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 (integer)
86
Solution
<p>$E=\frac{1}{2}(25+30+45)(100)^2 \quad \text{.... (i)}$</p>
<p>$$\text { Also, } \frac{9}{x} E=\frac{1}{2} \frac{1}{\left(\frac{1}{25}+\frac{1}{30}+\frac{1}{45}\right)}(100)^2 \quad \text{.... (ii)}$$</p>
<p>From (i) and (ii)</p>
<p>$x=86$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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