question_answer

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

A) 5 A                              

B) 0 A

C) 3 A                              

D) 4 A

Correct Answer: A

Solution :

The current in the circuit are assumed as shown in the figure.            Applying KVL along the loop ABDA, we get \[-6{{i}_{1}}-3{{i}_{2}}+15=0\,\,\,\,\,\,or\,\,\,\,\,2{{i}_{1}}+{{i}_{2}}=5\,\,\,\,\,\]         ? (i) Applying KVL along the loop BCDB, we get \[-3({{i}_{1}}-{{i}_{2}})-30+3{{i}_{2}}=0\,\,\,\,\,or\,\,\,-{{i}_{1}}+2{{i}_{2}}=10\]  ... (ii) Solving equations (i) and (ii) for \[{{i}_{2}}\], we get\[{{i}_{2}}\,\,=\,\,5A\].