Question

In the circuit shown in figure, the battery is an ideal one with $emf$ $\text{V}$. The capacitor is initially uncharged. Switch $S$ is closed at time $t=0$.

The final charge $Q$ on the capacitor is :

• A. $\frac{CV}{2}$
  
• B. $\frac{CV}{3}$
  
• C. $CV$
  
• D. $\frac{CV}{6}$
  

Answer 1

The correct option is A $\frac{CV}{2}$


When the capacitor is fully charged, current through the capacitor is equal to zero & it acts as a broken circuit, as shown below.


current in the $1^{st}$ loop, according to $KVL$
$V-iR/2-iR-iR/2=0$
$V-2iR=0$
$i=\left(V/2R\right)$

So, voltage drop across the capacitor is,
$\Delta V_{XY}=iR=\frac{V}{2R}\times R=\left(\frac{V}{2}\right)$
And, charge on the capacitor is, $q=C\Delta V_{XY}=C\cdot \frac{V}{2}=\frac{CV}{2}$

Hence, $\left(A\right)$ is the correct answer.