A $20 F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged capacitor of $30 F$ . The decrease in the energy of the system will be:

25 J

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200 J

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125 J

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150 J

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Solution

The correct option is D $150 J$
$Δ U = \frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} (V_{1} - V_{2})$
$= \frac{20 \times 30}{2 (20 + 30)} {(5 - 0)}^{2} = 150 J$
Hence,
option $(D)$ is correct answer.

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A 20F capacitor is charged to 5V and isolated. It is then connected in parallel with an uncharged capacitor of 30F. The decrease in the energy of the system will be:

The correct option is D 150 JΔU=C1C22(C1+C2)(V1−V2)=20×302(20+30)(5−0)2=150JHence,option (D) is correct answer.

A $20 F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged capacitor of $30 F$ . The decrease in the energy of the system will be:

The correct option is D $150 J$
$Δ U = \frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} (V_{1} - V_{2})$
$= \frac{20 \times 30}{2 (20 + 30)} {(5 - 0)}^{2} = 150 J$
Hence,
option $(D)$ is correct answer.