Question

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

Answer 1

Applying loop law in two closed loops we have,

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

$V_N-V_M=\frac{110}{3}volt$