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AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Physics
Chapter: Capacitance

AIIMS 2018 · 26 May · Physics · Capacitance

A fully charged parallel plate capacitor of 1 μ F capacity is...

A fully charged parallel plate capacitor of

1 \mu \mathrm{F}

capacity is discharging through a resistor. If the energy of the capacitor reduces to half in one second then the value of resistance will be

Options

A $\frac{2 \times 10^6}{\ln 2} \Omega$
B $\frac{2 \times 10^3}{\ln 2} \Omega$
C $\frac{10^6}{\ln 2} \Omega$
D $\frac{10^3}{\ln 2} \Omega$

Correct answer

A. $\frac{2 \times 10^6}{\ln 2} \Omega$

Step-by-step solution

As

Q=Q_0 e^{-t / \tau}

. When energy is

50 \%

then

Q=\frac{Q_0}{\sqrt{2}}

\begin{aligned} & \frac{Q_0}{\sqrt{2}}=Q_0 e^{-t / \tau} \\ & e^{t / \tau}=\sqrt{2} \\ & \frac{t}{\tau}=\ln (\sqrt{2}) \quad \tau=\frac{t}{\ln (\sqrt{2})} \\ & R C=\frac{2}{\ln 2} \\ & R=\frac{2}{C \ln 2}=\frac{2}{10^{-6} \times \ln 2}=\frac{2 \times 10^6}{\ln 2} \Omega\end{aligned}

$$

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