Question:
A capacitor of capacitance $\mathrm{C}=1 \mu \mathrm{F}$ 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 $50 \mathrm{~V}$ is :
$[$ Take $\ln 2=0.69]$
Correct Option: , 3
Solution:
$\mathrm{V}=\mathrm{V}_{0}\left(1-\mathrm{e}^{-\frac{\mathrm{t}}{R C}}\right)$
$50=100\left(1-e^{-\frac{t}{R C}}\right)$
$\mathrm{t}=0.69 \times 10^{-4} \mathrm{sec}$
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