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Questions  

A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2 , the percentage of its stored energy dissipated is

 

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a
75%
b
80%
c
0%
d
20%

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detailed solution

Correct option is B

Initially, the energy stored in  2μF  capacitor is Ui=12CV2=122×10-6V2=V2×10-6 Jinitially, the charge stored in  2μF capacitor isQi=CV=2×10-6V=2V×10-6coulomb. When switch S is turned to position 2 , the charge flows and both the capacitors share charges till a common potential VCis reached.VC= total charge  total capacitance =2V×10-6(2+8)×10-6=V5 volt  Finally, the energy stored in both the capacitors Uf=12(2+8)×10-6V52=V25×10-6 J% loss of energy, ΔU=Ui-UfUi×100%=V2-V2/5×10-6V2×10-6×100%=80%

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Similar Questions

Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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