Capacitor Series Circuit: Deriving Cs Relationship

[maulucci]

Homework Statement

Derive a relationship for Cs for a two capacitor series circuit with a resistor. Start by showing that the same charge separation q is present across the capacitor and each of the capacitors in series and that the voltage across Cs is equal to the sum of the potential differences across each capacitor.

Homework Equations

1/Cs=(1/C1)+(1/C2)

The Attempt at a Solution

not sure how to approach this

 

[CWatters]

If they are all in series what does that say about the current?

What is the definition of current?

 

[maulucci]

I tried to work it out and i was wondering if this was the correct way to show wht the question is asking for
q1=C1V and q2=C2V
q=q1+q2=(C1+C2)V
CsV=q=(C1+C2)V
Cs=C1+C2

 

[mfb]

q1=C1V and q2=C2V
q=q1+q2=(C1+C2)V
CsV=q=(C1+C2)V
Cs=C1+C2

This would apply to parallel capacitors, but not to capacitors in series.

The approach CWatters suggested gives you the correct equations for the charge.

 

[gneill]

I tried to work it out and i was wondering if this was the correct way to show wht the question is asking for
q1=C1V and q2=C2V
q=q1+q2=(C1+C2)V <----
CsV=q=(C1+C2)V
Cs=C1+C2

It is a series circuit, and as the problem states, "...the same charge separation q is present across the capacitor and each of the capacitors". But writing q = q1 + q2 not only assumes that the charges are different, but that the current can be different in different components of a series circuit!

First you should argue why the charge separation should be the same on both capacitors. Then you can use that fact and the q = C*V relationship to work out the rest.

{Note: If you happen to have a knowledge of calculus you could write the KVL equation for the circuit (Integral version) and answer all the questions of the problem from its characteristics}

 

[maulucci]

To show that they are equal charges q1=q2
And
q=Cs^-1V
q=(C1^-1+C2^-1)V
(C1^-1+C2^-1)V=Cs^-1V
Cs^-1=C1^-1+C2^-1

 

[mfb]

To show that they are equal charges q1=q2

You did not show it, you just assumed it.

q=Cs^-1V

Why ^(-1)?

q=(C1^-1+C2^-1)V

How did you get that?

Your solution will need V1, V2 in some way...

 

[maulucci]

Ok

V=q/C
V=V1+V2
q/C=q/C1+q/C2
C^-1=C1^-1+C2^-1

 

[mfb]

That is correct. You can argue that q1=q2 to satisfy charge conservation, or something similar.

 

[maulucci]

Ok thank you