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Asked by arup.isro 25th June 2019, 11:25 PM
Answered by Expert
Answer:
It is assumed that second uncharged capacitor also 0.1 μf.
 
Initial energy stored = (1/2)C V2 = (1/2)×0.1×10-6 ×100 = 5 μJ
 
If the charged capacitor is connected parallel to uncharged capacitor of equal capacitance,
charges are equally shared, potential difference is halved.
 
energy stored after connecting two capacitors = (1/2)×(2C)×(V/2)2 = (1/4) C V2 = 0.25× 0.1×10-6 × 100 = 2.5 μJ
 
ratio of energy stored in tweo capacitors to intial energy stored in one capacitor = 2.5/5 = (1/2)
 
 
Answered by Expert 26th June 2019, 1:32 PM
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