In the circuit shown in figure, find the current through the branch BD

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5 A

0 A

3 A

4 A

answer is A.

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Detailed Solution

The current in the circuit are assumed as shown in the fig.Applying KVL along the loop ABDA, we get – 6i1 – 3 i2 + 15 = 0 or 2i1 + i2 = 5 …..(i)Applying KVL along the loop BCDB, we get– 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 …..(ii)Solving equation (i) and (ii) for i2, we get i2 = 5 A

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