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CHAP. 3]

29

CIRCUIT LAWS

For a two-branch current divider we have i1 R2 ¼ i R1 þ R2 This may be expressed as follows: The ratio of the current in one branch of a two-branch parallel circuit to the total current is equal to the ratio of the resistance of the other branch resistance to the sum of the two resistances. EXAMPLE 3.8. A current of 30.0 mA is to be divided into two branch currents of 20.0 mA and 10.0 mA by a network with an equivalent resistance equal to or greater than 10.0 . Obtain the branch resistances. 20 mA R2 ¼ 30 mA R1 þ R2

10 mA R1 ¼ 30 mA R1 þ R2

R1 R2  10:0

R1 þ R2

Solving these equations yields R1  15:0 and R2  30:0 .

Solved Problems 3.1

Find V3 and its polarity if the current I in the circuit of Fig. 3-7 is 0.40 A.

Fig. 3-7 Assume that V3 has the same polarity as V1 . Applying KVL and starting from the lower left corner, V1  Ið5:0Þ  V2  Ið20:0Þ þ V3 ¼ 0 50:0  2:0  10:0  8:0 þ V3 ¼ 0 V3 ¼ 30:0 V Terminal b is positive with respect to terminal a.

3.2

Obtain the currents I1 and I2 for the network shown in Fig. 3-8. a and b comprise one node.

Applying KCL,

2:0 þ 7:0 þ I1 ¼ 3:0 Also, c and d comprise a single node.

I1 ¼ 6:0 A

Thus,

4:0 þ 6:0 ¼ I2 þ 1:0

3.3

or

or

Find the current I for the circuit shown in Fig. 3-9.

I2 ¼ 9:0 A

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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