Question

A parallel plate capacitor of $1\mu F$ capacity is discharging through a resistor. If its energy reduces to half in one $\text{second}$. The value of resistance will be

• A. $\frac{2}{ln\left(2\right)}M\Omega$
• B. $\frac{4}{ln\left(2\right)}M\Omega$
• C. $\frac{10}{ln\left(2\right)}M\Omega$
• D. $\frac{16}{ln\left(2\right)}M\Omega$

Answer 1

The correct option is A $\frac{2}{ln\left(2\right)}M\Omega$

As $Q=Q_o{e}^{-t/\tau }$
When energy is $50\%$ then $Q=\frac{Q_o}{\sqrt{2}}$
$\frac{Q_o}{\sqrt{2}}=Q_o{e}^{-t/\tau }$
$\Rightarrow e^{t/\tau }=\sqrt{2}$
$\frac{t}{\tau }=ln\left(\sqrt{2}\right)$
$\tau =\frac{t}{ln\left(\sqrt{2}\right)}$
$RC=\frac{2}{ln2}$
$R=\frac{2}{Cln2}=\frac{2}{{10}^{-6}\times ln2}$
$=\frac{2\times {10}^6}{ln2}\Omega$