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  • A capacitor is charged and then made to discharge through a resistance. The time constant is .

A capacitor is charged and then made to discharge through a resistance. The time constant is \mathrm{\tau}. In what time will the potential difference across the capacitor decreases by 10% ?
 

Option: 1

\mathrm{\tau \operatorname{In}(0.1) }


Option: 2

\mathrm{\tau \operatorname{In}(0.9)}


Option: 3

\mathrm{\tau \operatorname{In\left (\frac{10}{9} \right )}


Option: 4

\mathrm{\tau In \left (\frac{11}{10} \right )}


Answers (1)

best_answer

\mathrm{Q=Q_0 e^{-t / \tau} }and potential difference across C is proportional to Q. For the p.d. to fall by 10 %, Q must also fall by 10 %.

\begin{aligned} & \mathrm{Q=0.9 Q_0=Q_0 e^{\frac{t}{\tau}}} \\ & \text { or } \mathrm{e^{\frac{t}{\tau}}=\frac{10}{9} \text { or } \frac{t}{\tau}=\operatorname{In}\left(\frac{10}{9}\right)} \end{aligned}

Posted by

seema garhwal

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