Talha's Physics Academy
Charging and Discharging of a Capacitor Through a Resistor
Video Lecture
Watch the complete video lecture below to understand the exponential behavior, time constant, and process of charging and discharging a capacitor through a resistor.
RC Circuit Overview
Consider a circuit having a capacitance $C$ and a resistance $R$ connected in series with a DC voltage supply of potential difference $V$ through a switch, as shown in the figure. This type of circuit is known as an RC circuit.
(A) Charging of a Capacitor
When the switch is placed at position 1, the capacitor begins to store charge. At any time $t$ during the charging process, the charge $Q$ on the capacitor increases until it reaches its maximum final value $Q_0$, where:
$$Q_0 = C V$$
Experiments show that the charging process of a capacitor exhibits exponential behavior. Therefore, the equation for the charge at any time $t$ is given by:
$$Q = Q_0 \left(1 - e^{-t/RC}\right) \quad \text{--- (i)}$$
Time Constant ($\tau$): The product $\tau = RC$ is called the time constant or the capacitive time constant of the circuit, which has the dimensions of time ($[T]$).
The time constant determines how rapidly the capacitor charges or discharges:
- If $RC \ll 1$, the charge $Q$ will attain its final maximum value rapidly.
- If $RC \gg 1$, the charge will accumulate slowly.
If we set the time equal to the time constant ($t = \tau = RC$), equation (i) becomes:
$$Q = Q_0 \left(1 - e^{-1}\right) = Q_0 \left(1 - \frac{1}{e}\right)$$
$$Q = Q_0 \left(1 - \frac{1}{2.718}\right)$$
$$Q \approx 0.632\,Q_0$$
This indicates that in one time constant, a capacitor charges to approximately $63.2\%$ of its maximum final charge.
(B) Discharging of a Capacitor
When the switch is moved to position 2, the battery is disconnected from the circuit, and the charged capacitor begins to discharge itself through the resistor $R$. This discharging process follows an exponential decay equation:
$$Q = Q_0 e^{-t/RC} \quad \text{--- (ii)}$$
Where $Q_0$ represents the initial charge on the capacitor at the beginning of the discharge ($t = 0$).
If the time elapsed equals the time constant ($t = \tau = RC$), equation (ii) becomes:
$$Q = Q_0 e^{-RC/RC}$$
$$Q = Q_0 e^{-1} = \frac{Q_0}{2.718} \approx 0.368\,Q_0$$
This shows that during one time constant, the charge on a discharging capacitor drops to approximately $36.8\%$ of its initial value.
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