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7.2

FIRST-ORDER CIRCUITS

[CHAP. 7

In Problem 7.1, obtain the power and energy in the resistor, and compare the latter with the initial energy stored in the capacitor. pR ¼ vR i ¼ 25e125t ðWÞ ðt ðt wR ¼ pR dt ¼ 25e125t dt ¼ 0:20ð1  e125t Þ 0

ðJÞ

0

The initial stored energy is W0 ¼ 12 CV02 ¼ 12 ð40  106 Þð100Þ2 J ¼ 0:20 ¼ wR ð1Þ In other words, all the stored energy in the capacitor is eventually delivered to the resistor, where it is converted into heat.

7.3

An RC transient identical to that in Problems 7.1 and 7.2 has a power transient pR ¼ 360et=0:00001

ðWÞ

Obtain the initial charge Q0 , if R ¼ 10 . 2 ¼ 105 pR ¼ P0 e2t=RC or or C ¼ 2 mF RC ðt wR ¼ pR dt ¼ 3:6ð1  et=0:00001 Þ ðmJÞ 0

Then, wR ð1Þ ¼ 3:6 mJ ¼

7.4

Q20 =2C,

from which Q0 ¼ 120 mC.

The switch in the RL circuit shown in Fig. 7-21 is moved from position 1 to position 2 at t ¼ 0. Obtain vR and vL with polarities as indicated.

Fig. 7-21 The constant-current source drives a current through the inductance in the same direction as that of the transient current i. Then, for t > 0, i ¼ I0 eRt=L ¼ 2e25t 25t

vR ¼ Ri ¼ 200e

ðAÞ

ðVÞ 25t

vL ¼ vR ¼ 200e

7.5

ðVÞ

For the transient of Problem 7.4 obtain pR and pL . pR ¼ vR i ¼ 400e50t 50t

pL ¼ vL i ¼ 400e

ðWÞ ðWÞ

Negative power for the inductance is consistent with the fact that energy is leaving the element. this energy is being transferred to the resistance, pR is positive.

And, since

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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