# 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

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

$$4I_{2}=1$$

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

And$$I_{1}=2\chi \frac{1}{4}=\frac{1}{2}A$$