Three capacitors of capacitances , , and 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 . The value of x is

Question

Three capacitors of capacitances $25$ $μ F$ , $30$ $μ F$ , and $45$ $μ F$ 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 _______ . [2024]

Smart Achievers · Accepted Answer

(86) In series combination: $C_{1} = 25 + 30 + 45 = 100 μ F$

In parallel combination: $\frac{1}{C_{2}} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} \Rightarrow C_{2} = \frac{450}{43} μ F$

Energy $E = \frac{1}{2} C V^{2}$

$\frac{E_{2}}{E_{1}} = \frac{C_{2}}{C_{1}} = \frac{450 / 43}{100} = \frac{9}{86} \Rightarrow E_{2} = \frac{9}{86} E$

$\Rightarrow \frac{9}{86} E = \frac{9}{x} E \Rightarrow x = 86$

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Solution

Q. 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 9xE. The value of x is _______ . [2024]

Ans. (86) In series combination: C1=25+30+45=100μF In parallel combination: 1C2=125+130+145⇒C2=45043μF Energy E=12CV2 E2E1=C2C1=450/43100=986⇒E2=986E ⇒986E=9xE⇒x=86

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