Question

A fully charged parallel plate capacitor of 1μF 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

• A. $\frac{2\times 10^6}{In2}\Omega$
• B. $\frac{2\times 10^3}{In2}\Omega$
• C. $\frac{10^6}{In2}\Omega$
• D. $\frac{10^3}{In2}\Omega$

Answer 1

Answer: (A)

As $Q=Q_0{e}^{-t/\tau }.$ When energy is 50%
then Q$=\frac{Q_0}{\sqrt{2}}$
$\frac{Q_0}{\sqrt{2}}=Q_0{e}^{-t/\tau }$
$e^{t/\tau }=\sqrt{2}$
$\frac{t}{\tau }=In\left(\sqrt{2}\right)\tau =\frac{t}{In\left(\sqrt{2}\right)}$
$RC=\frac{2}{In2}$
$R=\frac{2}{CIn2}=\frac{2}{10^{-6}\times In2}=\frac{2\times 10^6}{In2}\Omega$