Question

Given circuit is in steady state. Potential energy stored in the capacitors is U. Now switch S is closed. Heat produced after closing the switch S is H. Find $\frac{U}{H}$.

Answer 1

Initially capacitance of the circuit is $C_1=\frac{C}{2}$ since two capacitors are in series.

$U=\frac{C_1{V}^2}{2}=\frac{C{V}^2}{4}$

When switch is closed-

Equivalent capacitance is $C_2=\frac{2C\times C}{2C+C}=\frac{2C}{3}$

As two capacitors in parallel are connected to a capacitor in series.

$U^{\prime }=\frac{1}{2}\frac{2C{V}^2}{3}=\frac{C{V}^2}{3}$

Heat, $H=U^{\prime }-U=\frac{C{V}^2}{12}$

$⟹\frac{U}{H}=3$