Question

A capacitor C₁ of capacitance 1 μF and a capacitor C₂ of capacitance 2 μF are separately charged by a common battery for a long time. The two capacitors are then separately discharged through equal resistors. Both the discharge circuits are connected at t = 0.
(a) The current in each of the two discharging circuits is zero at t = 0.
(b) The currents in the two discharging circuits at t = 0 are equal but not zero.
(c) The currents in the two discharging circuits at t = 0 are unequal.
(d) C₁ loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge.

Answer 1

(b) The currents in the two discharging circuits at t = 0 are equal but not zero.
(d) C₁ loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge.

Let the voltage of the battery connected to the capacitors be V. Both the capacitors will charge up to the same potential (V).
The charge on the capacitors C₁ is C₁V = (1 μF)×V
The charge on the capacitors C₂ is C₂V = (2 μF)×V
The charge on the discharging circuit at an instant t,
$Q=CV{e}^{-t/RC}$
The current through the discharging circuit,
$\frac{\mathrm{d}Q}{\mathrm{d}t}=-\frac{CV}{RC}{\mathrm{e}}^{-t/RC}=\frac{\mathit{V}}{\mathit{R}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{R}\mathrm{C}}$
At t = 0, the current through the discharging circuit will be $\frac{V}{R}$ for both the capacitors.

Let the time taken by the capacitor C₁ to lose 50% of the charge be t₁.
$$\begin{gathered} Q_1=\frac{C_{1}V}{2} \\ \frac{C_{1}V}{2}=C_{1}V{e}^{-t_1/RC} \\ \frac{1}{2}=e^{-t_1/RC} \end{gathered}$$
Taking natural log on both sides, we get:
$$\begin{gathered} -\ln 2=-\frac{t_1}{R{C}_1} \\ {t}_1=R{C}_1\ln 2 \end{gathered}$$
Similarly,
Time taken for capacitor C₂: $t_2=R{C}_2\ln 2$
As, C₁ < C₂, t₁ < t₂
Thus, we can say that C₁ loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge.