Two capacitors of capacitance of 6 μF and 12 μF are connected in series with a battery. The voltage across 6 μF capacitor is 2 V. The total voltage across the combination is

3 V

The correct answer is Option (4) → 3 V

Given:

$C_1 = 6\ \mu\text{F},\ C_2 = 12\ \mu\text{F}$

Voltage across $C_1$: $V_1 = 2\ \text{V}$

For series combination, charge $Q$ is the same on both capacitors:

$Q = C_1 V_1 = 6 \cdot 2 = 12\ \mu\text{C}$

Voltage across $C_2$:

$V_2 = \frac{Q}{C_2} = \frac{12}{12} = 1\ \text{V}$

Total voltage across combination:

$V_{\text{total}} = V_1 + V_2 = 2 + 1 = 3\ \text{V}$

Total voltage ≈ 3 V