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 ______.

Three capacitors are given as $$C_1 = 25\mu$$F, $$C_2 = 30\mu$$F, $$C_3 = 45\mu$$F, with a supply voltage $$V = 100$$ V.

When these capacitors are connected in parallel, the equivalent capacitance is

$$ C_p = 25 + 30 + 45 = 100 \mu\text{F} $$

and the energy stored in this configuration is

$$ E = \frac{1}{2}C_p V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times 10000 = 0.5 \text{ J} $$

In the case of series connection, the equivalent capacitance satisfies

$$ \frac{1}{C_s} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} $$

LCM of 25, 30, 45 = 450:

$$ \frac{1}{C_s} = \frac{18 + 15 + 10}{450} = \frac{43}{450} $$

$$ C_s = \frac{450}{43} \mu\text{F} $$

Thus the energy stored in series is

$$ E_s = \frac{1}{2}C_s V^2 = \frac{1}{2} \times \frac{450}{43} \times 10^{-6} \times 10000 = \frac{450}{43} \times 5 \times 10^{-3} $$

Finally, the ratio of the energy stored in series to that in parallel is

$$ \frac{E_s}{E} = \frac{C_s}{C_p} = \frac{450/43}{100} = \frac{450}{4300} = \frac{9}{86} $$

Therefore, $$E_s = \frac{9}{86}E$$, which implies that $$x = 86\,$$.