# Capacitance and Voltage Confusion

1. Oct 25, 2007

1. The problem statement, all variables and given/known data
This is for a lab I'm doing tomorrow. We will have a 10V circuit with 4 100µF capacitors connected in series. I need to calculate the voltage drop across each capacitor. I also need to create another circuit, but I'll worry about that after I'm sure I'm doing this one properly. :)

2. Relevant equations
V = q/C, V = V1 + V2 + V3 + V4

3. The attempt at a solution
I feel like I'm doing something wrong here.

at C1, V1 = q/(100*10^-6)
at C2, V2 = 2q/(100*10^-6)
at C3, V3 = 3q/(100*10^-6)
at C4, V4 = 4q/(100*10^-6)

Total V = 9q/(100*10^-6)

I used that to solve for q:
q = 1.111*10^-4C

Then I plugged q back into each equation.

V1 = 1.111V
V2 = 2.222V
V3 = 3.333V
V4 = 4.444V

Am I on the right track here?

2. Oct 25, 2007

### Staff: Mentor

From where is one getting the 1q, 2q, 3q, 4q?

The capacitors have the same nominal capacitance.

3. Oct 25, 2007

I have that because when capacitors are connected in series, the voltage is V = q(1/C1 + 1/C2 + 1/C3 + 1/C4), according to my text.

So at the first capacitor, V = q/C, then at the second V = q/C + q/C = 2q/C, and so on.. or am I understanding it incorrectly?

4. Oct 25, 2007

I apologize for bumping, but I still am not sure how to do this correctly.. any ideas?

5. Oct 25, 2007

### robbondo

capacitors in series have the same charge. So they all have charge q.

6. Oct 25, 2007

### saket

Find the fallacy on ur own! Going by ur reasoning.. at third it is.. 3q/C, and at fourth 4q/C. So, total voltage drop = q/c + 2q/C + 3q/C + 4q/C = 10q/C!
Going by your textbook, V = q(1/C1 + 1/C2 + 1/C3 + 1/C4). Substituting, C1 = C2 = C3 = C4 = C, u will get.. V = 4q/C!!

I think you have misinterpreted. Charge on each capacitor would remain same, because all of them were initially uncharged & they are in series. Also, individually they will have a voltage drop, Vi = qi / Ci.

7. Oct 26, 2007

### robbondo

maybe it would help to simplify the circuit using equivalent capacitance and then work backwards, it might take longer, but that method usually works for me.

8. Oct 26, 2007

Ok, so.. I used V = 4q/C, and got q = 2.5*10^-4C.

I then recalculated the voltage drops, and got these..
drop 1 = 2.5*10^-6V
drop 2 = 5.0*10^-6V
drop 3 = 7.5*10^-6V
drop 4 = 1.0*10^-5V

These still don't look correct (they are much smaller than anything I would be able to measure in the lab). Am I supposed to be finding q, then plugging it back into each voltage equation?

Unfortunately, I don't understand how I would simplify it using equivalent capacitance. We've spent maybe 15-20 minutes learning about these circuits so far - I'm having a very hard time trying to understand how it works without changing stuff around.

9. Oct 26, 2007

### saket

Since they are in series.. each one of the capacitors will have same 'q' appearing on them. +q on the positive plate and -q on the negative plate. Thus, voltage drop across each of the capacitors would be Vi = q / Ci. Here C1 = C2 = C3 = C4 = C. Thus, V1 = V2 = V3 = V4 = q/C.
Now, V = V1 + V2 + V3 + V4. Thus, each Vi = V/4 = 2.5 volts.

10. Oct 26, 2007

### Staff: Mentor

Try this site - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html#c2

As saket mentioned, the capacitors have the same charge as a consequence of having the same capacitance.

11. Oct 26, 2007