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

Correct answer: C. (10)/(9) As, we know that q=q_0 e^ -t / According to the question, V_C=(q)/(C)=(q_0)/(C) e^ -t / aligned & & (q_0)/(10) & =(q_0)/(C) e^ -t / & & e^ t / & =(10)/(9) & & t & = (10)/(9) aligned

JIPMER 2016

Exam: JIPMER
Year: 2016
Subject: Physics
Chapter: Capacitance

JIPMER 2016 · Physics · Capacitance

A capacitor is charged and then made to discharge through a...

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

\tau

. In what time will the potential difference across the capacitor decrease by

10 \%

Options

A $\tau \ln 0.1$
B $\tau \ln 0.9$
C $\tau \ln \frac{10}{9}$
D $\tau \ln \frac{11}{10}$

Correct answer

C. $\tau \ln \frac{10}{9}$

Step-by-step solution

As, we know that

q=q_0 e^{-t / \tau}

According to the question,

V_C=\frac{q}{C}=\frac{q_0}{C} e^{-t / \tau}

\begin{aligned} \Rightarrow & & \frac{q_0}{10} & =\frac{q_0}{C} e^{-t / \tau} \\ \Rightarrow & & e^{t / \tau} & =\frac{10}{9} \\ \Rightarrow & & t & =\tau \ln \frac{10}{9}\end{aligned}

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