A capacitor of capacitance C is charged to voltage V0 and allowed to discharge through a resistance R while charge another capacitance′αC′. What fraction of total energy is lost at steady state?
A
11+α
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B
11+1α
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C
2α1+2α
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D
2α1+α
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Solution
The correct option is B11+1α
Let Vf be the final voltage appearing across ab
Equating charge of two capacitors, VfαC=V0C−VfC Vf(1+α)=V0 Vf=V01+α
Initial energy =12CV20
Final energy =12(αC)V2f(1+α)=12C(1+α)(V01+α)2 =12CV2011+α
Loss in energy = inital energy - final energy =12CV20−12CV20(11+α)
loss in energy =12CV20(α1+α)
The fraction of energy lost=(α1+α)=11+1α