# Balancing Capacitor System

1. Oct 6, 2007

### mitleid

In the arrangement shown in the figure below, a potential difference $$\Delta$$V is applied, and C1 is adjusted so that the voltmeter between points b and d reads zero. This "balance" occurs when C1 = 3.00 uF, C3 = 8.00 uF and C4 = 15.0 uF. Calculate the value of C2.

I think the key here may be understanding the relationship between C1 and the voltmeter. Vbd is the cusp of this problem, I'm sure, but I don't quite get the relevance of it yet. I've looked at this a couple of different ways so far.

First that C1 and C4 are parallel, and their respective capacitances can be added together to get C', their equivalent capacitance. I assumed a similar thing can be done with C2 and C3. Since the voltmeter reads 0... well first of all, bd is parallel with the plane of the power supply - you would expect a voltage ACROSS the power supply, but not along it, so I'm not sure what this tells me really. Since there is no voltage here, of course there will be no charge associated with it either.

I'm encouraged to speculate the potential drops across C1 and C4 are equal to the sum of the potential drops across C2 and C3 - but I would have assumed this anyway without the voltmeter being there, reading 0, because their respective Ceq's are in series with one another.

2. Oct 6, 2007

### Mindscrape

In your problem statement do you mean that C2 is adjusted so that Vbd = 0?

C1 and C4 are not parallel. C1 series C2 and C3 series C4 are parallel, but not C1 and C4. Additionally, you don't want to start making equivalent circuits in this case because when you start to do that you are losing some of your information.

What you know is that the voltage across a and c is V. The drop between ad and ab is equivalent, which in turn means that C3 and C2 are at the same voltage, since c is grounded.

What would a KVL tell you? Do you know about impedances? What kind of equations do you know that might help you?

3. Oct 6, 2007

### mitleid

The problem states that C1 is adjusted, not C2 (though I was a bit confused about that too, perhaps its a typo). I see what you mean about the equivalent circuits, I wasn't quite sure about that.

Since ad and ab have equal potential differences, there is no need for the current to travel through bd, right? It'd be wasting its time, so to speak.

I don't know about KVLs or impedances yet. Equations that might help... that is serving to be the rather tricky part.

Q = C*V is kind of the basis for a majority of these homework problems. Can I get the charges to cancel out somehow in this problem? Let's see...

4. Oct 6, 2007

### mitleid

C4 series C3 have the same charge, as will C1 series C2. You said not to work with equivalent capacitances, but I don't really know what else to do to isolate C2.

Right now I have Q1 = C'V and Q2 = C''V, so I'm working with... 3 unknown variables :| I'm not sure this route is going to get me anywhere.

5. Oct 6, 2007

### Mindscrape

Yeah, there is a typo somewhere. Either C2 is fixed and you are trying to find C1, or C1 is fixed and you are trying to find C2.

That is strange that they didn't teach you Kirchoff's Voltage Law. Basically, what Kirchoff's voltage law says is that energy (or charge) is conserved, and, more importantly, that the sum of the voltages around any closed path must be zero. If you know any E&M this is similar to saying the curl of E = 0.

KVL would say, for example, that -∆V + Vad + Vcd = 0, and also that -∆V + Vab + Vbc = 0.

*Also, you are right that no current travels across the voltmeter, and it would be because there is no voltage drop, but even if there were a potential difference between those points a voltmeter has (assumed) infinite resistance and should not draw any current.

Last edited: Oct 6, 2007
6. Oct 6, 2007

### mitleid

Okay, well that makes sense. I'm still groping around in the dark here a bit, as I'm not sure how to get the voltages (Vab or Vbc) since I'm only given the capacitances...

Been sitting here about four hours working on this and the spherical problem I've posted, got almost 10 pages of nonsense from doing this homework... :P

I just need a hint... what is the link is between the capacitance and the Voltage? based on KVL and the fact that the voltage between C3 and C2 will be the same... i'm just coming up with more equations with the same missing variables.

7. Oct 6, 2007

### mitleid

I was thinking this was more complicated... C1/C2 is proportional to C4/C3, so I just plug in the numbers and cross-multiply to solve. I should probably get a firmer grasp on why that is before the exam, though...

8. Oct 6, 2007

### learningphysics

The voltage across C1 equals the voltage across C4 in order for the voltmeter to read 0.

Use the voltage divider rule for capacitors to get the voltage on C1 and C4...

Vc1 = Vc4

$$\frac{C2}{C1+C2}*V = \frac{C3}{C3 + C4}*V$$

solve for C2

9. Oct 6, 2007

### mitleid

Ahhh, I remember thinking almost that exact thing when I first started the problem, but I got a bit confused (obviously). Thanks for clearing that up.