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Find heat developed in connecting wire
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Asked by ashutosharnold1998 | 02 Oct, 2019, 01:14: AM
answered-by-expert Expert Answer
Initial charge Q on 50 μf capacitor when it is charged with 20 V battery ,
 
Q = C V = 50×10-6 × 20 = 1000 × 10-6 μC = 1 mC
 
when the chraged 50 μf capacitor is connected across 10 V battery,
 
charge retained by capacitor = 50×10-6 × 10 = 500 ×10-6  = 0.5 mC
 
Hence excess charge 0.5 mC will be flown to battery .
 
Since capacitor is at higher potential than bettery, work will be done by capacitor to pass the charges to battery.
 
workdone to pass the excess charge 0.5 mC to a 10 V battery , W = Q × V = 0.5×10-3 × 10 = 5 mJ
 
Energy loss by capacitor = (1/2) C [ Vi2 - Vf2 ]  = (1/2)×50×10-6 [ 202 - 102 ] = 7.5 × 10-3 J  = 7.5 mJ
 
Out of this 7.5 mJ energy lost by capacitor, 5 mJ energy is spent to pass the excess charges to battery.
 
Remaining 2.5 mJ is dissipated as heat in connecting wires.
 
 
 
Answered by Thiyagarajan K | 02 Oct, 2019, 02:06: PM

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