question_answer

In the circuit of following figure, the final voltage drop across the capacitor C is -

A) \[\frac{V{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}}\]                

B) \[\frac{V{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\]

C) \[\frac{V({{r}_{1}}+{{r}_{2}})}{{{r}_{2}}}\]              

D) \[\frac{V({{r}_{2}}+{{r}_{1}})}{{{r}_{1}}+{{r}_{2}}+{{r}_{3}}}\]

Correct Answer: B

Solution :

[B]

In steady state, the capacitor arm presents an infinite resistance.

\[\therefore \]Current through that arm = 0

So the potential difference across C is that across \[{{r}_{2}}\]

Current through \[{{r}_{2}}=\frac{V}{{{r}_{1}}+{{r}_{2}}}\]

P.D. across \[{{r}_{2}}=\frac{V{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\]

P.D. across \[C=\frac{V{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\]