Question

Three capacitors of capacitances 25 μF, 30 μF and 45 μF are connected in parallel to a supply of 100 V. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is \(\frac{9}{x} E\)  . The value of x is .................

Answer 1

Correct Answer: 86

In parallel combination: Potential difference is the same across all capacitors.

\[ \text{Energy} = \frac{1}{2}(C_1 + C_2 + C_3)V^2 \]

\[ = \frac{1}{2}(25 + 30 + 45) \times (100)^2 \times 10^{-6} = 0.5 = E \]

In series combination: Charge is the same on all.

\[ \frac{1}{C_{\text{equ}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} \]

\[ \frac{1}{C_{\text{equ}}} = \frac{18 + 15 + 10}{450} = \frac{43}{450} \implies C_{\text{equ}} = \frac{450}{43} \]

Energy:

\[ \text{Energy} = \frac{Q^2}{2C_1} + \frac{Q^2}{2C_2} + \frac{Q^2}{2C_3} \]

\[ = \frac{Q^2}{2} \left[ \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right] \]

\[ = \frac{(V \times C_{\text{equ}})^2}{2 \times C_{\text{equ}}} \times \frac{1}{C_{\text{equ}}} = \frac{V^2C_{\text{equ}}}{2} \]

\[ = \frac{(100)^2}{2} \times \frac{450}{43} \times 10^{-6} \]

\[ = \frac{4.5}{86} = \frac{9}{x} \times 0.5 \implies x = 86 \]