For the given network, find
the current through 4 ohm and 3 ohm. Assume that the cells have negligible internal resistance.
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Solution:
Applying Kirchhoff first lawAt junction F,
I I I 1 3 = 2 – 1 2 . 3 --- (1)
Applying Kirchhoff second law,
(i) loop EFCDE,
3 4 10 0 2 1
4 3 10 1 2 I --- (2)
(ii) loop FABCF
4 3 5 0 3 2 I I
4 3 5 3 2 I I --- (3)
From Eq. (1) and Eq. (2)
4 3 3 2 2 = 10
3 4 4 10 2 3 2
4 7 10 3 2 I I --- (4)
From Eq. (3) and Eq. (4)
10 5 2 I
I A 2 . 0 5
Negative sign indicates that I 2 current
flows from F to C
From Eq. (2) 4 3 0 5 10 1I .
I A 1 = 2 1. 2
∴ I I 3 1 I A
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