# Equivalent capacitance of a circuit

• jolly_math
In summary: Thank you for your help!In summary, the given circuit consists of five capacitors in a specific arrangement. Using symmetry, it can be simplified into an equivalent circuit with four capacitors, where ∆V2 = 0 due to the symmetry of the circuit. The reasoning behind this simplification can be better understood by redrawing the circuit and analyzing the symmetry of the branches.
jolly_math
Homework Statement
Find the equivalent capacitance between points x and y in the diagram below. Assume that C2 = 10 µF and that the other capacitors are all 4.0 µF.
Relevant Equations
in series: 1/C = 1/C1 + 1/C2
in parallel: C = C1 + C2

Why is it that, due to symmetry, ∆V2 = 0 and ∆V1 = ∆V4 = ∆V5 = ∆V3? I don't really understand the reasoning behind the simplification. Thank you.

Is this how you worked the problem?

##C_1## and ##C_2## are in series so we have an equivalent capacitor ##C_{12}##

##C_{12}## and ##C_4## are in parallel so we have an equivalent capacitor ##C_{124}##

##C_3## and ##C_5## are in parallel so we have an equivalent capacitor ##C_{35}##

##C_{124}## and ##C_{35}## are in series so we have an equivalent capacitor ##C_{12345}##

Callumnc1 said:
##C_1## and ##C_2## are in series so we have an equivalent capacitor ##C_{12}##
No, you can't do that. To be considered in series there must be no other circuit connection between the two.

member 731016 and DaveE
jolly_math said:
Why is it that, due to symmetry, ∆V2 = 0
It becomes obvious if you redraw the diagram. Remove 2 for the moment and draw the rest in a simple rectangle, then reinstate 2.

member 731016
haruspex said:
No, you can't do that. To be considered in series there must be no other circuit connection between the two.
Whoops, sorry @haruspex thank you for picking up on that!

haruspex said:
It becomes obvious if you redraw the diagram. Remove 2 for the moment and draw the rest in a simple rectangle, then reinstate 2.

I know my diagram is probably drawn incorrectly,

What are my mistakes thought?

Many thanks!

Why is it that, due to symmetry, ∆V2 = 0? If I remove V2, C1 and C4 (in parallel) are in series to C3 and C5 (in parallel). When I redraw V2, how I would be able to determine that V2 = 0? Thank you.

Callumnc1 said:

I know my diagram is probably drawn incorrectly,
View attachment 322555
What are my mistakes thought?

Many thanks!
Completely incorrect.

@haruspex was replying to OP, who at least realizes that ∆V2 = 0 .

OP does not understand the symmetry argument justifying ##\Delta V_2=0## .

Last edited:
member 731016
SammyS said:
Completely incorrect.

@haruspex was replying to OP, who at least realizes that ∆V2 = 0 .

Do you please know why ∆V2 = 0?

Many thanks!

Callumnc1 said:

I know my diagram is probably drawn incorrectly,
View attachment 322555
What are my mistakes thought?

Many thanks!
Take it in steps. First, remove C2, then straighten the rest of them diagram into a simple rectangle with two capacitors in series in the top line and the other two in the bottom.
Post those two.

member 731016 and SammyS
haruspex said:
Take it in steps. First, remove C2, then straighten the rest of them diagram into a simple rectangle with two capacitors in series in the top line and the other two in the bottom.
Post those two.

Sorry for the late reply. I will do that.

Many thanks!

haruspex said:
Take it in steps. First, remove C2, then straighten the rest of them diagram into a simple rectangle with two capacitors in series in the top line and the other two in the bottom.
Post those two.

Here it is:

Step 1:

Step 2:

Many thanks!

#### Attachments

• 1676939067032.png
3.1 KB · Views: 89
Callumnc1 said:

Here it is:

Step 1:
View attachment 322619
Step 2:
View attachment 322620

Many thanks!
Not correct.

In the upper branch, you should have ##C_4## in series with ##C_3##.

In the lower branch, you should have ##C_1## in series with ##C_5##.

scottdave and member 731016
SammyS said:
Not correct.

In the upper branch, you should have ##C_4## in series with ##C_3##.

In the lower branch, you should have ##C_1## in series with ##C_5##.

I think I see why they are not parallel now. It is because there along the wire there is another wire that splits off so not all the electrons join together again.

Here is my new diagram:

I think I see why the volage difference is zero now. Since capacitor two will cross between the two parallel branch's at either before or after the parallel capacitors as shown below: (I'm not sure whether you can tell where ##C_2## connects between the two branch's - Do you please know?)

So since ##C_{43}## is in parallel with ##C_{12}## then their potential difference across each will be the same. Therefore, they will read the same potential at the same distance down each wire.

Correct?

Many thanks!

Callumnc1 said:

I think I see why they are not parallel now. It is because there along the wire there is another wire that splits off so not all the electrons join together again.

Here is my new diagram:

Correct?

Many thanks!
Do not connect the series pairs. Make a drawing with the four capacitors as @haruspex and I suggested.

Then consider how/where to connect ##C_2## .

member 731016
SammyS said:
Do not connect the series pairs. Make a drawing with the four capacitors as @haruspex and I suggested.

Then consider how/where to connect ##C_2## .

Sorry for my mistake!

Here are the new diagrams,

And I think we connect ##C_2## here:

Many thanks!

scottdave
Callumnc1 said:

Sorry for my mistake!

Here are the new diagrams,
View attachment 322627
And I think we connect ##C_2## here:
View attachment 322628

Many thanks!
Right. Do you see how the symmetry proves there is no potential across C2?

scottdave and member 731016
haruspex said:
Right. Do you see how the symmetry proves there is no potential across C2?

Yes I think so, as along as the wires joining C2 to the circuit are perpendicular to the capacitors in series on each side then C2 and its wires will be equidistant from x so each side for the capacitor have the same potential and therefore have a zero potential difference

Many thanks!

Callumnc1 said:

Yes I think so, as along as the wires joining C2 to the circuit are perpendicular to the capacitors in series on each side then C2 and its wires will be equidistant from x. . . .

Depending on what frequencies you are working with, the geometry and distances of components can be an important factor.

member 731016
scottdave said:
Depending on what frequencies you are working with, the geometry and distances of components can be an important factor.
Thank you @scottdave !

scottdave said:
Depending on what frequencies you are working with, the geometry and distances of components can be an important factor.
I think this questions in this thread are being asked at a very basic level.

Last edited:
member 731016
Callumnc1 said:

Yes I think so, as along as the wires joining C2 to the circuit are perpendicular to the capacitors in series on each side then C2 and its wires will be equidistant from x so each side for the capacitor have the same potential and therefore have a zero potential difference

Many thanks!
If you mean that the wires are physically perpendicular, that has nothing to do with this situation.

The circuit, as modified by you, is electrically equivalent to the circuit given in the OP.

member 731016
SammyS said:
If you mean that the wires are physically perpendicular, that has nothing to do with this situation.

The circuit, as modified by you, is electrically equivalent to the circuit given in the OP.
Thank you for your help @SammyS !

SammyS said:
I think this questions in this thread are being asked at a very basic level.
I thought so too - @Callumnc1 mentioned the wires are perpendicular and components equidistant, so I thought that I would mention it.

member 731016
scottdave said:
I thought so too - @Callumnc1 mentioned the wires are perpendicular and components equidistant, so I thought that I would mention it.

scottdave

## What is equivalent capacitance in a circuit?

Equivalent capacitance is the total capacitance of a circuit that can replace a combination of capacitors without changing the overall effect on the circuit. It simplifies complex networks of capacitors into a single capacitor with the same electrical characteristics.

## How do you calculate the equivalent capacitance for capacitors in series?

For capacitors connected in series, the reciprocal of the equivalent capacitance (C_eq) is the sum of the reciprocals of the individual capacitances (C1, C2, C3, ...). The formula is 1/C_eq = 1/C1 + 1/C2 + 1/C3 + ....

## How do you calculate the equivalent capacitance for capacitors in parallel?

For capacitors connected in parallel, the equivalent capacitance (C_eq) is the sum of the individual capacitances (C1, C2, C3, ...). The formula is C_eq = C1 + C2 + C3 + ....

## What happens to the equivalent capacitance if you add more capacitors in series?

When you add more capacitors in series, the equivalent capacitance decreases. This is because the reciprocal of the equivalent capacitance is the sum of the reciprocals of all individual capacitances, leading to a smaller overall value.

## What happens to the equivalent capacitance if you add more capacitors in parallel?

When you add more capacitors in parallel, the equivalent capacitance increases. This is because the equivalent capacitance is the sum of all individual capacitances, leading to a larger overall value.

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