Question

A capacitor of capacitance 500 μF is connected to a battery through a 10 kΩ resistor. The charge stored in the capacitor in the first 5 s is larger than the charge stored in the next.
(a) 5 s
(b) 50 s
(c) 500 s
(d) 500 s

Answer 1

(a) 5 s
(b) 50 s
(c) 500 s
(d) 500 s

The charge (Q) on the capacitor at any instant t,
$Q=CV(1-e^{-t/RC})$,
where
C = capacitance of the given capacitance
R = resistance of the resistor connected in series with the capacitor
RC = (10 × 10³) × (500 × 10^$-$6) = 5 s
The charge on the capacitor in the first 5 seconds,
$Q_0=CV(1-e^{-5/5})=CV\times 0.632$
The charge on the capacitor in the first 10 seconds,
$$\begin{gathered} Q_1=CV(1-e^{-10/5}) \\ {Q}_1=CV(1-e^{-2})=0.864\times CV \end{gathered}$$
Charge developed in the next 5 seconds,
Q' = Q₁ $-$ Q₀
Q' = CV(0.864 $-$ 0.632) = 0.232 CV

The charge on the capacitor in the first 55 seconds,
$$\begin{gathered} Q_2=CV(1-e^{-55/5}) \\ {Q}_2=CV(1-e^{-11})=0.99\times CV \end{gathered}$$
Charge developed in the next 50 seconds,
Q' = Q₂ $-$ Q₀
Q' = CV(0.99 $-$ 0.632) = 0.358 CV

Charge developed in the first 505 seconds,
$Q_3=CV(1-e^{-500/5})=CV(1-e^{-100})\approx CV$
Charge developed in the next 500 seconds,
Q' = CV (1$-$ 0.632) = 0.368 CV

Thus, the charge developed on the capacitor in the first 5 seconds is greater than the charge developed in the next 5,50, 500 seconds.
Disclaimer : Out of the four given options, two options are same.