Question

A capacitor of capacitance $0.1\mu F$ is charged to certain potential and allow to discharge through a resistance of $10M\Omega$. How long will it take for the potential to fall to one half of its original value?

• A. $0.301s$
• B. $0.2346s$
• C. $1.386s$
• D. $0.693s$

Answer 1

The correct option is A $0.301s$

We know that for discharging of capacitor, where t is the time of discharge, $Q=Q_o{e}^{\left(\frac{-t}{RC}\right)}$

Where $Q_o$ is the initial charge in the capacitor. As we know, $Q=CV$.

So, $CV=C{V}_o{e}^{\left(\frac{-t}{RC}\right)}$

According to question, $V=\frac{V_o}{2}$.

Therefore, $\frac{V_o}{2}=V_o{e}^{\frac{-t}{RC}}=>\frac{1}{2}=e^{\frac{-t}{RC}}=>log0.5=\frac{-t}{RC}=>-0.301=\frac{-t}{10\times {10}^6\times 0.1\times {10}^{-6}}=>-t=-0.301\times (10\times 0.1)=>t=0.301s$