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 $....\,\times \,10^{-4}\,s.$
[Take $\ln 2=0.69]$
JEE MAIN 2021, Medium
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$V=V_{0}\left(1-e^{-\frac{t}{R C}}\right)$
$50=100\left(1-e^{-\frac{t}{R C}}\right)$
$t=0.69 \times 10^{-4} \;sec$
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