JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Current Electricity

A capacitor is connected to a 20 V battery through a resistance of

10Ω

. It is found that the potential difference across the capacitor rises to 2 V in

1 μ s

. The capacitance of the capacitor is ____

μ F

.
Given

ln (\frac{10}{9}) = 0.105

Answer: (B)

Solution

While charging, the capacitor voltage follows

V_{C} = V_{0} (1 - e^{- t / R C})

. Setting

V_{C} = 2 V

and

V_{0} = 20 V

gives

e^{- t / R C} = 0.9

, so

\frac{t}{R C} = ln \frac{10}{9} = 0.105

. With

t = 1 μ s

and

R = 10 Ω

C = \frac{t}{0.105 R} = \frac{1 0 ^{- 6}}{0.105 \times 10} = 0.95 μ F

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option