Three capacitors of capacitances 2 μF, 3 μF and 6 μF are connected in series. The combination is connected to a 12 V supply. The respective potential differences across them will be

6 V, 4 V and 2 V

The correct answer is Option (3) → 6 V, 4 V and 2 V

Given:

Capacitances: $C_1 = 2 \, \mu\text{F}, \, C_2 = 3 \, \mu\text{F}, \, C_3 = 6 \, \mu\text{F}$

Series connection, voltage: $V = 12 \, \text{V}$

For series capacitors: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

$\frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1 \Rightarrow C_{\text{eq}} = 1 \, \mu\text{F}$

Charge on series combination: $Q = C_{\text{eq}} V = 1 \cdot 12 = 12 \, \mu\text{C}$

Voltage across each capacitor: $V_i = \frac{Q}{C_i}$

$V_1 = \frac{12}{2} = 6 \, \text{V}$

$V_2 = \frac{12}{3} = 4 \, \text{V}$

$V_3 = \frac{12}{6} = 2 \, \text{V}$

Answer: $V_1 = 6 \, \text{V}, \, V_2 = 4 \, \text{V}, \, V_3 = 2 \, \text{V}$