Three capacitors of capacitances 25, 30 and 45 are connected in… | JEE Main 2024 (5 Apr, Shift 1)

Q34JEE Main 2024 · 5 APR · SHIFT 1Medium

Combination of Capacitors

Three capacitors of capacitances $25\ \mu\text{F}, 30\ \mu\text{F}$ and $45\ \mu\text{F}$ are connected in parallel to a supply of $100\text{ V}$ . 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 _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 86

Same supply: $U \propto C$ , so $\frac{E_s}{E} = \frac{C_s}{C_p}$ .
$C_p = 25 + 30 + 45 = 100\ \mu\text{F}$ .
$\frac{1}{C_s} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} = \frac{18+15+10}{450} = \frac{43}{450} \Rightarrow C_s = \frac{450}{43}\ \mu\text{F}$ .
$\frac{E_s}{E} = \frac{450}{43\times100} = \frac{9}{86}$ , so $x = 86$ .

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