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Question

Three capacitors each having capacitance C=2μF are connected to battery of emf 30V as shown in the figure. When the switch is closed. If heat generated in the circuit is x10mJ,then x is :

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A
6
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B
2
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C
3
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D
12
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Solution

The correct option is C 3
In the initial condition,
Cnet=(2C×C)/(2C+C)=2C/3
Charge provided by battery =2CV/3
Charge on capacitors =CV/3,CV/3,2CV/3
Potential across capacitors =V/3,V/3,2V/3
Energy of capacitors =CV2/18,CV2/18,2CV2/9
Total energy =CV2/3
In the final condition,
Cnet=C
Charge provided by battery =CV
Charge across capacitors =0,0,CV
Potential across capacitors =0,0,V
Total energy of capacitors =CV2/2
Therefore,
work done by battery =VδQ=V(CV2CV/3)=CV2/3
Now,
work done by battery = (Final energy - Initial energy) of capacitors + Heat
CV2/3=CV2/6+HEATHEAT=CV2/6=2×106×900/6=0.3mJ

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