Capacitance and Voltage Confusion

[jendead]

Homework Statement

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. :)

Homework Equations

V = q/C, V = V1 + V2 + V3 + V4

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?

 

[Astronuc]

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

The capacitors have the same nominal capacitance.

 

[jendead]

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?

 

[jendead]

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

 

[robbondo]

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

 

[saket]

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?

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!
Contradiction!

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, V_i = q_i / C_i.

 

[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.

 

[jendead]

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.

 

[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 V_i = q / C_i. Here C₁ = C₂ = C₃ = C₄ = C. Thus, V₁ = V₂ = V₃ = V₄ = q/C.
Now, V = V₁ + V₂ + V₃ + V₄. Thus, each V_i = V/4 = 2.5 volts.

 

[Astronuc]

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.

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.

 

[jendead]

I think I get it now.. thanks for the help and link :)

 

[verafloyd]

The voltage drop would be due only to electrical series resistance (ESR) of the capacitors. Other than that, in the situation you described, there is no reason for a voltage drop.