# Capacitor Series Circuit: Deriving Cs Relationship

• maulucci
In summary, to derive the relationship for Cs for a two capacitor series circuit with a resistor, you can start by showing that the same charge separation q is present across each capacitor. Using the equation q = C*V, you can then show that the voltage across Cs is equal to the sum of the potential differences across each capacitor, leading to the equation 1/Cs = 1/C1 + 1/C2. From this, it can be concluded that Cs = C1 + C2, assuming that the voltage across each capacitor is equal.
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

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

What is the definition of current?

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

maulucci said:
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.

maulucci said:
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}

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

maulucci said:
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...

Ok

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

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

Ok thank you

## 1. What is a capacitor series circuit?

A capacitor series circuit is a type of electrical circuit where two or more capacitors are connected in a series, meaning that the positive terminal of one capacitor is connected to the negative terminal of another. This results in a single path for the current to flow.

## 2. How do you derive the relationship between capacitors in a series circuit?

The relationship between capacitors in a series circuit can be derived using the equation Cs = C1 + C2 + C3 + ..., where Cs is the total capacitance and C1, C2, C3, etc. are the individual capacitances of each capacitor. This means that the total capacitance of a series circuit is equal to the sum of the individual capacitances.

## 3. What happens to the total capacitance when capacitors are connected in a series?

When capacitors are connected in a series, the total capacitance decreases. This is because the individual capacitances are added together, resulting in a smaller total capacitance compared to just one capacitor on its own.

## 4. Can you explain the concept of equivalent capacitance in a series circuit?

Equivalent capacitance in a series circuit refers to the single value of capacitance that would have the same effect as the combination of multiple capacitors in the circuit. It is calculated using the formula 1/Cs = 1/C1 + 1/C2 + 1/C3 + ..., where Cs is the equivalent capacitance and C1, C2, C3, etc. are the individual capacitances.

## 5. What is the purpose of using capacitors in a series circuit?

Capacitors in a series circuit can be used for a variety of purposes, such as filtering out unwanted frequencies, smoothing out voltage fluctuations, and storing electrical energy. They can also be used to create a larger effective capacitance than what is available with a single capacitor.

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