Calculating Heat Dissipation in a Circuit with Capacitors

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Homework Help Overview

The discussion revolves around calculating the heat generated in a circuit involving capacitors when a switch is moved from one position to another. The problem is situated within the context of electrical circuits and energy dissipation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the heat dissipated by analyzing the charges on the capacitors and the work done by the battery. Some participants question the correctness of the approach and the completeness of the equations provided, particularly regarding the units of the variables involved.

Discussion Status

Participants have offered varying levels of agreement with the original poster's approach, with some affirming its validity while others express concerns about specific mathematical details. There is an ongoing exploration of whether the result would remain consistent if the switch's initial position were reversed.

Contextual Notes

Some participants note the symmetry in the problem setup and discuss the implications of finite battery resistance on energy dissipation. There are mentions of potential missing information in the equations presented by the original poster.

Tanya Sharma
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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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