# Capacitor Part – 9

Charging and discharging of a capacitor:

Charging: Let us consider a capacitor C is connected to a battery of emf E through a resistance R. During charging of the capacitor the potential difference across the plates becomes equal to the emf of the battery. Let at any time t, I is the current through the resistance R and q is the charge of capacitor. Using KVL, VC + VR = E Â

Or, Â + IR = E Â

Or, Â + R = E Â

Or, R = E – Â

Or, CRdq = (EC â€“ q)dt Â

Or, Â = Â

Or, = Â

Or, – loge =

Or, 1 – Â = e– t/RC Â

Or, q = EC(1 – e– t/RC)

Or, q = q0(1 – e– t/RC) at t = , q = q0 = EC.

= CR is called capacitive time constant, i.e. the time in which charge of the capacitor is 0.632 times of its maximum value during charging.

The charging current is I = = Â = I0e– t/RC where I0 = Â = .

At t = 0, I = I0 i.e. initially it acts as short circuit or as a simple conducting wire.

At t = , I = 0 it acts as an open circuit.

Discharging: If a capacitor C with charge q0 is discharged through a resistance R then at any time t, the PD of the resistance is V = IR.

Using KVL, VC + VR = 0

Or, R Â + Â = 0

Or, Â = – Â

Or, q = q0e– t/RC [This is the charge of the capacitor during discharging.]

The capacitive time constant = CR is the time in which charge becomes Â  Â i.e. 0.368 times the initial value of charge q0.

The discharging current is I = – = – Â = I0e– t/RC where I0 = Â = .

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