MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Energy stored in a capacitor combination is given by $U = \frac{1}{2} C_{\text{eq}} V^{2}$.Explanation:Let $C_{1} = C$ and $C_{2} = 2C$.For the series combination, the equivalent capacitance is $C_{s} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \frac{C(2C)}{C + 2C} = \frac{2C}{3}$.For the parallel combination, the equivalent capacitance is $C_{p} = C_{1} + C_{2} = 3C$.The energy stored in the series combination is $U_{s} = \frac{1}{2} C_{s} V_{s}^{2}$.The energy stored in the parallel combination is $U_{p} = \frac{1}{2} C_{p} V_{p}^{2}$.Since the energies are equal, $U_{s} = U_{p}$.This gives $\frac{1}{2} \cdot \frac{2C}{3} V_{s}^{2} = \frac{1}{2} \cdot 3C V_{p}^{2}$.Cancelling common factors, we get $\frac{2}{3} V_{s}^{2} = 3 V_{p}^{2}$.So $\frac{V_{s}^{2}}{V_{p}^{2}} = \frac{9}{2}$.Taking the square root, $\frac{V_{s}}{V_{p}} = \frac{3}{\sqrt{2}}$.Thus, $V_{s} : V_{p} = 3 : \sqrt{2}$.Answer:The correct ratio is $3 : \sqrt{2}$, which is Option B.