Find the potential difference in volt between points M and N of the system shown in figure, if the emf is equal to $E=110V$  and the capacitance ratio $\frac{C_1}{C_2}$  is 2.

Applying loop law in two closed loops, we have

$110-\frac{q_2}{C}+\frac{q_1-q_2}{C}=0\text{or}{q}_2=\left(110C\right)$

And  $-110+\frac{q_1}{2C}+\frac{q_1-q_2}{C}=0$ 

Or   $q_1=\left(\frac{440C}{3}\right)$

Potential difference between points M and N is

$V_N-V_M=\frac{q_1-q_2}{C}$

$=\frac{110}{3}\text{volt}$

$=36.67\mathrm{Volt}$