Capacitors

JEE Main level, single correct, medium

167 of 180

Question 167 of 180


A capacitor of capacitance C=1μFC=1\mu F is suddenly connected to a battery of 100V100V through a resistance R=100Ω.R=100\Omega . The time taken for the capacitor to be charged to get 50V50V is:
(Take ln⁡2=0.69\ln 2=0.69)

Show answer and solution

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

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 .

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Capacitors, dielectrics and RC circuits simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Capacitors (Charging and Discarging of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.