Best Answer:
Assume that R and C are connected in series to a cell of emf E. Initially assume the circuit is open and no current is flowing and the capacitor is uncharged. Now when the circuit is closed, current will start to flow in the circuit. Now it is important to realize that a capacitor tries to "resist" the potential difference across it and so neither will the current immediately build up in the circuit, nor will get immediately charged. At any time instant 't', balancing the net potential drop across the capacitor and resistor to E we get:

iR + q/C = E, where i is the current flowing in the circuit (across the resistor) and q is the charge across the capacitor at time instant 't'. Now, i = dq/dt (since the charge that is changing across the capacitor will be flowing through the resistor since we assume that the connecting wires of the circuit are perfect conductors and there is no charge build up anywhere in the circuit):

We now solve the differential equation, by first dividing the equation by R and then multiplying the equation with the integration factor e^(t/RC)

on doing so, we see that the equation transforms into:

d(q(e^(t/RC) = dt*(E*e^(t/RC))

We integrate it between 0 to 't' to get the instantaneous charge in the capacitor:

q = CE(1 - e^(-t/RC))

we can easily see that after a very long time ('t' tending to infinity) q tends to CE (which is the standard equation for capacitance), the fully charged condition. The energy stored in the capacitor in this fully charged capacitor is: 0.5*CE^2

To calculate the energy dissipated in the resistor, we differentiate the expression for instantaneous charge to get the instantaneous current in the circuit:

i = dq/dt = (E/R)*e^(-t/RC)

The energy dissipated in the capacitor at time 't' in the time interval 'dt' is:

(i^2)*R*dt = ((E^2)/R)*e^(-2t/RC)*dt

integrating the eqation from 0 to infinity would give us the total energy dissipated through the resistor and this is seen to be 0.5*C*E^2

Thus energy dissipated through resistor is equal to energy stored in the capacitor.

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