# Finding how much energy disappears in two connected capacitors

#### Master Rai

Joined Jun 12, 2016
2
A plate of a 2 microFarad capacitor charged to 100V is connected by a fine wire to a plate of a 0.01 microFarad charged to 200V, the other plates of the two capacitors being joined together and connected to earth. How much charge passes along the wire and how much energy disappears.

The answer to the energy is 1.796E-4
I assumed that the circuit is basically two capacitors in series with no voltage source.
My first attempt was simply adding the energy in both capacitors, E = 1/2 (C1V1^2 + C2V2^2) = 1/2 (2u*100^2 + 0.01u*200^2) = 0.0102J, did not get the correct answer

I then took the series equivalent capacitance, C = 9.95 nF, then used the formula E = 1/2 (CV^2) again, but assuming that there is only a 100V potential difference across the circuit, E = 1/2 (9.95n * 100^2) = 4.975E-5 Joules, still didn't get the correct answer. Tried it with other values for V = 200V and 300V and still couldn't get the answer. What am I doing wrong?

#### MrChips

Joined Oct 2, 2009
23,956

Are the two capacitors wired in series or parallel?
What is the total capacitance of the two capacitors when connected to each other?
What is the final voltage across each capacitor when they are connected?

#### MrAl

Joined Jun 17, 2014
8,381
A plate of a 2 microFarad capacitor charged to 100V is connected by a fine wire to a plate of a 0.01 microFarad charged to 200V, the other plates of the two capacitors being joined together and connected to earth. How much charge passes along the wire and how much energy disappears.

The answer to the energy is 1.796E-4
I assumed that the circuit is basically two capacitors in series with no voltage source.
My first attempt was simply adding the energy in both capacitors, E = 1/2 (C1V1^2 + C2V2^2) = 1/2 (2u*100^2 + 0.01u*200^2) = 0.0102J, did not get the correct answer

I then took the series equivalent capacitance, C = 9.95 nF, then used the formula E = 1/2 (CV^2) again, but assuming that there is only a 100V potential difference across the circuit, E = 1/2 (9.95n * 100^2) = 4.975E-5 Joules, still didn't get the correct answer. Tried it with other values for V = 200V and 300V and still couldn't get the answer. What am I doing wrong?

Hello,

Has that answer result been checked to make sure it is correct?

#### Alec_t

Joined Sep 17, 2013
12,155
Are the capacitors assumed perfect, or do they have internal resistance? Is the wire resistance to be taken as zero or not?

#### crutschow

Joined Mar 14, 2008
27,457
If one plate of each of the capacitor are connected together and the other plates connected to ground, then the caps are in parallel.

You need to find the voltage and energy of both capacitors connected together in parallel.
You do that by determining what the voltage is with both in parallel for the total charge of the two capacitors (which does not change).
Then calculate the energy of them in parallel, and subtract that from the total energy of the two capacitors before they were connected.

#### crutschow

Joined Mar 14, 2008
27,457
Are the capacitors assumed perfect, or do they have internal resistance? Is the wire resistance to be taken as zero or not?
That has no effect on the calculation.

#### Zeeus

Joined Apr 17, 2019
615
If one plate of each of the capacitor are connected together and the other plates connected to ground, then the caps are in parallel.

You need to find the voltage and energy of both capacitors connected together in parallel.
You do that by determining what the voltage is with both in parallel for the total charge of the two capacitors (which does not change).
Then calculate the energy of them in parallel, and subtract that from the total energy of the two capacitors before they were connected.
did this but not the answer he said

doing this will get :
"but assuming that there is only a 100V potential difference across the circuit, E = 1/2 (9.95n * 100^2) = 4.975E-5 Joules, still didn't get the correct answer"

what you said, gives approx 100v

Unless didn't do well

#### MrAl

Joined Jun 17, 2014
8,381
Are the capacitors assumed perfect, or do they have internal resistance? Is the wire resistance to be taken as zero or not?
Hi there,

The resistance is irrelevant. It can be 0.1 Ohm, 1 Ohm, 10 Ohms, etc. That's why they dont have to state the value of that resistance.
The only thing that changes is the time to reach steady state.

#### Master Rai

Joined Jun 12, 2016
2
If one plate of each of the capacitor are connected together and the other plates connected to ground, then the caps are in parallel.

You need to find the voltage and energy of both capacitors connected together in parallel.
You do that by determining what the voltage is with both in parallel for the total charge of the two capacitors (which does not change).
Then calculate the energy of them in parallel, and subtract that from the total energy of the two capacitors before they were connected.
I tried this but I still can't seem to get it
Since the capacitors are in parallel, their equivalent capacitance is 2.01 uF, and I got voltage of the parallel connection between them by V = (C1V1 + C2V2)/(C1+C2) , I got 100.4975V
Then I got the Energy in the parallel connection, 1/2 (2.01u * 100.4975^2)

Then for the energy that disappeared, I obtained by E = 1/2 (2u * 100^2 + .01u * 200^2) - 1/2 (2.01u * 100.4975^2)
I tried playing the with polarities to see if I'd get closer to the answer, but I still can't get it
Did I miss anything?

#### crutschow

Joined Mar 14, 2008
27,457
assuming that there is only a 100V potential difference across the circuit
Why would you assume that?

#### Zeeus

Joined Apr 17, 2019
615
Hi there,

The resistance is irrelevant. It can be 0.1 Ohm, 1 Ohm, 10 Ohms, etc. That's why they dont have to state the value of that resistance.
The only thing that changes is the time to reach steady state.
Irrelevant? sure?

Why would you assume that?
Just quoted M. She said that in first post

#### djsfantasi

Joined Apr 11, 2010
7,763
Apply some common sense, please. What’s the voltage across the parallel caps when they reach steady state? No calculations allowed; just common sense.

#### WBahn

Joined Mar 31, 2012
26,398
A plate of a 2 microFarad capacitor charged to 100V is connected by a fine wire to a plate of a 0.01 microFarad charged to 200V, the other plates of the two capacitors being joined together and connected to earth. How much charge passes along the wire and how much energy disappears.

The answer to the energy is 1.796E-4
I assumed that the circuit is basically two capacitors in series with no voltage source.
My first attempt was simply adding the energy in both capacitors, E = 1/2 (C1V1^2 + C2V2^2) = 1/2 (2u*100^2 + 0.01u*200^2) = 0.0102J, did not get the correct answer

I then took the series equivalent capacitance, C = 9.95 nF, then used the formula E = 1/2 (CV^2) again, but assuming that there is only a 100V potential difference across the circuit, E = 1/2 (9.95n * 100^2) = 4.975E-5 Joules, still didn't get the correct answer. Tried it with other values for V = 200V and 300V and still couldn't get the answer. What am I doing wrong?
The answer given (where did it come from?) is wrong.

But you have bigger issues.

What you are doing wrong, when all is said and done, is not even attempting to apply fundamental concepts to solve the problem. You are doing solution by happening.

You seem to be trying random things without any reason or logic hoping that, at some point, you will just happen to get an answer that matches the one given. What you would if that occurred? Move on to the next problem without a second thought?

Why would you even compare the total energy on the capacitors when separate to the claimed energy lost when they are connected in parallel?

Why would you use the series capacitance equation when they aren't in series?

Why would you assume that there is 100 V difference across the series-connected circuit?

Why would you think that the total energy remaining on the series-connected capacitors would be the answer to how much energy was lost?

Why would you think that the voltage would be 200 V or 300 V?

BTW: One of those last two is the correct voltage (if they had been series connected)? Do you know why? What total energy would now be stored on the capacitors? How does it compare to the original total energy? Should it match? What if it doesn't?

#### MrAl

Joined Jun 17, 2014
8,381
Irrelevant? sure?

Just quoted M. She said that in first post

Hi,

Yes, the resistance does not change anything, it just makes things change faster or slower.
The energy in all cases is the same with any value resistor, because the energy changes either faster or slower but the same amount still occurs it just takes longer or goes faster depending on the value.
So theoretically, you could 0.000001 Ohm or 100megohms and it would still transfer and lose the same about of energy. For the low values it would be fast, for the high value it could take a long time, but once steady state is reached the energy is the same for all values of R.

#### MrAl

Joined Jun 17, 2014
8,381
Apply some common sense, please. What’s the voltage across the parallel caps when they reach steady state? No calculations allowed; just common sense.
Hi,

Can you elaborate? I had to calculate the final voltage.

#### MrAl

Joined Jun 17, 2014
8,381
I tried this but I still can't seem to get it
Since the capacitors are in parallel, their equivalent capacitance is 2.01 uF, and I got voltage of the parallel connection between them by V = (C1V1 + C2V2)/(C1+C2) , I got 100.4975V
Then I got the Energy in the parallel connection, 1/2 (2.01u * 100.4975^2)

Then for the energy that disappeared, I obtained by E = 1/2 (2u * 100^2 + .01u * 200^2) - 1/2 (2.01u * 100.4975^2)
I tried playing the with polarities to see if I'd get closer to the answer, but I still can't get it
Did I miss anything?

There is a consensus now that the given solution value is not correct.

#### crutschow

Joined Mar 14, 2008
27,457
Apply some common sense, please. What’s the voltage across the parallel caps when they reach steady state? No calculations allowed; just common sense.
You can certainly determine the approximate final voltage but not the exact final voltage without calculations (unless your common sense has a built-in calculator ).

#### djsfantasi

Joined Apr 11, 2010
7,763
You can certainly determine the approximate final voltage but not the exact final voltage without calculations.
Ok, I’ll give you that. But the approximate final voltage will give a a value for a reality check. Or I’m wrong.

#### WBahn

Joined Mar 31, 2012
26,398
Ok, I’ll give you that. But the approximate final voltage will give a a value for a reality check. Or I’m wrong.
It's always a good idea to do estimates for bounds checking. How tight you can make those bounds is a different matter.

I try to always do some hard bounds and then a soft estimate.

In this case, the hard bounds are 100 V and 200 V. I know that the final voltage absolutely has to be within that range. If I get anything even slightly out of that range, I know the answer is simply wrong. Another hard bound is that the energy lost cannot exceed the original energy on the higher voltage cap since it will transfer SOME energy to the other cap and so the final total energy will be larger than the original energy on the lower voltage cap.

For the soft bound, I consider that the high voltage cap is so much smaller than the other cap that, even charged to a significantly higher voltage, it will not change the voltage too much. It's more than two orders of magnitude smaller but the voltage is only a factor of two higher. As for the fraction of the energy on the higher cap that is lost, my gut feel was that I wouldn't be surprised if most of it was lost. With a bit of thought I was able to hard bound it at no more than 50% because the worst case would be for the voltage across the small cap to be cut in half with now change in the voltage across the larger cap. That would make 50% disappear. But even if the voltage didn't change at all, the total charge would be increased a bit, which would reduce that some. So I was expecting something approaching the 50% mark. It turns out that about a quarter of it was lost.

So my soft estimate was off by quite a bit, but that's okay. It's got me to thinking about what the relationship is between the fraction lost and the ratios of capacitor sizes and initial voltages. Something to tinker with.

#### BR-549

Joined Sep 22, 2013
4,938
I can't discern the setup.

Have you got a clearly labeled print?

I am always interested in disappearing energy.

• Zeeus