In the circuit diagram shown,$E = 5 \, V, r = 1 \, \Omega, R_2 = 4 \, \Omega, R_1 = R_3 = 1 \, \Omega$ and $C = 3 \, \mu F$. The magnitude of the charge on each capacitor plate is......$\mu C$.

In the given figure,the resistance of the coil of the galvanometer $G$ is $2\,\Omega$. The emf of the cell is $4\,V$. The ratio of the potential difference across $C_1$ and $C_2$ is:

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?

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 $i_1 / i_2$:

The time in seconds required to produce a potential difference of $20 \ V$ across a capacitor of $1000 \ \mu F$ when it is charged at a steady rate of $200 \ \mu C/s$ is

After closing the key in the given circuit:
$(a)$ Bulb $(2)$ will glow and maintain its brightness.
$(b)$ Brightness of bulb $(1)$ will gradually decrease and at steady state it goes completely dark.