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.
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