Calculating Heat Dissipation in a Circuit with Capacitors

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Homework Statement



How much heat is generated in the circuit after the switch is shifted from position 1 to 2?

Homework Equations


The Attempt at a Solution



Let the charge on the right cap be Q(+Q on bottom plate) , on middle be Q-Q1(-Q+Q1 on bottom plate) and Q1(-Q1 on bottom plate) on left plate .

Ceq = C(C+C0)/(2C+C0)

Q=VC(C+C0)/(2C+C0)

Q1=VC/(2C+C0)

Q-Q1 = VCC0/(2C+C0)

Now when the switch is at position1,charge on the bottom plate of left cap is -Q1 and after the switch is moved to position 2 charge is -Q i.e charge Q-Q1 has flown through the battery .In other words negative charge of magnitude Q-Q1 flows from positive to negative terminal of battery(current flows from negative to positive terminal) i.e work done by battery is (Q-Q1)V .

Work done by battery = Change of energy stored in the capacitors + heat dissipated .

So,heat dissipated = Work done by battery - Change of energy stored in the capacitors .

Change of energy stored in the capacitors will be zero .

Heat dissipated = Work done by battery = (Q-Q1)V = V2CC0/(2C+C0) .

Is the answer correct ?

Have I approached the problem correctly ?
 

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That looks right to me.
 
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Yes. Any finite battery resistance will dissipate the same amount of energy. Interesting problem.
(I did not check the delta stored energy).
 
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Thanks haruspex and rude man .

Do you agree that if the switch were initially at position 2 and moved from position 2 to 1 then in that case also the result obtained in the OP would remain same ?
 
I haven't checked all the math but the approach seems correct. The equation you provided for Q1 is missing something. The units don't match.
 
Tanya Sharma said:
Thanks haruspex and rude man .

Do you agree that if the switch were initially at position 2 and moved from position 2 to 1 then in that case also the result obtained in the OP would remain same ?

Yes, it's symmetric.
 
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Thanks everyone for your valuable inputs .
 

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