A series circuit consists of two capacitors, a resistor, and an ideal voltage source. The circuit is initially at steady state. Now a conducting wire is shorted across the capacitor C as shown by dotted line. After shorting the wire, select the correct statement/statements

 

Before shorting the wire at steady state the charges on each capacitors will be equal to 

$q_0=q_1=q_2=\left(\frac{2C}{3}\right)V$

After shorting the capacitor marked (2) will be useless in circuit. Hence 

$q_1^{\text{'}}=2CV=q_{f}and{q}_2^{\text{'}}=0$ 

The charge supplied by battery after shorting 

$\Delta q=q_f-q_0=2CV-\frac{2CV}{3}=\frac{4CV}{3}$

Hence work done by battery $W=\Delta qV=\frac{4C{V}^2}{3}$

Initial potential energy of system

$V_0=\frac{1}{2}\left(\frac{2}{3}C\right)V^2=\frac{1}{3}C{V}^2$

Final potential energy of system $U_f=\frac{1}{2}\left(2C\right)V^2=C{V}^2$

Hence change in PE $\Delta U-U_f-U_i$

Or $\Delta U-C{V}^2-\frac{1}{3}C{V}^2=\frac{2}{3}C{V}^2$

Work done by battery $W=\Delta U+Heat$

Heat developed $H=W-\Delta U$

Or $H=\frac{4}{3}C{V}^2-\frac{2}{3}C{V}^2=\frac{2}{3}C{V}^2$