Q128 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
Capacitors of capacities $C_1$ and $C_2$ are connected in series. If the combination is connected to a supply of V volt then the potential difference across capacitor $C_2$ is
- A$\dfrac{C_1 + C_2}{C_1 V}$
- Correct answerB$\dfrac{C_1 V}{C_1 + C_2}$
- C$\dfrac{C_1 + C_2}{C_2 V}$
- D$\dfrac{C_2 V}{C_1 + C_2}$
Solution
In series combination, the equivalent capacitance is $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}$.
The charge on each capacitor is the same and is given by $Q = C_{eq} V = \left( \dfrac{C_1 C_2}{C_1 + C_2} \right) V$.
The potential difference across capacitor $C_2$ is $V_2 = \dfrac{Q}{C_2}$.
Substituting the value of $Q$, we get $V_2 = \dfrac{C_1 V}{C_1 + C_2}$.
The charge on each capacitor is the same and is given by $Q = C_{eq} V = \left( \dfrac{C_1 C_2}{C_1 + C_2} \right) V$.
The potential difference across capacitor $C_2$ is $V_2 = \dfrac{Q}{C_2}$.
Substituting the value of $Q$, we get $V_2 = \dfrac{C_1 V}{C_1 + C_2}$.
Hence, the correct answer is option: 'B'.