A capacitor is connected to a 20 V battery through a resistance of 10$\Omega$. It is found that the potential difference across the capacitor rises to 2 V in 1 $\mu$s. The capacitance of the capacitor is __________ $\mu$F. Given : $\ln \left( {{{10} \over 9}} \right) = 0.105$
Solution
Given, the peak voltage of the battery, V<sub>0</sub> = 20V<br/><br/>The voltage of the battery, V = 2V<br/><br/>Time, t = 1 $\mu$s = 1 $\times$ 10<sup>$-$6</sup>s<br/><br/>Resistance of the capacitor, R = 10 $\Omega$<br/><br/>As we know that, <br/><br/>V = V<sub>0</sub>(1 $-$ e<sup>$-$t/RC</sup>)<br/><br/>Substituting the values in the above equation, we get<br/><br/>2 = 20(1 $-$ e<sup>$-$t/RC</sup>)<br/><br/>$\Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)$<br/><br/>$$ \Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95$$ $\mu$F<br/><br/>$\therefore$ The capacitance of the capacitor is 0.95 $\mu$F.
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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