GATE 2019 EE – Question 51
A 0.1 $\mu$F capacitor charged to 100 V is discharged through a $1\ \text{k}\Omega$ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1 V is ________.
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Correct answer: 0.45 to 0.47
Explanation
The time constant is $\tau=RC=1\text{ k}\Omega\times0.1\ \mu\text{F}=0.1$ ms. From $100e^{-t/\tau}=1$, $t=\tau\ln100=0.1\times4.605=0.46$ ms.