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Question

A capacitor 4 μF charged to 50 V is connected to another capacitor of 2 μF charged to 100 V with plates of like charges connected together. The total energy before and after connection in multiples of (102 J) is

A
1.5 and 1.33
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B
1.33 and 1.5
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C
3.0 and 2.67
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D
2.67 and 3.0
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Solution

The correct option is A 1.5 and 1.33
The total energy before connection
12×4×106×(50)2+12×2×106×(100)2
=1.5×102 J
When connected in parallel
4×50+2×100=6×VV=2003
Total energy after connection
=12×6×106×(2003)2=1.33×102 J

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