a 28 microfarad capacitor is charged to 100v and another 2 microfarad capacitor to 200v. then thew r connected in parallel. determine the total initial n final energies n account for d difference in d two.

Asked by Jhalak Jauhari | 4th Apr, 2013, 09:27: PM

Expert Answer:

C1 = 28 microF
V1 = 100 V
Hence Q1=C1xV1 = 2800 microC = 2.8 mC

C2= 2 microF
V2= 200 VSo,
Q2 = 400 microC = 0.4 mC

Total initial energy = c1VxV/2  + c2VxV/2
= 28x100x100/2   + 2x200x200/2 microJ
= .18 J
 
Now, when these are connected in parallel, their potential will become same.V1=V2 = V say
So, finallyQ1 = C1xVQ2 = C2xV

and Q1+Q2 = 2.8+.4 = 3.2 mC (due to conservation of charge)
so, C1V + C2V = 3.2 mC
Hence V = 3.2mC / (28+2 micro F) = 320/3 V 
So, energies stored in each capacitor=E1= C1Vxv/2E2= C2VxV/2

Finally , Total energy of the system = VxV(C1+C2)/2
= 320x320x(30microF)/2x3x3 = 0.17066 J

Difference in energy = -0.00934 J
This difference is due to changing the potential of the capacitors from the initial values, hencecausing the redistribution of charges means redistribution of energy.

Answered by  | 28th Apr, 2013, 06:05: PM

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