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A capacitor of 150pF charged by 220V supply .Battery then disconnected and charged capacitor is connected to another uncharged capacitor of 50pF .calculate diff. beyween final energy stored in combined system and initial energy..ans given is 0.75×10^-6

A capacitor of 150pF charged by 220V supply .Battery then disconnected and charged capacitor is connected to another uncharged capacitor of 50pF .calculate diff. beyween final energy stored in combined system and initial energy..ans given is 0.75×10^-6

Grade:12

1 Answers

Arun
25763 Points
3 years ago

Here, the loss in energy (difference between initial and final energy) is given by the following relation

ΔU = C1C2.(V1 - V2)2 / 2(C1 + C2)

here,

C1 = 150 pF = 150 x 10-12 F

C2 = 50 pF = 50 x 10-12 F

V1 = 220 volts

V2 = 0 volts

so, we get 

ΔU = [(150 x 10-12 x 50 x 10-12) x (220 - 0)2] / [2x(150 x 10-12 + 50 x 10-12)] 

which can be calculated...

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