A series combination of N_1 capacitors (each | vedclass

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(A) For the series combination of $N_1$ capacitors of capacity $C_1$,the equivalent capacitance is $C_{eq1} = \frac{C_1}{N_1}$.The energy stored in th

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$\frac{C_2 N_1 N_2}{9}$

Vedclass · Answer

(A) For the series combination of $N_1$ capacitors of capacity $C_1$,the equivalent capacitance is $C_{eq1} = \frac{C_1}{N_1}$.The energy stored in this combination is $E_1 = \frac{1}{2} C_{eq1} (3V)^2 = \frac{1}{2} \left( \frac{C_1}{N_1} \right) 9V^2 = \frac{9 C_1 V^2}{2 N_1}$.For the parallel combination of $N_2$ capacitors of capacity $C_2$,the equivalent capacitance is $C_{eq2} = N_2 C_2$.The energy stored in this combination is $E_2 = \frac{1}{2} C_{eq2} V^2 = \frac{1}{2} N_2 C_2 V^2$.Given that the total energy stored in both combinations is the same $(E_1 = E_2)$:$\frac{9 C_1 V^2}{2 N_1} = \frac{N_2 C_2 V^2}{2}$.Canceling $\frac{V^2}{2}$ from both sides,we get $\frac{9 C_1}{N_1} = N_2 C_2$.Solving for $C_1$,we get $C_1 = \frac{C_2 N_1 N_2}{9}$.

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