16.8 Charging and Discharging a Capacitor

16.8 Charging and Discharging a Capacitor

Many electric circuits consist of both capacitors and resistors. Fig. 16.11 shows a resistor-capacitor circuit called R-C-circuit. When the switch S is set at terminal A, the R-C combination is connected to a battery of voltage $V_{o}$ which starts charging the capacitor through the resistor $R$.

(Fig. 16.11)

(Fig. 16.12)

The capacitor is not charged immediately, rather charges build up gradually to the equilibrium value of $q_{o} = CV_{o}$. The growth of charge with time for different resistances is shown in Fig. 16.12. According to this graph $q = 0$ at $t = 0$ and increases gradually with time till it reaches its equilibrium value $q_{o} = CV_{o}$. The voltage $V$ across capacitor at any instant can be obtained by dividing $q$ by $C$, as $V = q/C$.

How fast or how slow the capacitor is charging or discharging, depends upon the product of the resistance $R$ and the capacitance $C$ used in the circuit. As the unit of product $RC$ is that of time, so this product is known as time constant and is defined as the time required by the capacitor to deposit $0.63$ times the equilibrium charge $q_{o}$. The graphs of Fig. 16.12 show that the charge reaches its equilibrium value sooner when the time constant is small.

(Fig. 16.13)

Fig. 16.13 (a) illustrates the discharging of a capacitor through a resistor. In this figure, the switch S is set at point B, so the charge $+q$ on the left plate can flow anti-clockwise through the resistance and neutralize the charge $-q$ on the right plate.

The graphs in Fig. 16.13 (b) shows that discharging begins at $t = 0$ when $q = CV_{o}$ and decreases gradually to zero. Smaller values of time constant $RC$ lead to a more rapid discharge.

Example 16.9

The time constant of a series RC circuit is $\tau = RC$. Verify that an ohm times farad is equivalent to second.

Solution: Ohm's law in terms of potential difference $V$, current and resistance $R$ can be written as,

$V = IR$

Putting $I = q/t$, this equation transforms into the equation,

$V = \dfrac{q}{t}R \quad \text{or} \quad R = \dfrac{V \times t}{q}$

According to equation $q = CV$, $C = q/V$

Multiplying this equation with above equation gives,

$RC = \dfrac{V \times t}{q} \times \dfrac{q}{V} = \tau$

Hence $1$ ohm $\times$ $1$ farad $= 1$ second

Where ohm is the unit of resistance $R$.