A capacitor of capacitance $\mathrm{C}=1 \mu \mathrm{~F}$ is suddenly connected to a battery of 100 volt through a resistance $\mathrm{R}=100 \Omega$. The time taken for the capacitor to be charged to get 50 V is: [Take $\ln 2=0.69$]

(D) $0.69 \times 10^{-4} \mathrm{~s}$

(A) $3.33 \times 10^{-4} \mathrm{~s}$

(B) $1.44 \times 10^{-4} \mathrm{~s}$

(C) $0.30 \times 10^{-4} \mathrm{~s}$

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Current Electricity

A capacitor of capacitance

C = 1 μ F

is suddenly connected to a battery of 100 volt through a resistance

R = 100Ω

. The time taken for the capacitor to be charged to get 50 V is:
[Take

ln 2 = 0.69

]

Answer: (D)

While charging through a resistance the capacitor voltage builds up as

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

with

V_{0} = 100

V. Putting

V = 50

V gives

e^{- t / R C} = \frac{1}{2}

, so

t = R C ln 2

. With

R = 100 Ω

and

C = 1 μ F

the time constant is

R C = 100 \times 1 0^{- 6} = 1 0^{- 4}

s, and

t = 1 0^{- 4} \times 0.69 = 0.69 \times 1 0^{- 4} s

The capacitor reaches half the battery voltage after

0.69 \times 1 0^{- 4}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option