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Question

Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The capacitors are connected after charging. The decrease in energy of the combined system is :

A
14C(V21V22)
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B
14C(V21+V22)
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C
14C(V1V2)2
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D
14C(V1+V2)2
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Solution

The correct option is D 14C(V1V2)2
When connected total charge
Q=C(V1+V2)
when connected in parallel
x=CV
Qx=CV.
xQx=1
x=Q2
V=Q2C
V=V1+V22
decrease in energy=12CV21+12CV221cC(V1+V22)2
ΔE=14C(V1V2)2
64451_11490_ans_a907207fd37440b498cf4d2bcf1e5b34.png

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