During the charging of a capacitor,the variation of the potential $V$ of the capacitor with time $t$ is shown as:

The time constant of the given circuit is:

Switch $S$ of the circuit shown in the figure is in position $1$ for a long time. At instant $t = 0$,it is thrown from position $1$ to $2$. Find the thermal power $P_1(t)$ generated in resistance $R_1$.

In the circuit shown in the figure,$C_1 = 2C_2$. Switch $S$ is closed at time $t = 0$. Let $i_1$ and $i_2$ be the currents flowing through the branches containing $C_1$ and $C_2$ respectively at any time $t$. Then the ratio $i_1 / i_2$:

$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

In an $RC$ circuit as shown below,both switches are open initially. Now switch $S_1$ is closed and $S_2$ is kept open. ($q$ is the charge on the capacitor and $\tau = RC$ is the capacitive time constant). Which of the following statements is correct?