Charge redistribution on capacitors

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Homework Help Overview

The discussion revolves around charge redistribution in a circuit involving capacitors, specifically focusing on the behavior of charges before and after a switch is closed. Participants are analyzing the implications of charge changes on capacitor C and the work done by the battery.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the charge values on capacitors A, B, and C before and after the switch is closed, questioning the energy supplied by the battery based on these changes. There are attempts to calculate total energy in the circuit and considerations of energy dissipation as heat.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some express confusion over specific options and the role of resistance in the circuit. There is no clear consensus on which option is incorrect, as different participants have varying views on the correctness of the statements presented.

Contextual Notes

Participants note the absence of resistance in the circuit diagram, which raises questions about energy dissipation. There is also mention of a potential misunderstanding regarding the energy stored in capacitors and the implications of the equations used.

Jahnavi
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Homework Statement


capacitor.jpg


Homework Equations

The Attempt at a Solution


I will name the top left capacitor A , bottom left capacitor B and right capacitor C .

Before the switch is closed , charges on A and B will be 20 μC each and that on C will be 40 μC .

After the switch is closed ,both A and B will be shorted . Charge on C will be 60μC .

A and B will get discharged i.e there will be no charge on A and B .

On doing further calculations I am getting all four options correct .

The correct answer i.e incorrect statement is 3) .

But if charge on C changes from 40μC to 60μC , doesn't this mean 20 μC has flown through the battery and battery has done 0.6 mJ work or 0.6 mJ energy is supplied by the battery ?
 

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Jahnavi said:
But if charge on C changes from 40μC to 60μC , doesn't this mean 20 μC has flown through the battery and battery has done 0.6 mJ work or 0.6 mJ energy is supplied by the battery ?
Find the total energy in the circuit before and after the switch is closed. The difference between the two energies would be the energy supplied by the battery.
 
cnh1995 said:
Find the total energy in the circuit before and after the switch is closed. The difference between the two energies would be the energy supplied by the battery.

I don't think this is correct :smile:

You are forgetting energy dissipated as heat .
 
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Also, the second option is confusing (or should I say 'tricky') because there is no resistance shown in the circuit. I'm afraid that makes this statement incorrect.
cnh1995 said:
The difference between the two energies would be the energy supplied by the battery.
Can you correct this statement?
Hint: The '1/2' in the equation for the energy stored in the capacitor.
 
cnh1995 said:
Also, the second option is confusing (or should I say 'tricky') because there is no resistance shown in the circuit. I'm afraid that makes this statement incorrect.

It doesn't make a difference whether a resistance is shown in the circuit or not :smile: .

Heat is dissipated in the wires .
 
Jahnavi said:
Heat is dissipated in the wires .
And the wires are shown to be ideal (although you can assume them to have some resistance).
 
Anyways , I am finding all four options correct .

Which option do you find incorrect ?

Answer given is 3)
 
Jahnavi said:
Anyways , I am finding all four options correct
Yes, I misread option C as 30mJ. I agree that all the options are correct.
(I still find dc capacitive circuits very tricky!).
 
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