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,

\[Q_1  = CV(1 -  e^{- 10/5} )\]

\[ Q_1  = CV(1 -  e^{- 2} ) = 0 . 864 \times CV\]

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,

\[Q_2  = CV(1 -  e^{- 55/5} )\]

\[ Q_2  = CV(1 -  e^{- 11} ) = 0 . 99 \times CV\]

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.