Question

A capacitor of capacitance $100\mu F$ is connected across a battery of emf $6.0V$ through a resistance of $20k\Omega$ for $4s$. The battery is then replaced by a thick wire. What will be the charge on the capacitor, $4s$ after the battery is disconnected?

• A. $710\mu F$
• B. $830\mu F$
• C. $640\mu F$
• D. $518\mu F$

Answer 1

The correct option is D $518\mu F$

In the RC circuit, if the source of voltage is replaced by a thick wire, then capacitor will start discharging.

The amount of charge after 4s of discharging is $Q=Q_o(1-e^{\frac{-t}{RC}})$

We know that, $Q_0=C{V}_0-------\left(1\right)$

From (1)... $Q=C{V}_0(1-e^{\frac{-4}{20\times 100\times {10}^3\times {10}^{-6}}})=C{V}_0(1-e^{-2})=6\times {10}^{-4}(1-e^{-2})=518\mu F$