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(D) For the series combination of $n_1$ capacitors of capacitance $C_1$ connected to a $4V$ source:The equivalent capacitance is $C_s = \frac{C_1}{n_1

Vedclass · Accepted Answer

$\frac {16C_1}{n_1n_2}$

Vedclass · Answer

(D) For the series combination of $n_1$ capacitors of capacitance $C_1$ connected to a $4V$ source:The equivalent capacitance is $C_s = \frac{C_1}{n_1}$.The energy stored is $U_s = \frac{1}{2} C_s (4V)^2 = \frac{1}{2} \left( \frac{C_1}{n_1} \right) (16V^2) = \frac{8 C_1 V^2}{n_1}$.For the parallel combination of $n_2$ capacitors of capacitance $C_2$ connected to a $V$ source:The equivalent capacitance is $C_p = n_2 C_2$.The energy stored is $U_p = \frac{1}{2} C_p V^2 = \frac{1}{2} (n_2 C_2) V^2$.Given that $U_s = U_p$:$\frac{8 C_1 V^2}{n_1} = \frac{1}{2} n_2 C_2 V^2$.Solving for $C_2$:$C_2 = \frac{16 C_1}{n_1 n_2}$.

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