Question

For the circuit shown in the figure, the current I will be

(1) 0.5 A 

(2) 0.75 A 

(3) 1 A 

(4) 1.5 A

Answer 1

Ans: (3) 1 A

Sol.

By KVL in a closed loop ABCDA,

V_A – I × 4 – I × 1 + 4 – I × 1 + 2 = V_A

–6I + 6 = 0 

I = 1 A

Answer 2

I=(2+4)÷(4+1+1) 

I=6V÷6omega

I=1A

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