A charged capacitor $C_{1}$ is discharged through a resistance R by putting switch $S$ in position $1$ of circuit shown in figure. When discharge current reduces to $1_{o}$ , The switch is suddenly shifted to position $2$ . Calculate the amount of heat liberated in resistance $R$ starting from this instant. Also, calculate current $i$ through the circuit as a function of time.

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Solution

Initially switch is at position 1, capacitor will discharge.We know that during discharging charge at any tume:-
q= $q {= q}_{0} e^{\frac{- t}{R C}} \to 1$
We know that $i = \frac{d q}{d t} = \frac{q_{0}}{R C} e^{\frac{- t}{{R C}_{1}}}$
at any instant $i = i_{0} i_{0} - \frac{q_{0}}{{R C}_{1}} e^{\frac{- t}{{R C}_{1}}} \to (2)$
charge on that time when $i_{0} {R C}_{1}$
$i_{0} R c_{1} = q_{0} e^{\frac{- t}{R C}}$
After switching 0 to position 2 this charge will distribute among $C_{1}$ and $C_{2}$ .
Initially energy = $U_{i} = \frac{q^{2}}{2 C_{1}}$
Final energy = $U_{f} = \frac{q^{2}}{2 (C_{1} - C_{2})}$
Heat liberated = $U_{i} - U_{f} = \frac{q^{2}}{2} (\frac{1}{C_{1}} - \frac{1}{C_{1} {+ C}_{2}}) \frac{q^{2}}{2} \frac{C_{2}}{C_{1} (C_{1} + C_{2})}$
we know $q = i_{o} R C_{1}$
Heat liberated = $\frac{i_{0}^{2} R^{2} C_{1}^{2}}{2} \times \frac{C_{2}}{C_{1} (C_{1} + C_{2})}$
$= \frac{i_{0}^{2} R^{2} C_{1} C_{2} J o u l s}{2 (C_{1} + C_{2})}$

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