How Does Switching Polarity Affect Capacitor Energy and Heat Dissipation?

[varunKanpur]

I am not able to solve this problem

I am thinking that C_o would be charging in reverse polarity switch is shifted to 2 position, so change in energy of capacitor will be released as heat energy, i.e twice the energy of stored capacitor.

 

[willem2]

Nice problem :).

The energy that dissipates will be equal to the energy delivered by the battery + (the energy stored in the capacitors before you flip the switch) - (the energy stored in the capacitors after you flip the switch).

The energy in the capacitors before and after you flip the switch is equal, because the end situation after the switch is flipped is just a reflection of the situation before the switch is flipped. The voltage across C₀ will be in the other direction, but that won't matter for its energy.

So you only need the energy delivered by the battery which is QV, where Q is the amount of charge it delivered.

Now you can find the amount of charge that the battery delivered just by watching what happens to the charge on the left C, because when you disconnect C0 from point 1, there will be no current anywhere in the circuit, and after you connect C0 to point 2, all of the current that is going through the battery is also going through the left C.

So you just have to find the charge on the left capacitor before and after you flip the switch, and multiply it by V.

 

[varunKanpur]

Now you can find the amount of charge that the battery delivered just by watching what happens to the charge on the left C, because when you disconnect C0 from point 1, there will be no current anywhere in the circuit, and after you connect C0 to point 2, all of the current that is going through the battery is also going through the left C.

So you just have to find the charge on the left capacitor before and after you flip the switch, and multiply it by V.

Thank for the quick reply. When I will connect C_o to point 2 then a fraction of current will flow through to the left C and rest through C_o.

The work done by the battery = To the heat generated.
When the switch is closed the polarity of capacitor changes.
Before Switching, Charge is -Q| |Q
But after switching, Charge would be Q| |-Q
Then the Charge delivered by the battery is 2Q.
According to my calculation, Q=CC_oV/(2C C_o)

Therefore the answer should be 2CC_oV²/(2C C_o)

 

[willem2]

Thank for the quick reply. When I will connect C_o to point 2 then a fraction of current will flow through to the left C and rest through C_o.

The point is that when the switch makes a connection at point 2, the left C is in series with the battery, so the charge delivered by the battery is equal to the change in charge of the left C.

This is not the case for C₀. You calculated the charge on before and after the flip correctly, but some of the current going through C₀ will go through the rightmost C, and not the battery.

before the flip, the leftmost C has charge

$$\frac { V C^2} {2C + C_0}$$

and after it:

$$\frac { V C (C+C_0)} {2C + C_0}$$

The difference is the answer you need.

This is half your answer, and that's because half of the charge needed to decharge C₀ and charge it in the opposite direction, will be delivered by the righthand C and not by the battery.

 

[NascentOxygen]

I am not able to solve this problem

It's indeed a good question! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

With this being a multi-choice question, I'd be seeking a reasonably quick (or insightful) method of deciding on an answer.

Examining the options, we can immediately eliminate 2 simply because they don't have units consistent with energy. This then leaves a 50% chance of guessing the correct option.

 

[NascentOxygen]

The point is that when the switch makes a connection at point 2, the left C is in series with the battery, so the charge delivered by the battery is equal to the change in charge of the left C.

This is not the case for C₀. You calculated the charge on before and after the flip correctly, but some of the current going through C₀ will go through the rightmost C, and not the battery.

before the flip, the leftmost C has charge

$$\frac { V C^2} {2C + C_0}$$

and after it:

$$\frac { V C (C+C_0)} {2C + C_0}$$

The difference is the answer you need.

http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon14.gif That looks right. The difference is a measure of the quantity of charge that has been supplied by the battery to effect the changes brought about by the switching. No nett energy has been added to the bank of capacitors, so the energy supplied by the battery must have all gone into heat as the losses.

The extra energy supplied by the battery = V.ΔQ
where ΔQ is that difference in charge determined above.

Well done!