question_answer

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be: [JEE Main 2017]

A) \[CE\frac{{{r}_{2}}}{(r+{{r}_{2}})}\]

B) \[CE\frac{{{r}_{1}}}{({{r}_{1}}+r)}\]

C) CE

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

Correct Answer: A

Solution :

[a] It steady state, current through AB = 0

\[\Rightarrow \]\[{{V}_{AB}}={{V}_{CD}}\]

\[\Rightarrow \]\[{{V}_{AB}}=\left( \frac{\varepsilon }{r+{{r}_{2}}}\times {{r}_{2}} \right)-{{V}_{CD}}\]

\[\Rightarrow \]\[{{Q}_{c}}=C{{V}_{AB}}\]

\[=CE\left( \frac{{{r}_{2}}}{r+{{r}_{2}}} \right)\]