Question

Find the potential difference between the points $M$ and $N$ of the system shown in figure, if the emf is equal to $E=110V$ and the capacitance ratio $C_2/C_1$ is $23$ than.

Answer 1

$E=110V$
$\frac{C_2}{C_1}=23$
Let $q_1,q_2\&q_3$ be the charges in the branches as shown
Applying KV first Law at point N
$\sum q=0$
$\Rightarrow q_1-q_2-q_3=0$
$\Rightarrow q_1=q_2+q_3\to 1$
The plates of capacitor is positive where charge is entering
Applying KV second law in loop $I$
$\sum E=\sum \frac{q}{C}$
$\Rightarrow E=\frac{q_3}{C_2}+\frac{q_1}{C_1}\Rightarrow C_{2}E=q_3+\frac{C_2}{C_1}{q}_1$
$\Rightarrow C_{2}E=q_3+23q_1\to 2$
Applying KV second law in loop $II$
$\Rightarrow E=+\frac{q_2}{C_2}-\frac{q_3}{C_2}$
Negative sign is taken as liio is entering from negative plate
$C_{2}E=q_2-q_3\to 3$
From equation $1$ and $3$
$\Rightarrow C_{2}E=q_1-q_2-q_3$
$\Rightarrow C_{2}E=q_1-2q_3\to 4$
From equation $2$ and $4$
$\Rightarrow C_{2}E=q_3+23(C_{2}E+2q_3)$
$\Rightarrow C_{2}E=47q_3+23C_{2}E$
$\Rightarrow \frac{q_3}{C_2}=-\frac{22}{47}E=-\frac{22}{47}\times 110=60.5$
Potential drops between points M and N are
$V_{MN}=-\frac{q_3}{C_2}=60.5V$