Four equal capacitors are connected to a $10 \, V$ battery as shown in the figure. The potentials of $A$ and $B$ are:

$A$ capacitor of capacitance $5\,\mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5\,\Omega$. The amount of charge on the capacitor plate is......$\mu C$.

Find the potential difference $V_a - V_b$ between the points $a$ and $b$ shown in the figure.

Two capacitors of capacitances $1\ \mu F$ and $3\ \mu F$ are charged to the same voltages $5\ V$. They are connected in parallel with oppositely charged plates connected together. Then:

Assertion: Charges are given to plates of two plane parallel plate capacitors $C_1$ and $C_2$ (such that $C_2 = 2C_1$) as shown in the figure. Then the key $K$ is pressed to complete the circuit. Finally,the net charge on the upper plate and the net charge on the lower plate of capacitor $C_1$ is positive.
Reason: In a parallel plate capacitor,both plates always carry equal and opposite charge.

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is