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JEE Main 202127 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2021 · 27 July · Morning Shift · Physics · Capacitance

A capacitor of capacitance C = 1 is suddenly connected to a battery...

A capacitor of capacitance C = 1 is suddenly connected to a battery of 100V through a resistance R = 100Ω . The time

A capacitor of capacitance C=1\muFC = 1\muF is suddenly connected to a battery of 100V100V through a resistance R=100Ω.R = 100Ω . The time taken for the capacitor to be charged to get 50V50V is:

(Take ln⁡2=0.69\ln 2 = 0 . 69)

Options

  1. A1.44×10−4s1 . 44 \times {10}^{- 4}s
  2. B3.33×10−4s3 . 33 \times {10}^{- 4}s
  3. C0.69×10−4s0 . 69 \times {10}^{- 4}s
  4. D0.30×10−4s0 . 30 \times {10}^{- 4}s

Correct answer

C. 0.69×10−4s0 . 69 \times {10}^{- 4}s

Step-by-step solution

V=V0(1−e−tRC)V = {V}_{0}\left(1 - {e}^{- \dfrac{t}{RC}}\right)
50=100(1−e−tRC)50 = 100\left(1 - {e}^{- \dfrac{t}{RC}}\right)
t=0.69×10−4sec.t = 0 . 69 \times {10}^{- 4}sec .
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