What is the value of C2 in the given circuit with capacitors C1, C3, and C4?

[maiad]

Homework Statement

In the arrangement shown in the figure below, a potential difference DeltaV is applied, and C1 is adjusted so that the voltmeter between points b and d reads zero.
http://capa.physics.mcmaster.ca/figures/sb/Graph26/sb-pic2667.png

This "balance" occurs when C1 = 3.56 µF. If C3 = 9.50 µF and C4 = 12.3 µF, calculate the value of C2

Homework Equations

CΔV=Q

The Attempt at a Solution

I know that capacitor 1-2, 3-4 is connected in series and the charge on capacitors 1-2, 3-4 are the same because the voltmeter reads zero which mean there's no transfer of electrons between the plates so the charge of the capacitors connected in series should be the same. I came up with the following attempt:

1: C(equivalent 1-2)=(1/C1)+(1/C2)
2:C(equivalent 3-4)=(1/C3)+(1/C4)
3:C(total)= C(equivalent 1-2)+C(equivalent 3-4)
4:Qtotal=Q1+Q2
5:Qtotal= ΔV(C(equivalent 1-2)+C(equivalent 3-4))
When i got step 5, i realized everything just cancels out... lol
so where did I go wrong in my approach?

 

[gneill]

If the voltmeter reads zero, what does that say about the potential difference between nodes b and d?

 

[maiad]

o i see...the voltmeter is zero cause the potential difference is the same in both wires?

 

[gneill]

o i see... the the is zero cause the potential is the same...

Correct. That being so, what can you say about the potentials across capacitors C4 & C1, and capacitors C3 & C2?

 

[maiad]

Well the potential are the same but when we say potential we're talking about V and not ΔV right?

 

[gneill]

Well the potential are the same but when we say potential we're talking about V and not ΔV right?

It is the voltage that exist on the capacitors, measured from terminal to terminal.

So if the voltages across those capacitor pairs are the same, if you know the voltage on one you automatically know the other.

Now consider just the two capacitors on the left side of the bridge, C3 and C4. What must be the sum of the voltages of those capacitors?

 

[maiad]

is it equal to that of the potential difference applied?

 

[gneill]

is it equal to that of the potential difference applied?

Clearly, yes, since KVL around the battery-C4-C3-battery loop demands it

Since you know the values of C3 and C4 you should therefore be in a position to determine the voltages on both C3 and C4.

 

[maiad]

But the only equation with i know is ΔV=Q/C but Q is unknown?

 

[gneill]

But the only equation with i know is ΔV=Q/C but Q is unknown?

The charges on capacitors connected in series, such as C3 and C4, are equal. It's also the same as the charge on the equivalent capacitance of C3 in series with C4. So if you calculate that equivalent capacitance and place the source voltage across that, you'll know the charge that must be on both capacitors. The charge then gives you the individual voltages.

Another approach, and one I'd recommend learning about, is the voltage divider relationship for capacitors. In the same way that you can determine the voltage that appears across two resistors in series when you known the total voltage across them, you can determine the voltages that appear on individual series-connected capacitors. For two capacitors in series, the voltage divides in inverse ratio to their capacitance values. So for two capacitors, say Ca and Cb, with total voltage V across them, the voltage on Ca is:

$Va = V \frac{Cb}{Ca + Cb}$

 

[maiad]

My friend broke it don into a simple ratio equation, C2/C3 = C1/C4 and solved for C2 and it works! but I'm not sure why lol

 

[gneill]

My friend broke it don into a simple ratio equation, C2/C3 = C1/C4 and solved for C2 and it works! but I'm not sure why lol

It works because of the voltage ratios imposed on the capacitors by nodes b and d being at the same potential (which we discussed above), in combination with the charges on series capacitors being equal (which we discussed above). You would have got there eventually via the voltage divider equation route I was taking above.