In the circuit shown in figure the capacitance of each capacitor is equal to $C$ and resistance $R$ . One of the capacitors was charge to a voltage $V_{0}$ and then at the moment $t = 0$ was shorted by means of the switch $S$ . Find,
(a) the current in the circuit as a function of time $t$
(b) the amount of generated heat

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Solution

Consider the connections as shown in figure 1 by which the capacitor C charges through resistance R.
The time required by capacitor to get charged to 63% of its capacity is called as time constant of capacitor and is given by $τ = R C$ .

The current flowing through the circuit is given by
$I = \frac{V \cdot e^{- t / τ}}{R} . . . (i)$ .

As the battery is disconnected and another capacitor is connected in place of battery (in right arm in circuit) as shown in the figure 2, the capacitor starts charging. The discharging current is given by $I = \frac{V_{C} \cdot e^{- t / τ}}{R} . . . (i i)$ .
The circuit will oscillate and continuous charging and discharging of both capacitors will take place

The heat developed in the circuit is given by $H = P t = I^{2} R t$
Where I is given by equation (ii)

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