4.4 Discharging of a Capacitor through Resistor:

To discharge a capacitor through a resistor, let us consider a capacitor of capacitance $(C)$ is initially charged to a potential difference $V_{0}$ and charge $q_{0}$, then

$q_{0} = C V_{0}$ .......... (i)

Now, the charged capacitor is joined to a resistor of resistance $(R)$ in series as shown in figure.

Figure 1: Circuit diagram for discharging Capacitor

At a time $(t)$ after the discharging through $(C)$ has begun, let the $V_{C}$ is the potential across the capacitor and $q$ be the charge. If current $(I)$ flows through $R$, then we have

$V_{C} = V_{R}$ .......... (ii)

or, $\frac{q}{c}$ = $IR$

or, $\frac{q}{c}$ = $-\frac{dq}{dt}$$R$    [∵ $I$ = $-\frac{dq}{dt}$;  '-$ve$' sign shows that $q$ decreases with increasing $t$]

or, $\frac{dq}{q}$ = $-\frac{1}{RC}$

Integrating both sides, we get

or, $\int_{q}^{q_{0}} \frac{dq}{q} = \int_{0}^{t} - \frac{1}{RC}$

or, $[\ln q]$$_{q_{0}}^{q}$ = $- \frac{1}{RC}$$[t]$$_{0}^{t}$

or, $\ln q - \ln q_{0}$ = $-\frac{t}{RC}$

or, $\ln$ ($\frac{q}{q_{0}}$) = $- \frac{t}{RC}$

or, $\frac{q}{q_{0}}$ = $e^{-\frac{t}{RC}}$

∴ $q = q_{0}$$e^{-\frac{t}{RC}}$ .......... (iii)

This is called the decay of charge equation. Clearly, $q$ decreases exponentially with time (t) as shown in figure.

Figure 2: Variation of charge with time in a discharging Capacitor

If t = RC in above equation, then from the equation (iii), we get

or, $q = q_{0}$$e^{-\frac{RC}{RC}}$ = $q_{0}\;e^{-1}$ =  $\frac{q_{0}}{e}$ $= 0.37\; q_{0} = 37 \%$ of $q_{0}$

Thus the discharging time constant may be defined as the time at which the charge on the capacitor during discharging becomes about 37 % of the initial charge.

Return to Main Menu