Question

Initially, the capacitor was uncharged, Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

• A. after time travel $\tau$, charge on the capacitor is $CV/2$
• B. after time travel $2\tau$, charge on the capacitor is $CV/(1-e^{-2})$
• C. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.
• D. after time interval $2\tau$, charge on the capacitor is $CV/(1-e^{-1})$

Answer 1

The correct option is B after time travel $2\tau$, charge on the capacitor is $CV/(1-e^{-2})$

After time interval $2\tau$ charge in capacitor

$\frac{cv}{1-e^{\frac{-2\tau }{t}}}=\frac{cv}{1-e^{-2}}$