# Exercise with Capacitors

1. Feb 29, 2016

### Kernul

1. The problem statement, all variables and given/known data
Three capacitors $C_a = C, C_b = 2C, C_c = 3C$ are positioned like in the figure. The A capacitor's electrode has a $V_a = 20 V$, while the B capacitor's electrode has a $V_b = 80 V$. What is the electric potential $V_3$ of the C capacitor's electrode?

2. Relevant equations
$Q = C \Delta V$

3. The attempt at a solution
So, since $Q = C \Delta V = C (V_1 - V_2)$ we can write the electric charge for all three the capacitors.
$Q_a = C_a (V_a - V_c) = C (20 - V_c)$
$Q_b = C_b (V_b - V_c) = 2C (80 - V_c)$
$Q_c = C_c (V_c - 0) = 3C V_c$
Knowing from the picture that A and B are parallel we have a total electric charge as:
$Q = Q_a + Q_b = C (20 - V_c) + 2C (80 - V_c) = C (180 - 2V_c)$
What we end up with is a series so the total electric charge doesn't change. This means that:
$Q = Q_c$
so
$C (180 - 2V_c) = 3C V_c$
$V_c = 36 V$

Is this correct? Or did I mess up something/my method wasn't correct?

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2. Feb 29, 2016

### Staff: Mentor

Capacitors A and B cannot be in parallel since they apparently have different potentials across them.

Try re-drawing the circuit, inserting voltage sources to provide the potentials at A and B. Then assume that each source has provided some amount of charge:

Given the assumed directions of charge flow, what can you say about the relationship between the Q's (hint: KCL works for the total charge that moves through a given node).

3. Mar 1, 2016

### epenguin

I think it is wrong actually. I think it is an easy mistake to make. But it is also easy to check. You can do this by calculating the charge on each capacitor which you do not report. Then look at the charges on the 'internal' plates of the T. With your Vc = 36, do they add up to 0? I don't think so.

So the lesson is in most questions it is easier to check a result than to get it in the first place (so students make undue numbers of mistakes by not doing the easy part after they have done the hard part). I get Vc = 30, and as far as I can see the charges add up the right way.

I think the cause of the mistake is a sign mistake in one of the terms. I have only ever done calculations of this kind since a week or two ago, but I find the most convenient is to take the junction of the T as reference point, calculate all voltage differences from that junction, and calculate charges on the plates letting the algebra take care of the signs of charges, and require they add up algebraically to 0. This was illustrated here #8 where a T (aka Y) is part of a more complicated circuitand, and a bit here #9.

4. Mar 1, 2016

### Kernul

Oh! I got it. I found 30 V too now doing like this:
$-Q_a - Q_b + Q_c = 0$
$-C_a(V_a - V_c) - C_b(V_b - V_c) + C_c(V_c - 0) = 0$
$-C(20 - V_c) - 2C(80 - V_c) + 3C V_c = 0$
$-20 C + V_c C - 160 C + 2 V_c C + 3C V_c = 0$
$(1 + 2 + 3) V_c C = 180 C$
$V_c = 30 V$
Thank you!

5. Mar 1, 2016

### epenguin

Good if that's right - but watch out. I don't know exactly what you've done, maybe you have abbreviated some steps. But it looks like you are still taking the charges is absolute magnitude is, and then deciding what signs later. It seems to me that doing this will get it wrong half the time. Because I think half the time you'll know the signs of two charges, but if they are opposite it will not be that obvious what the sign of the third is. Okay we could work out a rule for that maybe, but I've found it easier to start with one junction point, use the voltage changes from there without committing to any sign, and let the algebra find it for you.

Anyway remember to do that check calculation at the end.

6. Mar 11, 2016

### Kernul

Anyway, using the voltage changes isn't what I have done? I simply used the equation $Q = C \Delta V$ so there are the voltage changes(differences). They have signs because of the way the current goes. (negative when entering and positive when exiting)

7. Mar 16, 2016

### epenguin

Well what I say is that came right because you got the signs right this time. Previously it didn't because you didn't. I also calculating this problem got them right once, then wrong, then right. It is easy to get the sign wrong. I had some difficulty deciding what the sign of being charge on the third capacitor was, I think it will be easy for you to get the same wrong again if you do similar calculation in six months or 18 months time. You could also have difficulty in a more complicated circuit of capacitors and other elements, perhaps in time-varying problems.

So I decided it was easier for me at least to not have to think out what what sign the charge was but let it be an algebraic quantity and then I could always write
Q1 + Q2 + Q3 = 0
and take all potentials relative to one point, the junction point. Then the algebra takes care of the signs in the calculation. I think this more rigid or algorithmic approach would suit most students as well as anyone else.

However do it your way if you find it preferable - a more important lesson than any of this was to do the check calculation when you have the result, i.e. not retracing your derivation but seeing whether the result works, is really a solution. This extra 10% of work will reduce the error rate of your results, which might often be much or all you are valued on, drastically.

8. Mar 16, 2016

### Kernul

Oh! Now I got what you meant with "let the algebra take care of the signs". Okay, I'll take your advice and calculate like you did. ($Q_1 + Q_2 + Q_3 = 0$) And obviously check at the end.
Thanks!