) 14-4.3

The S-domain circuit is given below

⎛ 1 ⎞ (2 + S ) ⎜ ⎝ 0.1S ⎟⎠ H(S) = Transfer function = ⎛ 1 ⎞ (2 + S ) + ⎜ ⎝ 0.1S ⎟⎠ =

2+S 0.1S + 0.2 S + 1 2

As ω = 4, transfer function =

2 + J4 0.1( J 4)2 + 0.2( J 4) + 1

= 4.47 –63.43° ∴ V0 = H(S) Ig(S)

= 44.7 cos (4t – 63.43º)

Solution Manual (14) – Page No. 14

CHAPTER 15 (Solution Manual)

15-4.1

The transformed circuit is shown below.

Transforming ABC delta into equivalent star, we have the following circuit.

s¥ Za =

=

1 s

1 s + +1 s

s s + s +1 2

Solution Manual (15) – Page No. 1

1 1 s Zb = = 2 1 s + s +1 s + +1 s

Zc =

s ¥1 s2 = 2 1 s + s +1 s + +1 s

The driving point impedance Z(s) = Zd || Ze + Zc

15-4.2

=

2 s 5 + 5s 4 + 7s3 + 6s 2 + 3s + 1 (s 2 + s + 1)(s3 + 2s 2 + 3s + 2)

=

(2 s 2 + 3s 2 + 2 s + 1)(s2 + s + 1) (s 2 + s + 1)(s3 + 2s 2 + 3s + 2)

Z(s) =

2s3 + 3s2 + 2s + 1 s3 + 2s2 + 3s + 2

y(s) =

s3 + 2s2 + 3s + 2 2s3 + 3s2 + 2s + 1

Transformed network is shown below

s +1 Ê 1ˆ V2 (s) = Á1 + ˜ I1 (s ) = I (s ) Ë s¯ s 1

(1)

È 1 ˘ Í1 ¥ s 1˙ +1+ ˙ V1 (s) = I1 (s) Í s˙ Í1 + 1 ÎÍ s ˚˙

È s2 + 3s + 1 ˘ V1 (s) = I1 (s) Í ˙ Î s(s + 1) ˚

(2)

From equations (1) and (2),

Solution Manual (15) – Page No. 2

s +1 I1 ( s ) V2 ( s) G21 ( s) = = 2 s V1 ( s) s + 3s + 1 I ( s) s( s + 1) 1 G21 (s) =

15-4.3

s2 + 2s + 1 s2 + 3s + 1

The transformed network is shown below.

Applying KCL at nodes A and B,

I1 (s) = 3I (s) + and

VA (s) - VB (s) 2s

VA (s ) - VB (s ) VB = + I (s ) 2 2s S

(1) (2)

Substituting Eq. (2) in Eq. (1), we get

I1 (s) = 3I (s) +

VB ¥ s I ( s) 2

Also, VB = 10 I(s)

5 I1 (s) = 3I (s) + 10I (s) + I (s) 2 I1(s) = (3 + 5s + 1) I(s) = (5s + 4) I(s)

(3)

Applying KVL to the outer loop, Vin = 5I1(s) + VA Vin = 5I1(s) + VA(s) From Eq. (2),

(4)

VA (s) - VB (s) +SVB (s) = + I (s ) 2s 2 +

VA (s) È 1 s˘ = VB (s) Í + ˙ I (s) 2s Î 2s 2 ˚ È1 + s2 ˘ = 10 I (s) Í ˙ + I ( s) Î 2s ˚ VA(s) = (10 + 10s2 + 2s) I(s) Substituting VA(s) in Eq. (4),

Solution Manual (15) – Page No. 3

Vin = 5I1(s) + (10s2 + 2s + 10) I(s) From Eq. (3), I(s) =

1 I1 ( s ) 5s + 4

Vin( s ) = 5I1 (s) + (10s2 + 2s + 10)

15-5.1

1 I1 (s) 5s + 4

Zd =

Vin (s) Ê 5 + 10S 2 + 2S + 10 ˆ 25s + 20 + 10s 2 + 2s + 10 = ˜¯ = I1 (s) ÁË 5S + 4 5s + 4

Zd =

10 s 2 + 27s + 30 5s + 4

In the network function, P(s) = 5(s + 5) and (S) = (s + 2) (s + 7) By taking partial functions, V(s) can be written as

V ( s) =

A B + s+2 s+7

Therefore, the time-domain response is V(t) = Ae–2t + Be–7t

Consider the pole at –2.

M52 = distance between zero to the pole at –2 = 3 The angle between the line joining the pole at –2 to the zero φ52 = 0° Similarly, M72 = 5 φ 72 = 0° A=H

M 52 e jf52 5 ◊ 3 ◊ e jo = =3 5 ◊ e jo M 72 ◊ e jf72

B=H

M 57 e jf57 5 ◊ 2 ◊ e j180 = =2 5 ◊ e j180 M 27 ◊ e jf27

Substituting A, B values, we have

Solution Manual (15) – Page No. 4

V(t) = 3e–2t + 2 e–7t Analytically, V(s) =

5(s + 5) (s + 2) (s + 7)

By taking partial fractions,

A B + s+2 s+7

V (s) =

5(s + 5) = A(s + 2) + B(s + 7)

A = V (s) (s + 2) S=-2 =

5(-2 + 5) =3 -2 + 7

B = V (s) (s + 7) s =-7 =

5(-7 + 5) =2 (-7 + 2)

V ( s) =

3 2 + s+2 s+7

By taking inverse Laplace transform, we get V(t) = 3e–2t + 2e–7t 15-5.2

The transformed network is shown below.

The impedance function 1ˆ Ê Z (s) = (5 + 10 s) || Á 5 + ˜ Ë 5s ¯ Ê 25s + 1ˆ (5 + 10 s) Á Ë 5s ˜¯ (5 + 10 s)(25s + 1) = = 1 5s(10 + 10 s) + 1 5 + 10 s + 5 + 5s

1ˆ Ê 1ˆ Ê 250 Á 5 + ˜ Á s + ˜ Ë 2 ¯ Ë 25 ¯ Z (s) = (s + 0.02) (s + 0.98) Here, p(s) = (s + 0.5) (s + 0.04) p(s) = (s + 0.02) (s + 0.98) Zeros are s = –0.5, s = –0.04 Poles are located at s = –0.02, s = –0.98 Pole-zero diagram

Solution Manual (15) – Page No. 5

15-5.3

By taking partial fractions, we get

V ( s) =

K1 K + 2 s+2 s+3

By taking inverse transform, V(f) = K1 e -2 f + K 2 e -3 t Pole zero plot

K1 =

=

K2 =

H ◊ M02 e jf02 M32 e jf32 4 ◊ 2 e j 180 = -8 1◊ e j0

H ◊ M03 e jf03 4 ◊ 3 e j 180 = = 12 jf03 1 ◊ e j180 M23 e

Substituting K1 and K2 values, v(f) = -8 e -2 t + 12 e -3 t 15-5.4

The transformed circuit is shown below.

Solution Manual (15) – Page No. 6

(a) By applying Kirchhoff’s laws, we get V1(s) = I1 (s) 2s +

1 [I (s) - I2 (s)] S 1

(1)

1 (2 +s) I 2 (s) + [ I 2 (s) - I1 (s)] = 0 5

(2)

1˘ 1 È I 2 (s ) Í 2 + s + ˙ = I1 (s ) s˚ s Î

I 2 ( s) =

1 I1 (s) s + 2s + 1 2

(3)

1È 1 ˘ From Eq. (3), V1(s) = I1 (s) 2s + Í I1 (s) - 2 I ( s) sÎ s + 2s + 1 1 ˙˚ È 1 1 ˘ = I1 (s) Í2s + - 2 s s(s + 2s + 1) ˙˚ Î

È 2s3 + 4s2 + 3s + 2 ˘ = I1 (s) Í ˙ s2 + 2 s + 1 ˚ Î (b)

Also, V2(s) = I2(s) × s

G21(s) =

V2 (s) = V1 (s)

(4)

I 2 ( s) ¥ s È 2 s 3 + 4 s 2 + 3s + 2 ˘ I1 ( s ) Í ˙ s2 + 2s + 1 ˚ Î

s I ( s) 2 2 s + s +1 1 = È 2 s 3 + 4 s 2 + 3s + 2 ˘ I1 ( s ) Í ˙ s2 + 2s + 1 ˚ Î

Y21 (s) =

I 2 ( s) V1 (s) V = 0 2

(c)

2 ˘ È Í ˙ V1(s) = I1 (s) Í2s + s ˙ 1 Í 2+ ˙ s˚ Î È 4s2 + 2 s + 2 ˘ = I1 (s) Í ˙ Î 2s + 1 ˚

(5)

By applying current division rule,

1 1 I2(s) = - I1 (s ) s = I1 (s) 1 2s + 1 +2 s

(6)

From equations (5) and (6),

Solution Manual (15) – Page No. 7

I 2 ( s) = V1 (s)

1 2s + 1 È 4s2 + 2s + 2 ˘ I1 (s) Í ˙ Î 2s + 1 ˚

Y21 (s) = -

- I1 (s)

1 4 s + 2s + 2 2

I ( s) a 21 (s) = 2 I1 (s) V = 0 2

From eq.(6), a 21 = 15-5.5

I 2 ( s) -1 = I1 (s) (2 s + 1)

Transformed network is shown below

(a)

b)

c)

2 ¥s 1 The driving-point impedance Z11(s) = s + + 2s 2 s +s s =

2s 1 2 s 2 + s 2 + 2 + 2 s 2 (s 2 + 2) + + 2s = s +2 5 s(s 2 + 2)

=

2 s 4 + 7 s 2 + 2 2 s 4 + 7s 2 + 2 = s(s 2 + 2) s3 + 2 s

2

The driving point admittance Y11(s) =

s3 + 2 s 2s 4 + 7s2 + 2

Ê1 ˆ + 2s˜ s(s 2 + 2) ¯ V2 (s) ÁË s G21 (s) = = 4 V1 (s) 2 s + 7s 2 + 2 =

(2 s 2 + 1)(s 2 + 2) 2 s 4 + 5s 2 + 2 = 4 2 s 4 + 7s 2 + 2 2 s + 7s 2 + 2

I ( s) x21 (s) = 2 I1 (s) V = 0 2

Solution Manual (15) – Page No. 8

Then the circuit is redrawn as given.

I1(s) = –I2(s) x21 (s) =

15-7.1

I ( s) =

I 2 ( s) =1 I1 ( s )

5s 5s = 2 (s + 3)(s + 2 s + 2) (s + 3)(s + 1 + j1 )(s + 1 - j1)

Here, p(s) = s θ(s) = (s + 3) (s + 1 + j1) (s + 1 –j1) By taking partial fractions, we get I ( s) =

A B C + + s + 3 s + (1 + j1) s + (1 - j1)

i(t) = A ◊ e-3t + B ◊ e- (1 + j1)t + C ◊ e(1 + j1)t Pole zero diagram

A=

=

H ◊ Mo3 e jfo 3 jf

jf

M p11 e p11 ◊ M p21 e p21 5 ◊ 3 ◊ e j180 5e

j 206◊56

5e

j153.43

=

15 - j180 ◊e = -3 5

Solution Manual (15) – Page No. 9

jf

B=

H M op 2 e op 2 M p1 p2 e

if p1 p2

M 3 p2 e

jf3 p2

=

5 ◊ 2 e j 225 2 ◊ e j 270 ◊ 5e j 333.44

= 1.58 e–j378.44 = 1.58[0.95 – j 0.316] jf

C=

H M op1 e op1 if

jf

M p2 p1 e p 2 p1 M 3 p1 e 3 p1

=

5 ◊ 2 e j135 2 ◊ e j 90 ◊ 5 e j 26.56

= 1.58 ej18.44 = 1.58[0.95 + j 0.316] Substituting the constants A, B, and C, i(t) = –3e–3t + 1.58(0.95 – j0.316) e–(1 + j1)t + 1.58(0.95 + j0.316) e(–1 + j1)t or i(t) = –3e–3t + 1.58 e–j378.44 e–(1 + j1)t + 1.58 e j18.44 e(–1 + j1)t Analytically, I ( s) =

5s (s + 3)(s + 1 + j1)(s + 1 - j1)

I ( s) =

A B C + + s + 3 s + 1 + j1 s + 1 - j1

A = I (s) (s + 3) s =-3 = =

5(-3) (-3 + 1 + j1) (-3 + 1 - j1)

-15 -15 = = -3 (-2 + j1)(-2 - j1) 4 + 1

B = I ( s ) (s + 1 + j1) s = -1- j1

=

=

5(-1 - j1) -5(1 + j1) = (-1 - j1 + 3)(-1 - j1 + 1 - j1) (2 - j1)(- j 2)

-5 2 e j 45 5 ◊e

- j 26.56

2e

j 270

= 1.58 e-378.44

= 1.58(0.95 – j0.316) C = B* = 1.58(0.95 + j0.316) Result is verified.

15-8.1

Routh array for this polynomial is given below. s4 s3 s2 s1 s0

1 1

2 2

12

0 ÆŒ 12 2 Œ-12 Œ 12

Solution Manual (15) – Page No. 10

As, the first element in the S2 row is zero, assume small positive values. And

2 Œ-12 Æ Negative value. Œ

The system is unstable.

15-8.2

(a) The Routh array for the polynomial is s3

6

2

s s1

2 2 -1

s0

2

5

Number of sign changes in the first column = 2 (i)

Positive real parts = 2

(ii)

Zero real parts = 0

(iii)

Negative real parts = 1

(b) The Routh array for the polynomial is s6

1

13

20 8

s5

5

21

16

s4 s3

8.8 11.45

16.8 11.45

8

s2

8

8

1

s

0ƌ

0

8

s

No sign changing in the first column. S1 row is zero. Auxillary polynomial 8S2 + 8 = 0

(i)

Positive real parts = 0

(ii)

Zero real parts = 2

(iii)

Negative real pars = 4 s=± j 3

(c) The Routh array for the polynomial is s6 s5 s4 s3 s2 s1 s0

1 -1 +2 +12 10 0ƌ 30

-2 -5 30 4 21 16 30 36 30

Solution Manual (15) – Page No. 11

Sign changes in the first column = 2 S1 row has all zeros.

Auxiliary polynomial, 10s2 + 30 = 0 s=± j 3

(i) Positive real parts = 2 (ii) Zero real parts = 2 (iii) Negative real parts = 2

Solution Manual (15) – Page No. 12

CHAPTER 16 (Solution Manual) 16-2.1

From the circuit, Y11 = YA + YC;

Y22 = YB + YC;

Y12 = –YC = Y21

Let I1 and I2 be the input and output currents. ΔY =

Y11

Y12

Y21 Y22

= [YA YB + YA YC + YB YC]

Z-parameters

16-2.2

Z11 =

Y22 ; ∆Y

Z12 =

Y12 Y = Z21; Z22 = 11 ∆Y ∆Y

Z11 =

YB + YC ; ∆Y

Z12 =

YC Y +Y = Z21; Z22 = A C ∆Y ∆Y

Z11 = Za + Zb = 2 – j3 Z21 = Zb + Z12 = – j3 Z22 = Zb + Zc = j2 Equivalent circuit

16-2.3

From the given equivalent circuit, Z11 = 500 Ω;

Z12 = 3000 90º

Z21 = 600 Ω;

Z22 = 4 kΩ;

ZL = 500 Ω

We know that V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 Open-circuit input impedance Z11 = 500 Ω Open-circuit output impedance Z22 = 4 kΩ If the load impedance is considered, the input impedance is given as 16-2.4

V1 = 25 V

⎡ Z Z ⎤ V1 = I1 ⎢ Z11 – 12 21 ⎥ Z L + Z 22 ⎦ ⎣ I1 = 5.952 I2 = –I1

Z 21 = –1.1904 A Z L + Z 22

VL = –I2 × 15 = 17.856

Solution Manual (16) – Page No. 1

∆Z + Z11 Z L Z 22 + Z L

Input impedance =

V1 25 = = 4.2 Ω I1 5.952

16-2.5

V1 = Z11 I1 + Z12 I2

We have

V2 = Z21 I1 + Z22 I2 Open-circuiting 22' terminals, I2 = 0, V2 =

6 V1 16

(1)

Applying nodal analysis at V1, I1 –

⇒

V –V 27 V2 = 1 2 10 90

I1 – 0.1125 V1 =

V1 16

[from (1)] ⇒

V1 = 5.71 I1

⇒

Z11 = 5.71

⇒

16 V 6 2 = 5.71 I1

⇒

V2 = 2.14 I1

∴

Z12 = 2.14

Open-circuiting the 11' terminals Applying nodal analysis at V2, I2 = At V1,

V2 – V1 V2 + 10 6

V1 – V2 –27 V2 = 10 90

From (2) ⇒ I2 = ⇒

(2) ⇒ V1 = –2V2

(3)

3V2 V2 [from (3)] + 10 6

V2 = 2.14 I2

∴ Z22 = 2.14

–1 V1 ⇒ 2 = 2.14 I2

⇒

V1 = – 4.28 I2

Solution Manual (16) – Page No. 2

⇒ Z12 = – 4.28 16-2.6

The Y-parameters of the network are Y11 = 1 +

1 3 – 2 2

Y21 = Y12 =

1 3 –1 ; Y22 = 1 + = 2 2 2

3 –1 2 2 =2 ΔY = –1 3 2 2 A' = 16-2.7

–Y11 = 3; Y12

B' =

1 = 2; Y12

C' =

∆Y = 4; Y12

D' =

Y11 = Y1 + Y2 = 10–3 (0.5 – j0.2)ϑ Y21 = –Y2 = Y21 = j0.2 × 10–3 ϑ Y22 = Y3 + Y2 = 10–3 (j0.22 – j0.2) = j0.02 × 10–3 ϑ

16-2.8

We have I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2 Y12 =

I1 I Y = 2 =0 V2 V = 0 21 V1 V = 0 1

2

Short-circuiting 11', V1' =

V2 100

Y12 =

I1 V2

Solution Manual (16) – Page No. 3

Z 22 =3 Z2

⇒

–V1' –V2 –1 = = 5 = V2 5 × 100 × V2 500

∴

Y12 =

–1 500

Short-circuiting 22',

⎡V ⎤ I2 = – ⎢ 1 – 0.1I1 ⎥ ⎣ 5S ⎦ V ⎤ ⎡V = – ⎢ 1 – 0.1 1 ⎥ 5S ⎦ ⎣ 5S =

–V1 [0.9] 5S

∴

Y21 =

– 0 .9 5S

16-2.9

We have, I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2

⇒

Y11 =

I1 I Y = 2 V1 V = 0 21 V1 V = 0 2

2

V1 = 5I1 ⇒

Y11 =

1 = 0.2 ϑ 5

2 ⎤ ⎡ I2 = – ⎢ I1 – I1 ⎥ 5 ⎦ ⎣

=

⇒

Y21 =

–3 –3 V1 I = × 5 1 5 5 –3 5

Short-circuiting the 11' terminals, V2 5 + V2 = –2 I 1 5 20 5

V2 +

Solution Manual (16) – Page No. 4

⇒

I1 = –0.725 V2

∴

Y12 = –0.725 ϑ

⇒ ⇒

V2 5 + V2 = I – 2 I 2 5 20 5 1

V2 +

V2 = 0 ∴ Y22 = 0 I2

16-2.11

Y– transformation

I1 = Y11 V1 + Y12 V2 I2 = Y21 V2 + Y22 V2 When V2 = 0

Y11 =

I1 , V1

Y21 =

I2 V1 I2 = – I1/2 V1 = 8 I1

Y11 =

I1 = − 0.5 V1

Y21 =

I 2 − I1/2 −1 = = (0.125) = 0.0625 V1 V1 2

Solution Manual (16) – Page No. 5

When V1 = 0, Y12

I1 I , Y22 = 2 V2 V2

2 32 V2 = 16 ´ I 2 = I2 3 3 I 3 Y22 = 2 = V2 32 Y12

I1 I 2æ3 ö = – 2 / 3 2 = – çç ÷÷ V2 V2 3 è 32 ø

=–

1 16

é 1 –1 ù ê 8 16 úú Y-parameter = ê ê–11 ú êë 16 3 32 úû 16-2.12

Y – Parameter

I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2

When V2 = 0, Y11 =

V1 = 2.I1/3

I1

V1 ,

I

Y21 = 2

V1

I2 = –I1/3

I Y11 = 1 = 3 2 V1 I2 I 1 1 =– 1 =– 3 =–1 2 3 V1 3 2 V1 I I , Y22 = 2 When V1 = 0 Y12 = 1 V2 V2 Y21 =

V2 =

( )

2 I2 3

Solution Manual (16) – Page No. 6

I2 3 = 2 V2 – I2 I1 I ö 3 =–1æ ç 2÷=–1 3 =– 1 Y12 = = ç 2 3 èV2 ÷ø 3 2 V2 V2

Y22 =

( )

Y-parameter

–1 ù 2 úú 3 úú 2 û

é ê 32 ê ê–1 êë 2

Z-parameter V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 - T-transformation

When I2 = 0

V1

Z11 =

I1

, Z21 =

V2

I1 = 3

V1

4

I1 , Z11 = 3 V2 = 1

V

When I1 = 0, Z12 = 1

V2 = 3 V1 = 1

4 4

I2

I2 , Z22 = 3

, Z22 =

V2

4

4

I1 , Z21 = 1

I2

4

I2 , Z12 = 1

é3 ê 4 Z-parameter ê ê1 ë 4

4 1 ù 4ú ú Ans. 3 ú 4û

16-2.13

Y-parameter I1 = Y11 V1 + Y12 V2

Solution Manual (16) – Page No. 7

4

I2 = Y21 V1 + Y22 V2

When V2 = 0, Y11 = I1 / V1 , Y21 = I 2 / V1 KVL in the loop (1) V1 = 3 (I2 – 2 I1)

(1)

KVL in the loop(2) 3 (I2 – 3 I1) + 3 (I2 – 2 I1) = 0

I 2 = 5 I1 2

(2)

By Eq. (1) and (2),

Y11 = I1/V1 = 2 / 3 Y21 = I 2/V1 = 5 / 3

When V1 = 0, Y12 = I1/V2 , Y22 = I 2/V2

KVL in Loop (1) V2 = 3 (I2 – 2 I1)

(3)

Loop(2) 3 I1 + 3 (I2 – 2 I1) = 0 I2 = I1

(4)

By using Eqs. (3) and (4),

Y12 = I1/V 2 = -1/ 3, Y22 = I 2/V2 = -1/ 3

é2 ê 3 y -parameters are ê ê5 êë 3 16-3.1

–1 ù 3 úú –1 ú 3 úû

The Z-parameters of the circuit are given by Z11 =

1 + 2; 2S

Z22 =

1 +2 2S

Z12 = Z21 = 2 Transmission parameters are A=

Z11 1 + 4S = ; Z 21 4S

B=

∆Z 1 + 8 S = Z 21 8S 2

C=

1 1 = ; Z 21 2

D=

Z 22 1 + 4 S = Z 21 4S

Solution Manual (16) – Page No. 8

16-3.2

Let A1, B1, C1, D1, be transmission parameters of the first section and A2, B2, C2, and D2 be the transmission parameters of the second section.

Z-parameters of the first section Z11 = 6 Ω; From which, A1 =

Z22 = 7 Ω;

6 ; 5

B1 =

Z12 = Z21 = 5 Ω

17 ; 5

C1 =

1 7 ; D1 = 5 5

Z-parameters of the second section, Z11 = 12 Ω; A2 =

6 ; 5

Z22 = 14 Ω; B2 =

34 ; 5

Z12 = Z2 = 10 Ω C2 =

1 ; 10

D2 =

7 5

Overall transmission parameters of the cascaded network ⎡ A C ⎤ ⎡ A1 ⎢ B D ⎥ = ⎢C ⎣ ⎦ ⎣ 1

16-3.3

B1 ⎤ ⎡ A2 D1 ⎥⎦ ⎢⎣C2

B2 ⎤ ⎡1.78 12.92 ⎤ = D2 ⎥⎦ ⎢⎣ 0.38 3.32 ⎥⎦

The given network can be converted into

Z11 = 4.5 Ω = Z22 Z12 = Z21 = 3.5

ΔZ =

4.5 3.5 3.5 4.5

=8

A=

Z11 1 1 = = 1.28 C = Z 21 3.5 Z21

B=

Z ∆Z = 2.28 D = 22 = 1.28 Z 21 Z21

KVL in the loop (3) 4I1 + 2I1 – 2 (I1 – I’) = 0 8I1 + 2I1 = 0

I1 =

16-3.3

Z-Parameter V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 When I2 = 0, Z11 = V1/I1 = Z21 = V2/I1

1 I1 4

KVL in the loop (1) 2(I – I1) + 3I1 – V1 = 0 2 (I1 – 1/4 I1) + 3I1 + V1 9 I1 = V1 2 V Z11 = 1 = 9/2 Ω I1

Solution Manual (16) – Page No. 9

KVL in the loop (2) 2I1 + 3I1 – V2 = 0

æ1 ö 2 çç I1 ÷÷ + 3I1 = V2 è4 ø 7 I1 = V2 2 V Z 21 = 2 = 7 W 2 I1 V Z12 = 1 Where I1 = 0, I2

, Z 22 =

V2 I2 KVL in the loop (3) 4 Ia + 2 Ia – 2 (I2 – Ia) = 0 8 Ia = 2 I2

Ia =

1 I2 4

Loop (1) 2 Ia + 3 I2 – V1 = 0 KVL in the loop (2) 2 (I2 – Ia) + 3 I2 – V2 = 0)

æ 1 ö 2 çç I 2 – I 2 ÷÷ = + 3I 2 = V2 4 ø è 9 I 2 = V2 2 V 9 Z 22 = 2 = W I2 2

æ1 ö 2 çç I 2 ÷÷ + 3 I 2 = V1 è4 ø 7 I 2 = V1 2 V Z12 = 1 = 7 W 2 I2

é9 7 ù ê 2 2ú Z-Parameters ê ú 9 ú ê7 2û ë 2 ABCD parameters in terms of Z-parameters are

Z11 Z Z - Z12 Z 21 , B = 11 22 Z 21 Z 21 Z 1 , C= D = 22 Z 21 Z 21 A

Solution Manual (16) – Page No. 10

9 16 2 9 , B= , C= , D= 7 7 7 7 Z12 = Z21 = 4 Ω; Z22 = 6 Ω Z11 = 4 Ω; So A =

16-4.1

16-4.2

h11 =

∆Ζ ; Z 22

h12 =

Z12 ; Z 22

h21 =

Z12 ; Z 22

h22 =

1 Z 22

h11 =

4 Ω; 3

h12 =

2 ; 3

h21 =

–2 ; 3

h22 =

1 6

g11 =

–Z12 Z 1 1 ∆Z = ϑ; g12 = = –1; g21 = 21 = 1; g22 = =2 Z11 4 Z11 Z11 Z11

We know that V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 From the given equivalent circuit, h11 = 1.5K;

h12 = 0.003;

h21 = 40;

(a) Short-circuit input impedance = h11 =

h22 = 30K;

ZL = 1 kΩ

V1 = 1.5 kΩ I1 V = 0 2

If the load impedance is considered then

(b) Open-circuit output admittance h22 =

V1 ∆ h Z L + h11 = I1 1 + h22 Z L

I2 = 30 kΩ V2 I = 0 2

Output impedance =

1 30 × 10 –3

= 0.33 × 10–3 kΩ

16-4.3

We have, V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 h11 =

V1 I h21 = 2 I1 V = 0 I1 V = 0 2

2

3

Given ω = 10 rad/s.

Z = 1 kΩ || 4PF || 2PF = 1 kΩ ||

(– j 2500 )(– j 5000 ) – j 2500 – j 5000

= 1 kΩ || –1666.67j

Solution Manual (16) – Page No. 11

= ∴

(103 )(–1666.67) = 857.49 –3 j = 0.857 –3 j kΩ 1000 – 1666.67 j h11 = 0.857 –31º kΩ

I4PF || 2PF = I1

103 103 – 1666.67 j

= 0.514 59º I1

(– j 2500 ) – j 2500 – j 5000

I2PF = 0.514I1 59º × = 0.17I1 59º I2 = – 0.17 59º I1 + = – 0.17 59º I1 +

V1 100 1 [857.49 –3i ]I1 100

I2 = 7.289 – 4.556j I1

∴

= 8.57 –32.1i ∴

h21 = 8.57 –32.1i

h12 =

V1 I = 0, h22 = 2 =0 V2 I = 0 V2 I = 0 1

1

1 kΩ || –j2500 =

(1000 )(– j 2500 ) 1000 – j 2500

= 928.477 –21.8º Voltage across 4PF || 1 kΩ =

⇒

V1 = V2 0.171 59.04º

∴

V1 = 0.17 59º V2

∴

h12 = 0.17 59º

V2 928.477 –21.8º – j 5000 + 862.1 – 344.806 j

V2 –V = 1 + I2 – j 5000 + 862.1 – 344.806 j 100

⇒

V2

⎡ ⎤ 1 1 ⎢ 862.1 – 5344.806 j + 100 0.17 90º ⎥ = I2 ⎣ ⎦ ⇒

I2 = 1.88 61.1º × 10–3 V2

Solution Manual (16) – Page No. 12

∴

16-5.1

h22 = 1.88 61.1º mϑ

(a) At the port 22', V2 = –I2 2500 We know that V1 = h11 I1 + h12 V2

(1)

I2 = h21 I1 + h22 V2

(2)

I2 = h21 I1 – h22 I2 2500 I2(1 + h22 2500) = h21 I1 I2 =

250 I 1.375 1

From Eq. (1), V1 = 1500I1 – 2 × 10–3 (2500I2) At the port 11', V1 = VS – I1(1000) VS – I1(1000) = 1500I1 – I1 =

5 × 250 I 1.375 B

100 × 10 –3 = 62.857 × 10–6 A 1591

V1 = 100 × 10–3 – 62.875 × 10–6 × 1000 = 37.143 mV V2 =

V1 – h11 I1 = –28 V h12

(b) Z-parameters, Z11 =

∆h –0.275 = –1833.3 Ω = h22 150 × 10 –6

Z12 =

h12 = 13.33 Ω h22

Z21 =

–h21 = 1.666 × 106 Ω h22

Z22 =

1 = 6.666 × 103 Ω h22

16-7.1

By source transformation Z-parameter for the given network is

Solution Manual (16) – Page No. 13

é R1 + R2 Z =ê êëR2 – 0.2 R3

R2 ù ú R2 + R3 úû

Y-parameter in terms of Z-parameter are

é Z 22 – Z12 ù , ê ú ê DZ DZ ú Y =ê ú ê – Z12 , Z11 ú êë DZ DZ úû é R2 + R3 – R2 ù ê ú ê DZ DZ ú Y =ê ú ê0.2 R3 – R2 R1 + R2 ú ê ú DZ DZ û ë Equivatent

network will be

2 R3 – 0.8 R3 DZ Y2 = – Y12 = – R2 / DZ Y1 = Y11 + Y21 =

Y3 = Y22 + Y12 = R1 / DZ

16-10.1 The Z-parameters of the given lattice network are Z11 =

(Z a + Z b )(Z d + Z c ) Za + Zb + Zc + Zd

where Za, Zd are called series arms, and Zb, Zc are called the diagonal arms of the lattice.

Since the network is symmetric, Za = Zd; Zb = Zc Z11 =

Za + Zb 9 = 2 2

Z12 = Z21 =

Z b – Z a –1 = 2 2

Z22 = Z11 A = –9, B = – 40; C = –2; D = –9 Z1 = Z12 =

B = 4.1 C

Za = Zd = 10 Ω Zc = Zb = 20 Ω Z11 = Z22 =

Za + Zb = 15 2

Z12 = Z21 =

Zb + Za =5 2

Solution Manual (16) – Page No. 14

A = 3;

B = 40; C =

Z11 = Z12 =

1 ; D=3 5

8 = 2.828

Solution Manual (16) – Page No. 15

CHAPTER 17 (Solution Manual) 17-3.2

From the given figure, C = 0.16 µF L = 10 mH; 2

The cut-off frequency, fC = fC =

C = 0.32 µF

1 π LC

1

π 0.32*10*10 –9

= 5,626.98 Hz

fC = 5.63 kHz The nominal impedance, k = k=

L C

10 mH = 176.78 Ω 0.32 HF

17-4.1

⎛ f ⎞ The phase shift in the pass band, β = 2 sin n–1 ⎜ ⎟ ⎝ fC ⎠ ⎛ 10 k ⎞ β = 2 sin n–1 ⎜ ⎟ = 2.678º ⎝ 5.63 k ⎠ fc = 1 kHz; K = 800 Ω = RL 1 C= = 0.09 µF 4π ×1000 × 800 800 = 0.06 H L= 4 π × 1000 The T and π-sections of the high-pass filters are shown in the figure.

17-5.1

fC = 1.5 kHz

L=

K

π fC

=

600

π ×1.5 ×103

= 0.127 H

1 1 = = 0.35 µF π fC K 3.14 ×1.5 ×103 × 600 The filter is shown in the figure below.

RL = 600 Ω = K C =

Solution Manual 17 # page no. 1

17-5.4

fC = 10 × 103 Hz; K = 600 Ω; m = 0.3 K 600 = = 4.774 mH; L= 4π fC 4π ×10 4 1 1 = = 0.0013 µF 4π fC K 4π ×10 4 × 600 The π-section elements are 2L = 0.031 H Shunt element m C 4m Series element L = 6.19 mH; = 3.33 × 10–9 F m 1– m2 The T-section elements are C = 6.66 × 10–9 F Series element m L 4m Shunt element = 0.015 mH; C = 1.3 × 10–9 F 2 m 1– m

C=

17-5.5

17-6.1

K=

L = C

fC =

1 4π LC

4 × 10 −3 0.5 ×10 –6

=

= 89.44 Ω

1

4π 4 ×10 × 0.5 ×10 –6 Given k = 600 Ω f1 = 3000 Hz For the prototype bandpass filter, L L Z1Z2 = k2 = 2 = 1 C1 C2 L1 =

–3

= 1.779 kHz f2 = 6000 Hz (1)

k 600 = = 63.66 mH π( f2 – f1 ) π(6000 – 3000)

f2 – f1 3000 = = 0.0221 µF 4π kf1 f2 ) 4π × 600 ×18 ×106 From Eq. (1), L2 = k2C1 = (600)2 × 0.0221 × 10–6 = 7.956 mH L 63.66 ×10 –3 = 0.177 µF C2 = 21 = k (600)2

C1 =

Solution Manual 17 # page no. 2

The resonant frequency of the shunt arm (or) series arm is f0 = f1 * f2 = 4, 242.64 Hz

17-7.1

R1 = 275 Ω R2 = 450 Ω ⎛ N 2 –1 ⎞ R1 = Ro ⎜ ; ⎜ 2 N ⎟⎟ ⎝ ⎠ R1 ( N –1) = R2 2N

⎛ N –1 ⎞ R2 = Ro ⎜ ⎟ ⎝ N +1 ⎠

( N –1)2 = 0.611 2N N = 2.872 (or) 0.3482 N2 – 3.22 N + 1 = 0, N = 0.3482 is not possible, because gain will be achieved. ∴ N = 2.872 2N R1 = 217.87 Ω Ro = ( N 2 –1) ∴ characteristic impedance, Ro = 217.87 Ω dB = 20 log10 N = 9.164 dB ∴ attenuation per section = 9.164 dB

17-8.1

Given

D = 10 log N D = 20, f = 400 Hz N = 102 = 100

Also,

N=1+

ω 2 L21 R02

⇒ L1 =

R0 N –1 ω

600 99 = 2.375 H 2 π × 400 L L As 1 = R02 , C1 = 12 C1 R0

L1 =

Solution Manual 17 # page no. 3

⇒

C1 =

2.375

(600)2 Now, at 1000 MHz, ω 2 L21 N=1+ R02

N=1+

= 6.597 µF

(2 × 1000 × 106 )2 (2.375)2 (600)2 14

17-8.2

= 6.1856 × 10 Attenuation, D = 10 log N = 148 dB Design of full-shunt equalizer; R0 = 600 Ω; attenuation = 20 dB; f1 = 600 Hz, f2 = 1200 Hz. (a) f = 600 Hz D = 10 log N 20 = 10 log N N = 100 R2 X2 N = 1 + 12 = 1 + 02 R0 X2 X1 = R0

N –1

R0 N –1 600 100 –1 = 1.5835 H = 2 πf 2. π .600 R0 1 600 = 60.3022 X2 = = = ωC1 N –1 100 –1

L1 =

1 = 4.3988 µF 2. π .600(60.3022) f = 1200 Hz

C1 = (b)

600 100 –1 = 0.7917 H 2. π .1200 1 C1 = = 2.1994 µF 2. π .1200(60.3022)

L1 =

17-8.3

We have 1/2

⎡ ( R + R1 )2 + W 2 L21 ⎤ N= ⎢ 0 2 2 2 ⎥ ⎢⎣ ( R0 – R1 ) ω L1 ⎥⎦ But attenuation at 50 Hz = 20 dB ∴ 20 = 10 log10 N ⇒ N = 102 = 100 2

So,

(100)2 =

⎛ R0 + R1 ⎞ 2 ⎜⎝ L ⎟⎠ + ω 1

2

⎛ R0 – R1 ⎞ 2 ⎜⎝ L ⎟⎠ + ω 1

2

104 =

⎛ R0 + R1 ⎞ 2 ⎜⎝ L ⎟⎠ + (2π × 50) 1

2

⎛ R0 – R1 ⎞ 2 ⎜⎝ L ⎟⎠ + (2π × 50) 1

Solution Manual 17 # page no. 4

By substituting

a=

104 =

R0 + R1 R – R1 and b = 0 L1 L1

a 2 + (100π )2

b 2 + (100π )2 a – 10 b = (100π)2 [9999] ∴ a2 – 104 b2 = 986.862 × 10–6 (1) 2 2 a + (6000π ) 104 = 2 b + (6000π )2 ⇒ a2 – 3.9808 b2 = 1.05909 × 109 (2) From (1) and (2), b = ± 85 a = ± 32.544 × 103 R0 + R1 ∴ = a = 32.544 × 103 (3) L R0 – R1 = b = 85 (4) L Adding the above two equations, we have 2 R0 = 32629 and R0 = 500 Ω L ∴ L = 0.03065 = 30.65 mH 2 R1 = 32459 and substituting L = 30.65 mH (3) – (4) ⇒ L ⇒ R1 = 497.434 Ω 2

4

2

L1

= 2

30.65 × 10 –3 500 × 500

Also,

C1 =

⇒

C1 = 12.26 nF

R0

R02 = R1 R2 500 × 500 = 502.579 Ω 497.434 The loss at 500 Hz is given by ⇒

R2 =

1/2

⎡ ( R + R1 )2 + ω 2 L21 ⎤ M= ⎢ 0 2 2 2⎥ ⎣⎢ ( R0 – R1 ) + ω L1 ⎦⎥

R0 = 500 Ω, R1 = 497.434 Ω, R2 = 502.579 Ω L1 = 30.65 mH

ω = 2π × 500; M = (33.33)

1/2

= 5.77

Loss in dB = 10 log10 M = 10 log10 5.77 α = 7.612 dB

Solution Manual 17 # page no. 5

CHAPTER 18 (solution Manual) 18-1.1

3

2

(a) P(s) = s + 2s + 4s + 2 0(s) = s3 + 4s e(s) = 2s2 + 2 since the quotient terms are positive, P(s) is Hurwitz. (b) P(s) = s4 + s3 + 4s2 + 2s + 3 0(s) = s3 + 2s e(s) = s4 + 4s2 + 3

s 2 s2 + 2)s3 + 4 s( s3 + 3 2 2 3s)2s 2 + 2( s 3 2 2s 3 2)3s( s 3s 2 0

s3 + 2 s)s2 + 4 s 2 + 3(s s2 + 2 s2 s 2 s2 + 3)s3 + 2 S ( 2 3s s3 + 2 S 2 )2 s + 3(4s 2 2 2s s s 3) ( 2 6 s 2 0 since the quotient terms are positive, P(s) is Hurwitz. (c) P(s) = s4 + 2s3 + 2s2 + 6s + 10 e(s) = s4 + 2s2 + 10 0(s) = 2s3 + 6s s 2 S 3 + 6 s)s 4 + 2 s2 + 10( 2 s 4 + 3s2 – s2 + 10)2s3 + 6 s(–2 2 s3 – 20 s s – 14 s) – s 2 + 10( 14 – s2 –14 s 10) – 14 s( 10 –14 s 0 since the two quotient terms are negative, P(s) is not Hurwitz.

18-2.2

Z (s) =

2s 4 + 3s 3 + 5s 2 + 5s +1 ( s 2 +1)(2s 2 + 2s +1)

(a) For Z(s) to be positive real, Q(s) should be a Hurwitz polynomial.

Q(s) =

m( s ) n( s )

Q(s) = (s2 + 1) (2s2 + 2s + 1)

Solution Manual 18 • Page No. 1

= 2 s4 + 2s3 + 3s2 + 2s + 1

2 s 3 + 2 s )2 s 4 + 3s 2 +1( s 2s 4 + 2s 2 s 2 + 1)2 s 3 + 2 s (2 s 2s3 + 2s ´ Continued fraction expansion terminates prematurely. This means that the last dursor w(s) = s2 + 1 is one of the factors of Q(s) Q(s) = (s2 + 1) (2s2 + 2s + 1) = 0 Its poles are s =

j, lies on the righ j axis and left

j axis

1 1 – ± j 2 2 (2) (M1 M2 – N1 N2) |s=jw H 0 for all values of

A ( 2) = (2s4 + 3s2 + 1) (2s4 + 5s2 + 1) – (2s3 + 2s) (3s3 + 5s)

= 4s8 + 10s6 + 3s4 – 2s2 + 1|s=j

= 4 8 – 10 6 + 3 4 + 2 2 + 1 > 0 for all values of . since A( 2) with real positive roots & A( 2) > 0 for all values of .

Thus, Z(s) is a minimum positive real function.

18-4.1

Z (s) = (

s +1)(s + 3) £ ( s + 2)

First form

Solution Manual 18 • Page No. 2

By direct division of the numerator by the denominator, we have

Z (s) =

s 2 + 4s + 3 2s + 3 =1+ 2 s + 2s s ( s + 2)

3 1 Z ( s ) =1+ 2 + 2 s s +2 =1+

1 1 4 + 2 S 1 S +1 2 3

Foster second Form

Y ( s) =

s( s + 2) ( s +1) ( s + 3)

Y (s) =

1 2 s 1 2s + s +1 s + 3

1 s s +1 Y1 = 2 , Z 1 = 1 s s +1 2

= 2+

1 1 s 2

1 s s +3 Y2 = 2 , Z 2 = 1 s s +3 2 =2+

18-4.2

Z(s) =

1 1 S 6

109 s(s2 + 16 × 1012 )

(s + 1012 )(s2 + 36 × 1012 ) The partial fraction expansion of Z(s) is 2P s 2 P2 s Z(s) = 2 1 12 + 2 s + 10 s + 36 + 1012 2

Solution Manual 18 • Page No. 3

10 9 s( s 2 + 16 × 1012 )

P1 =

12

(s + 10 )( s – j 6 × 10 ) S = − j 6 × 106 2

12

6

1 = 2.33 × 10–9 F; 2P1

C1 = L1 =

2

109 s( s 2 + 16 × 1012 )

P2 = ∴

=

(s – j10 )(s + 36 × 10 ) S = − j106 6

2P1

ω12

= 0.43 mH;

15 × 109 70

=

C2 =

2P1

L2 =

ω 22

10 × 109 35

1 = 1.75 × 10–9 F 2P2

= 0.016 mH

Hence, the first Foster form is given by

To get the second Foster form, P P1 P* + + HS Y(s) = 0 + s s + j 4 × 106 s – j 4 × 106 H = 10–9 P0 =

( s + 1012 )(s + 36 × 1012 ) 10 ( s + 16 × 10 ) 9

= 2.25 × 103 S=0

(s + 10 )( s + 36 × 10 ) 2

P1 =

12

12

12

10 9 s( s – j 4 × 106 )

1 L0 = = 0.44 mH; P0

= 9.375 × 103 S = – J 4 × 10

6

C0 = H = 109 F

2 P1 1 = 0.053 mH; C1 = = 1.17 × 10–9 F 12 2P1 16 × 10 Hence, the second Foster form is given by

L1 =

18-4.3 Z ( s ) =

( s 2 + 1) ( s 2 + 9) s ( s 2 + 4)

Foster I Form

Solution Manual 18 • Page No. 4

Z (s) = s +

6s 2 + 9 A BS + C =s+ + 2 2 s ( s + 4) s s +4

9

15 s 4 =s+ + 4 s s2 + 4

Foster II Form

Y (s) =

s ( s 2 + 4) ( s 2 +1) ( s 2 + 9)

Y (s) A B = 2 + 2 s s +1 s + 9 3 5 Y (s) = 28 + 2 8 s s +1 s + 9

3 s 5 s Y ( s ) = 28 + 28 s +1 s + 9

Y(s) = Y1 + Y2

18-4.4

Zl =

s 2 +1 8 1 = s+ 3 s 3 3 s 8 8

Z2 =

s2 + 9 8 1 = s+ 5 s 5 5 s 8 72

Taking continued fraction expansion of Z1(s) =

2 s 3 + 8s s2 + 1

Solution Manual 18 • Page No. 5

s2 + 1)2s3 + 8s(2s 2 s3 + 2 s s 6s)s2 + 1( 2 6 s 1)6s(6s 6s 0

Z(s) = 2s +

1 s 1 + 6 6s

Hence, the first Cauver form is shown here.

1 Z(s) = 2 (s + 1) / (2s3 + 8s) 1 8s + 2 s3 ) 1 + s 2 ( 8 1 + s2 / 4 3 2 32 s )8s + 2s3 ( 4 3s 8s 3 3 2 3 2s ) s ( s 4 8 3 2 s 4 0 Hence, the second Cauer network is shown below

Z(s) =

s3 + 4 s

2 s 4 + 20s 2 + 18 Taking continued fraction expansion, s3 + 4 s)2 s 4 + 20 s2 + 18(2s 2 s 4 + 8s 2 s 12s2 + 18)s3 + 4 s( 12 6 s3 + s 4 10 48 S )12s2 + 18( s 4 10 12 s2 10 5 18 ) s( s 4 36 10 s 4 0

Solution Manual 18 • Page No. 6

∴

Z(s) =

1 1

2s +

s 1 + 1 12 24 s+ 5 5 s 36 Hence, the first caver network is shown below.

To get the second caver network, 18 4 s + s3 )18 + 20s2 + 2s 4 ( 4s 9 2 18 + s 2 31 2 8 s + 2 s 4 )4 s + s3 ( 2 31s 16 4 s + s3 31 15 3 31 2 961 s ) s + 2s4 ( 31 2 30 S 31 2 s 2 15 15 2 s 4 ) s3 ( 31 62s 15 3 s 31 0 1 ∴ Z(s) = 4 1 + 8 1 2s + 961 1 31s + 30s 15 62s Hence, the second caver network is shown below.

18-4.5 Z ( s) =

K ( s + Z1 ) ( s + Z 2 ) ( s + P1 )( s + P2 )

As per the pole-zero diagram,

Solution Manual 18 • Page No. 7

poles are –1 and –3. Zeros are 0 and –2 Z (s) =

K ( s + 1) ( s + 3) s ( s + 2)

For the value of K, Z (–4) = 3/8 is given Z (−4) =

K (−4 + 1) (−4 + 3) = 3/8 (−4) (−4 + 2)

K=1 Z ( s) =

( s + 1)( s + 3) s ( s + 2)

[Impedance function]

Caver – I Form Z (s) =

s 2 + 4s + 3 [By continued fraction] s 2 + 2s s 2 + 2s ) s 2 + 4s + 3 (1 ← Z1 s 2 + 2s 2s + 3) s 2 + 2s (1 / 2 s ← Y2 s 2 + 3 / 2s 1 / 2 s ) 2s + 3 (4 ← Z 3 2s 3)1 / 2s (1 / 6s ← Y4 1 / 2s X

Caver – II Form

Z (s) =

3 + 4s + s 2 2s + s 2

Solution Manual 18 • Page No. 8

2s + s 2 )3 + 4s + s 2 ( 3

2s

¬ Z1

3+ 3 s 2 – – 5 s + s 2 )2s + s 2 ( 4 ¬ Y 2 2 5 2 2s + 4 s 5 – – 1 s 2 ) 5 s + s 2 ( 25 ¬ Z 3 5 2 2S 5 s 2 – s 2 ) 1 s 2 ( 1 ¬ Y4 5 5 1 s5 5 ´

18-4.6

2( s + 1)( s + 3) ( s + 2)( s + 6) 2(4 s + 9) Z(s) = 2 – (s + 2)( s + 6) Therefore, we have to expand Z ( s) 2(s + 1)( s + 3) A B C = = + + s s( s + 2)( s + 6) s s + 2 s + 6 1 1 5 where A = ; B= ; C= 2 4 4 1 5 s s 1 Z(s) = + 4 + 4 2 s+2 s–6 R since Pn = Rm and σn = M Ln

Z(s) =

1 1 Ω, L1 = H 4 8 5 5 R2 = Ω, and L2 = H 4 24 Hence, the first Foster form is shown below

R1 =

Solution Manual 18 • Page No. 9

To find out the second caver network, 1 12 + 8s + s2 )6 + 8s + 2 s2 ( 2 1 2 6 + 4s + s 2 3 2 3 4 s + s )12 + 8s + s2 ( 2 s 9 12 + s 2 7 3 8 s + s 2 )4 s + s2 ( 2 2 7 8 4 s + s2 7 5 2 7 49 s ) s + s2 ( 14 2 5s 7 s 2 5 5 s2 ) s2 ( 14 14 5 2 s 14 0 ∴

18-4.7

Z(s) =

Z(s) =

1 1 + 3 1 2 + 1 S 8 + 1 7 49 + 5S 5 / 14

s2 + 4s + 3

s2 + 8s + 12 To get the second caver network,

Solution Manual 18 • Page No. 10

12 + 8s + s 2 )3 + 4 s + s2 (

1 4

s2 4 3s 2 6 2s + )12 + 8s + s2 ( 4 3 9 12 + s 2 7 3s 2 4 ( s + s 2 )2 s + 2 4 7 4 2s + s2 7 5 2 7s 98 s ) + S2 ( 28 2 5s 7s 2 5 5 s2 ) s2 ( 28 28 5 2 s 28 0 Hence, the second caver network is shown below.

3 + 2s +

18-4.9

(s + 1)(s + 3)(s + 5) s(s + 2)(s + 4)(s + 6) Taking partial fractions, A B C D + + Z(s) = + s s +2 s+4 s+6 5 3 3 A= ; B= ; C= ; 16 16 16 5 / 16 3 / 16 3 / 16 5 / 16 ∴ + + + Z(s) = s s+2 s+4 s+6 1 1 Pn = ; σm = Cm Rm Cm

Z(s) =

C1 =

16 F; 3

16 F; 3 16 C3 = – F; 5

C2 =

2=

1

16 R1 × 3 3 R2 = Ω 64 5 R3 = Ω 96

∴ R1 =

D=

3 Ω 32

Solution Manual 18 • Page No. 11

5 16

First Foster form, s(s + 2)(s + 4)(s + 6) Y(s) = (s + 1)(s + 3)(s + 5) Y (s) (s + 2)(s + 4)(s + 6) A B C = = + + s (s + 1)(s + 3)(s + 5) s + 1 s + 3 s + 5 15 3 3 A= , B= , C= 8 4 8 15 3 3 s s s 8 + 4 + 8 ∴ Y(s) = s +1 s + 3 s + 5 1 1 Rm = , Cm = Pm σ m Rm 8 Ω, 15 4 R2 = Ω, 3 8 R3 = Ω, 3

R1 =

15 F 8 1 C2 = F 4 3 C2 = F 40

C1 =

To get the first caver network,

Solution Manual 18 • Page No. 12

s3 + 9s2 + 23s + 15)s 4 + 12 s3 + 44s 2 + 48s (3 s 4 + 9s3 + 23s2 + 15s 3s3 + 21s 2 + 33s )s3 + 9s2 + 23s + 15( s3 + 7s2 + 11S

1 3

3 2 s2 + 12s + 15)3s3 + 21s2 – 33s ( S 2 45 3s3 + 18s 2 + s 2 2 2 21 3s + s )2s 2 + 12 s – 5( 2 3 2 s 2 + 7s 21 3 s( s 2 5 2 3s + 9 s 3 10 S )5s + 15( 2 3 5s 3 3 15) s ( 2 10 3 s 2 0

5s + 15)3s2 +

∴

Z(s) = 1

1

1 1 1 + 3 1 3 s+ 2 1 2 + 1 3 3 s+ 10 1 5 + s 3 10 Hence, the first caver network is given below. s

+

To get the second caver network, we take the partial fraction expansion as shown below.

Solution Manual 18 • Page No. 13

0.3125 s 15 + 13.75s + 3.75s2 0.3125s3

48s + 44 s2 + 12s3 + s 4 )15 + 23s + 9s 2 + s3 (

9.25s + 5.25s2 + 0.6875s3 )48s + 44 s2 + 12 s3 + s 4 ( 5.189 48s + 27.24s 2 + 3.567s3 16.76s 2 + 8.433s3 + s 4 )9.25s + 5.5s2 + 0.6875s3 ( 9.25s + 4.66 s2 + 0.552 s3

0.532 s

0.59s 2 + 0.1355s3 )16.76s2 + 8.433s3 ( 28.41 16.76s 2 + 3.849s3 0.1237 4.584 s3 )0.59s 2 + 0.1355s3 ( s 0.59s2 0.1355s3 )4.584 s3 ( 33.83 4.584s3 0 Hence, the second caver network is shown below.

Solution Manual 18 • Page No. 14

CHAPTER 19 (Solution Manual) 19.1

19.2

19.3

OBTAINING THE VOLTAGE ACROSS 40 OHMS I1 0 1 5A R1 1 0 25 V1 2 1 60 R2 2 0 20 I2 3 2 4 R3 3 0 40 V2 1 3 100 • DC V1 LIST 60 • PRINT DC V(R3) • END OUTPUT: **** DC TRANSFER CURVES ************* V1 V(R3) 6.000E+01 –6.087E+01 RESULT: Voltage across 40 Ω = –60.87 V ----------------------------------------------------------OBTAINING THE OPERATING POINT TO PRINT CURRENT THROUGH 4V VOLTAGE SOURCE V1 1 0 4 R1 2 1 5 I1 0 2 2 R2 2 0 20 R3 2 3 10 V2 3 0 36 • OP • END OUTPUT; NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE (1) 4.0000 (2) 18.2860 (3) 36.0000 VOLTAGE SOURCE CURRENT NAME CURRENT V1 2.857E+00 V2 –1.771E+00 RESULT: Current through 4V SOURCE (V1) = 2.857 A OBTAINING THE CURRENT IN THE RESISTOR AND TOTAL POWER IN THE CIRCUIT V1 1 0 10 R1 1 2 1K I1 0 2 5 A R2 2 0 200 R3 2 0 100 • DC V1 LIST 10 • PRINT DC I(R2) • END OUTPUT: V1 1.000E+01

I(R2) 1.566E+00

Solution Manual (19) • Page No. 1

19.4

DC ANALYSIS WITH DEPENDENT SOURCES V1 1 0 25 R1 1 2 2K R2 2 4 1K VX 4 0 0 H1 3 2 VX 0.1 R3 3 0 1K • DC V1 LIST 25 • PRINT DCI(R2) • END OUTPUT: V1 2.500E+01

19.5

V(2) 5.000E+00

AC CIRCUIT ANALYSIS VS 1 0 AC 100 15 R1 1 2 100 L1 2 0 2M • AC LIN 1 50 50 • PRINT ACIM(R1) IP(R1) IR(R1) II(R1) • END OUTPUT: FREQ 5.000E+0.1

19.6

I(R2) 5.000E–03

IM(R1) 1.000E+00

IP(R1) 1.464E+01

IR(R1) 9.675E-01

II(R1) 2.527E-01

RESONANT CONDITION V1 1 1 0 AC 120 0 R 1 2 20 OHMS L 2 3 5MH C 3 0 80UF • AC LIN 100 1 1000 • PROBE • END OUTPUT: From the graph obtained, MAX(I(L)) = 6 A at a frequency of 251.6 Hz.

19.7

COUPLED CIRCUIT ANALYSIS VS 1 0 AC 230 0 R1 1 2 25 L1 2 0 15MH L2 3 0 15MH R2 3 4 25 C 4 5 10UF RL 5 0 40 K12 L1 L2 – 0.6667 • AC LIN 1 50 50 • PRINT Ac IM(R1) IP(R1) IM(R2) IP(R2) • END OUTPUT: FREQ 5.000E+01

IM(R1) 9.037E+00

IP(R1) –1.074E+01

IM(R2) 8.865E-02

Solution Manual (19) • Page No. 2

IP(R2) –2.245E+01

19.8

RL TRANSIENT I1 0 1 PWL (0 0 1N 2M 1 2M) R1 1 0 40 R2 1 2 40 L 2 0 5MH OUTPUT: TIME 0.000E+00 1.000E-04 2.000E-04 3.000E-04 4.000E-04 5.000E-04 6.000E-04 7.000E-04 8.000E-04 9.000E-04 1.000E-03

19.9

V(1) 0.000E+00 4.804E-02 4.160E-02 4.032E-02 4.006E-02 4.001E-02 4.000E-02 4.000E-02 4.000E-02 4.000E-02 4.000E-02

V(2) 0.000E+00 1.607E-02 3.200E-03 6.372E-04 1.269E-04 2.526E-05 5.030E-06 1.002E-06 1.994E-07 3.971E-08 7.862E-09

SOURCE-FREE TRANSIENT F1 0 1 VX 0.8 R1 1 0 60 L1 1 2 5M IC = 10 MA VX 0 2 0 • TRAN 50U 0.5M UIC • PRINT TRAN I(L1) • PROBE • END OUTPUT: **** TRANSIENT ANALYSIS ***** TIME I(L1) 0.000E+00 1.000E-02 5.000E-05 3.401E-03 1.000E-04 1.510E-03 1.500E-04 3.889E-04 2.000E-04 1.315E-04 2.500E-04 4.448E-05 3.000E-04 1.504E-05 3.500E-04 5.086E-06 4.000E-04 1.720E-06 4.500E-04 5.816E-07 5.000E-04 1.958 E-07

19.10

RL TRANSIENT UP TO 5 TIME CONSTANTS R1 1 0 25 L1 1 2 2H IC = 40 MA VX 0 2 0 F1 0 1 VX 0.4 • TRAN 50M 0.4 UIC • PRINT TRAN I(L1) • PROBE • END

Solution Manual (19) • Page No. 3

OUTPUT: TIME 0.000E+00 5.000E-02 1.000E-01 1.500E-01 2.000E-01 2.500E-01 3.000E-01 3.500E-01 4.000E-01

I(L1) 4.000E-02 1.666E-02 6.944E-03 2.893E-03 1.204E-03 5.002E-04 2.085E-04 8.686E-05 3.607E-05

Solution Manual (19) • Page No. 4