MHTCET 15 Apr 2026 Shift 2 Paper

Concept:In a series combination of capacitors, the charge stored on each capacitor is identical, while the total voltage divides across them.Explanation:Let the equivalent capacitance of $C_1$ and $C_2$ connected in series be $C$.For two capacitors in series, we have:$$C = \frac{C_1 C_2}{C_1 + C_2}$$When this combination is connected to a supply of $V$ volts, the total charge $Q$ stored in the combination is:$$Q = C V = \frac{C_1 C_2}{C_1 + C_2} V$$Since the charge is the same on every capacitor in series, the charge on $C_2$ is also $Q$.The potential difference across capacitor $C_2$ is given by:$$V_2 = \frac{Q}{C_2}$$Substituting the expression for $Q$:$$V_2 = \frac{1}{C_2} \times \frac{C_1 C_2}{C_1 + C_2} V$$Simplifying:$$V_2 = \frac{C_1 V}{C_1 + C_2}$$This matches option B directly.Answer:The potential difference across capacitor $C_2$ is $\frac{C_1 V}{C_1 + C_2}$.Correct option: B.