In the given circuit, if the potential at A is 50 V, then potential at B is:

$\frac{50}{3}V$

The correct answer is Option (1) → $\frac{50}{3}V$

Solution:

Given: Potential at A = $50\ \text{V}$

The capacitors $C_2$ and $C_3$ are connected in parallel, so their equivalent capacitance is:

$C_{23} = C_2 + C_3 = 4 + 2 = 6\ \mu\text{F}$

Now, $C_1$ ($3\ \mu\text{F}$) is in series with $C_{23}$ ($6\ \mu\text{F}$).

For capacitors in series, the charge $Q$ on each is the same, and total voltage is given by:

$50 = \frac{Q}{C_1} + \frac{Q}{C_{23}}$

Substitute the values:

$50 = Q\left(\frac{1}{3} + \frac{1}{6}\right)$

Therefore, $50 = Q \cdot \frac{1}{2}$

$\Rightarrow Q = 100\ \mu\text{C}$

The potential at point B is the voltage across $C_{23}$:

$V_B = \frac{Q}{C_{23}} = \frac{100}{6} = \frac{50}{3}\ \text{V}$

Hence, the potential at B is: $V_B = \frac{50}{3}\ \text{V} \approx 16.67\ \text{V}$