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Question

As shown in figure two identical capacitors are connected to a battery of V volts in parallel. When capacitors are fully charged, their stored energy is U1. If the key K is opened and material of dielectric constant εr=3 is inserted in each capacitors their stored energy is now U2. Then U1U2 will be:


A
35
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B
53
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C
45
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D
54
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Solution

The correct option is A 35

According to the question,
Initial total energy U1=UA+UB
=12CV2+12CV2
U1=CV2...(i)

Now switch is open, then energy of capacitor A,
EA=12KCV2 {K = dielectric constant}
EA=12×3CV2
EA=32CV2.....(ii)

The energy of capacitor B,
EB=12KC(V)2
(V=VK)
EB=12KC(VK)2
EB=(12CV2).13....(iii)

From equation (ii) and (iii):
U2=EA+EB
=(3+13).12CV2
U2=103.12CV2...(iv)

Compare equation (i) with (iv):
U1U2=35

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