Q.

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

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a

0.5 A

b

0.75 A

c

1 A

d

1.5 A

answer is C.

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Detailed Solution

By KVL in a closed loop ABCDA,VA – I × 4 – I × 1 + 4 – I × 1 + 2 = VA–6I + 6 = 0 I = 1 A
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