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Q.
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
As $Q=Q_{0}\,e^{-t/τ}.$ When energy is 50%
then Q$ =\frac{Q_{0}}{\sqrt{2}}$
$\frac{Q_{0}}{\sqrt{2}}=Q_{0}\,e^{-t/τ}$
$e^{t/τ}=\sqrt{2}$
$\frac{t}{τ}=\,In \,\left(\sqrt{2}\right)\quad\quad\tau=\frac{t}{In\,\left(\sqrt{2}\right)}$
$RC=\frac{2}{In\,2}$
$R=\frac{2}{C\,In\,2}=\frac{2}{10^{-6}\times In\,2}=\frac{2\times10^{6}}{In\,2}\Omega$