MHTCET 16 May 2024 Shift 1 Paper

Concept:Energy stored in a capacitor combination depends on its equivalent capacitance and the applied voltage.Explanation:For $n_1$ capacitors each of capacitance $C_1$ in series, the equivalent capacitance is:$$C_{\text{eq1}} = \frac{C_1}{n_1}$$This combination is charged by a $6\,\text{V}$ source, so its energy is:$$U_1 = \frac{1}{2}\left(\frac{C_1}{n_1}\right)(6)^2$$For $n_2$ capacitors each of capacitance $C_2$ in parallel, the equivalent capacitance is:$$C_{\text{eq2}} = n_2 C_2$$This combination is charged by a $2\,\text{V}$ source, so its energy is:$$U_2 = \frac{1}{2}(n_2 C_2)(2)^2$$Given that the total energy of both combinations is the same:$$U_1 = U_2$$Substitute the expressions:$$\frac{1}{2}\left(\frac{C_1}{n_1}\right)(36) = \frac{1}{2}(n_2 C_2)(4)$$Simplify:$$\frac{18 C_1}{n_1} = 2 n_2 C_2$$Solving for $C_2$:$$C_2 = \frac{9 C_1}{n_1 n_2}$$Answer:$$C_2 = \frac{9 C_1}{n_1 n_2}$$Therefore, the correct option is D.