Question

In the given circuit diagram, find the heat generated on closing the switch $S$. (Initially the capacitor of capacitance $C$ is uncharged)

• A. $2C{V}^2$
• B. $\frac{3}{2}C{V}^2$
• C. $\frac{1}{2}C{V}^2$
• D. $C{V}^2$

Answer 1

The correct option is C $\frac{1}{2}C{V}^2$

Before closing switch $S,2C$ and $3C$ connected in series across the cell. $C$ is not in the circuit .

$\therefore \frac{1}{C_{eff}}=\frac{1}{2c}+\frac{1}{3c}$

$C_{eff}=\frac{6C}{5}$

$\therefore$ Initial energy $\left(E_i\right)=\frac{1}{2}\left(C_{eff}\right)V^2$

$=\frac{1}{2}\left(\frac{6C}{5}\right)V^2$

$=\frac{3C{V}^2}{5}$

After closing switch $S$ , capacitor $C$ and $\frac{6C}{5}$ will be in parallel

$\therefore C{\text{'}}_{eff}=\frac{6C}{5}+c$

$=\frac{11C}{5}$

∴ Final energy $\left(E_f\right)=\frac{1}{2}\left(C{\text{'}}_{eff}\right)V^2$

$=\frac{1}{2}\left(\frac{11C}{5}\right)\left(V^2\right)$

$=\frac{11C{V}^2}{10}$

after closing the switch, charge flow through cell $=$ charge on capacitance $C=q_1$
$q_1=CV$

work done by cell $W_{cell}=q_{1}V=C{V}^2$

From energy balance

$E_i+W_{cell}=E_f+$heat loss

Heat loss $=W_{cell}+E_i-E_f$

$=C{V}^2+\frac{3C{V}^2}{5}\frac{-11C{V}^2}{10}$

$=\frac{1}{2}C{V}^2$
Hence, option (c) is correct.