A circuit consisting of a capacitor \( C \), a resistor of resistance \( R \), and an ideal battery of emf \( V \), as shown in the figure, is known as an RC series circuit.
As soon as the circuit is completed by closing key \( S_1 \) (keeping \( S_2 \) open), charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference \( V_C = \frac{q}{C} \) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (\( Q = VC \)). During this process of charging, the charge \( q \) on the capacitor changes with time as:
\[
q = Q\left( 1 - e^{-\frac{t}{RC}} \right)
\]
The charging current can be obtained by differentiating it and using:
\[
I = \frac{dq}{dt} = \frac{V}{R} e^{-\frac{t}{RC}}
\]
Consider the case when \( R = 20 \, \text{k}\Omega \), \( C = 500 \, \mu\text{F} \), and \( V = 10 \, \text{V} \).
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The time constant \( \tau = RC \) defines the rate at which a capacitor charges in an RC circuit. The charge increases exponentially with time, and so does the current, which decreases exponentially.
The time constant \( \tau = RC \) represents the time it takes for the capacitor to charge up to about 63% of its full charge. For the given values of \( R = 20 \, \text{k}\Omega \) and \( C = 500 \, \mu\text{F} \): \[ \tau = RC = (20 \times 10^3) \times (500 \times 10^{-6}) = 10 \, \text{seconds} \] Now, the expression for the charge on the capacitor at any time \( t \) is: \[ q(t) = Q \left( 1 - e^{-\frac{t}{RC}} \right) \] where \( Q = VC \) is the maximum charge on the capacitor: \[ Q = 10 \, \text{V} \times 500 \times 10^{-6} \, \text{F} = 5 \times 10^{-3} \, \text{C} \] Thus, the charge at any time \( t \) is: \[ q(t) = 5 \times 10^{-3} \left( 1 - e^{-\frac{t}{10}} \right) \, \text{C} \] To find the charging current \( I(t) \), we differentiate \( q(t) \): \[ I(t) = \frac{dq}{dt} = \frac{5 \times 10^{-3}}{10} e^{-\frac{t}{10}} = 5 \times 10^{-4} e^{-\frac{t}{10}} \, \text{A} \] Thus, the charging current at any time \( t \) is \( 5 \times 10^{-4} e^{-\frac{t}{10}} \, \text{A} \).
