Question

In the circuit shown, a potential difference of $60V$ is applied across $AB$. The potential difference between the points $M$ and $N$ is :

• A. $10V$
• B. $15V$
• C. $20V$
• D. $30V$

Answer 1

The correct option is D $30V$

The charge distribution on different plates is shown in the figure. Suppose charge $Q_1+Q_2$ is given by the positive terminal of the battery, out of which $Q_1$ resides on the positive plate of the capacitor $1$ and $Q_2$ on that of $2$. The remaining plates will have changes as shown in the figure.
Let us take the potential at the point be $zero$. The potential at A will be $60V$. Let the potentials at points $M$ and $N$ be $V_M$ and $V_N$ respectively. Writing the equations, $Q=CV$ for the four capacitors, we get,
$Q_1=C\times 60V=60C$ --- (i)
$Q_2=2C(60V-V_M)C$ --- (ii)
$Q_2=C(V_M-V_N)C$ --- (iii)
$Q_2=2C\times V_{N}C$ --- (iv)
From (ii) and (iii) we have,
$2C(60-V_M)=C(V_m-V_N)$
$⟹3V_M-V_N=120$ --- (v)
From (iii) and (iv) we have,
$C(V_M-V_N)=2C{V}_N$
$⟹V_M=3V_N$ --- (vi)
Substituting (vi) in equation (v) we get,
$3\left(3V_N\right)-V_N=120$
$⟹V_N=15V$
Hence $V_M=3\times 15=45V$
The potential difference between the points $M$ and $N$ is given by
$V_M-V_N=45-15=30V$