JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 24)
Three capacitors of capacitances $$25 \mu \mathrm{F}, 30 \mu \mathrm{F}$$ and $$45 \mu \mathrm{F}$$ are connected in parallel to a supply of $$100 \mathrm{~V}$$. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is $$\frac{9}{x} \mathrm{E}$$. The value of $$x$$ is _________.
Answer
86
Explanation
$$E=\frac{1}{2}(25+30+45)(100)^2 \quad \text{.... (i)}$$
$$\text { Also, } \frac{9}{x} E=\frac{1}{2} \frac{1}{\left(\frac{1}{25}+\frac{1}{30}+\frac{1}{45}\right)}(100)^2 \quad \text{.... (ii)}$$
From (i) and (ii)
$$x=86$$
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