A capacitor of capacitance {C}=1 \\mu{F} is suddenly connected to a battery of 100 volt through a resistance {R}=100 \\Omega. The time taken for the capacitor to be charged to get 50 {V} is : [ Take \ln 2=0.69] [27 Jul 2021 Shift 1]

Correct Answer is : 0.69×10−4s

JEE Main 27 July 2021 Shift 1 Solved Paper

Section: Physics

Question : 6 of 90

Marks: +1, -0

A capacitor of capacitance C=1µF 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 50V is :
[ Take ln‌2=0.69]

[27 Jul 2021 Shift 1]

1.44×10−4s
3.33×10−4s
0.69×10−4s
0.30×10−4s

Validate

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