Question

A capacitor of capacity $C$ is charged to a potential difference $V$ and another capacitor of capacity $2C$ is charged to a potential difference $4V.$ The charging batteries are disconnected and the two capacitors are connected with reverse polarity (i.e., positive plate of first capacitor is connected to negative plate of second capacitor). The heat produced during the redistribution of charge between the capacitors will be :

• A. $\frac{25C{V}^2}{3}$
• B. $2C{V}^2$
• C. $\frac{50C{V}^2}{3}$
• D. $\frac{125C{V}^2}{3}$

Answer 1

Answer: (A)

As the two capacitors are connected with reverse polarity, So total charge

$Q_t=Q_2-Q_1=\left(2C\right)\left(4V\right)-CV=7CV$

After connecting them they are in parallel so total capacity is $C_t=C+2C=3C$

We know that after connecting the common potential is

$V_c=\frac{totalcharge}{totalcapacity}=\frac{Q_t}{C_t}=\frac{7CV}{3C}=\left(7/3\right)V$

Heat produced $H=$ energy loss$=U_i-U_f$

$H=U_i-U_f=[\frac{1}{2}C{V}^2+\frac{1}{2}\left(2C\right)(4V{)}^2]-\frac{1}{2}{C}_t{V}_c^2$

$=\frac{33}{2}C{V}^2-\frac{1}{2}\left(3C\right)(7/3V{)}^2=C{V}^2[\left(33/2\right)-\left(49/6\right)]=(25/3)C{V}^2$