Page 40 - Schaum's Outline of Theory and Problems of Electric Circuits

CHAP. 3]
For a two-branch current divider we have CIRCUIT LAWS 29

i 1 R 2
¼
i R 1 þ R 2
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 R 2 10 mA R 1 R 1 R 2
¼ ¼ 10:0
30 mA R 1 þ R 2 30 mA R 1 þ R 2 R 1 þ R 2
Solving these equations yields R 1 15:0
and R 2 30:0
.

Solved Problems

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

Fig. 3-7

Assume that V 3 has the same polarity as V 1 . Applying KVL and starting from the lower left corner,
V 1 Ið5:0Þ V 2 Ið20:0Þþ V 3 ¼ 0
50:0 2:0 10:0 8:0 þ V 3 ¼ 0
V 3 ¼ 30:0V
Terminal b is positive with respect to terminal a.

3.2 Obtain the currents I 1 and I 2 for the network shown in Fig. 3-8.
a and b comprise one node. Applying KCL,
2:0 þ 7:0 þ I 1 ¼ 3:0 or I 1 ¼ 6:0A
Also, c and d comprise a single node. Thus,
4:0 þ 6:0 ¼ I 2 þ 1:0 or I 2 ¼ 9:0A

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