GATE 2016 EE (SET 1) Q18

18.In the given circuit, the current by the battery, in ampere, is _______

Solution:

Applying KCL at node A,

-I_{1}+I_{2}+I_{2}=0

2I_{2}=I_{1}\rightarrow 0

And applying KVL in loop ABCD,

1-I_{1}-I_{2}-I_{2}=0

I_{1}+2I_{2}=1\rightarrow 2

From 1 and 2

2I_{2}+2I_{2}=1

4I_{2}=1

I_{2}=\frac{1}{4}A

AndI_{1}=2\chi \frac{1}{4}=\frac{1}{2}A

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