Question

Find the amount heat generated in the circuit shown in the figure after the switch is shifted from position $1$ to position $2$.

Answer 1

When the switch is in position $1$, the combination has $C$ and $C_0$ in parallel and $C$ in series for which the equivalent capacitance is
$Ceq=\frac{C(C+C_0)}{2C+C_0}$
The total charge on the combination is
$Q=E{C}_{eq}=\frac{EC(C+C_0)}{2C+C_0}$
The total charge on the three capacitors can be obtained as
$q_3=E{C}_{eq}=\frac{EC(C+C_0)}{2C+C_0}$
$q_2=\frac{EC(C+C_0)C_0}{(2C+C_0)(C+C_0)}=\frac{ECC}{2C+C_0}$
$q_1=\frac{EC(C+C_0)C}{(2C+C_0)(C+C_0)}=\frac{E{C}^2}{2C+C_0}$
When the switch is in position $2$, the charge distribution on the three capacitors is obtained as
$q_3^{\prime }=\frac{E{C}^2}{2C+C_0},q_2^{\prime }=q_2$ and $q_1^{\prime }=\frac{EC(C+C_0)}{2C+C_0}$
Now, heat produced $=$ loss in stored electrical energy $+$ extra energy drawn from the battery. Since the equivalent capacitance $C_{eq}$ remains unchanged in both the position of the key, the loss in stored energy is zero.
Hence, Heat produced $=$ energy drawn from the battery
$=E.\Delta q=E(q_1^{\prime }-q_1^{\prime })=E(q_3-q_3^{\prime })$
$=E[\frac{EC(C+C_0)}{2C+C_0}-\frac{E{C}^2}{2C+C_0}]=\frac{E^{2}C{C}_0}{2C+C_0}$.