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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 Solved Paper-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 :

 It steady state, current through AB = 0 \[\Rightarrow \]\[{{V}_{AB}}={{V}_{CD}}\] \[\Rightarrow \]\[{{V}_{AB}}=\left( \frac{\varepsilon }{r+{{r}_{2}}}\times {{r}_{2}} \right)-{{V}_{CD}}\] \[\Rightarrow \]\[{{Q}_{c}}=C{{V}_{AB}}\] \[=CE\left( \frac{{{r}_{2}}}{r+{{r}_{2}}} \right)\]