• # question_answer In the given circuit diagram, potential difference between A and C is $\frac{V}{4},$ then current in branch BC is- A) $\frac{V}{2R}$ B) $\frac{V}{4r}$ C) zero       D) $\frac{3V}{4r}$

 ${{I}_{AC}}=\frac{V}{4R}=I$ ${{I}_{CD}}=\frac{3V}{4R}=3{{I}_{AC}}=3I$ Hence using KCL at junction C, ${{I}_{BC}}=3I-I=2I=\frac{V}{2R}.$