• # 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) $\frac{V({{r}_{2}}+{{r}_{1}})}{{{r}_{1}}+{{r}_{2}}+{{r}_{3}}}$

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 $C=\frac{V{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}$

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