What is the Potential Difference Between c and d in This Circuit?

[gneill]

I had already posted.
As in here
https://www.physicsforums.com/threads/equivalent-capacitance-circular-arrangement.845579/
I was asked to find equivalent capacitance in between A and B but there is no voltage source between those two points.
I think I am missing something.

It has been stated previously: a voltage source will not change the capacitance between nodes, and it is not required to determine the capacitance between nodes. That problem does not ask you to find a potential difference between nodes, just capacitance.

 

[SammyS]

But I have seen some problems in which I was asked to determine equivalent potential between two points and there was no voltage source.

I had already posted.
As in here
https://www.physicsforums.com/threads/equivalent-capacitance-circular-arrangement.845579/
I was asked to find equivalent capacitance in between A and B but there is no voltage source between those two points.
I think I am missing something.

Equivalent capacitance is not equivalent potential.

 

[gracy]

Equivalent capacitance is not equivalent potential.

I wrote that mistakenly.But my question is valid.

 

[SammyS]

I wrote that mistakenly.But my question is valid.

I don't think your question is valid.

You said: "But I have seen some problems in which I was asked to determine equivalent capacitance between two points and there was no voltage source."

Then you posted a link to a thread in which you were asked to find equivalent capacitance and in which there was no voltage source..

 

[gracy]

But I have seen some problems in which I was asked to determine equivalent capacitance between two points and there was no voltage source."

Then you posted a link to a thread in which you were asked to find equivalent capacitance and in which there was no voltage source.

Yes,what is wrong in here?

 

[gracy]

Basically I was saying I don't think we need a voltage source in order to determine the equivalent capacitance as ch1995 said not @gneill .Quotation is wrong.
As I have seen some problems in which I was asked to determine equivalent capacitance between two points and there was no voltage source.@gneill asked me to post those problems that's what I did.

 

[SammyS]

Yes,what is wrong in here?

Oh you edited that thread, Thread # 26. which originally said,

"But I have seen some problems in which I was asked to determine equivalent potential between two points and there was no voltage source."

 

[gracy]

Yes,I wrote potential instead of capacitance .See my post 33.

 

[SammyS]

I wrote that mistakenly.But my question is valid.

It would certainly help to have pointed out what the mistake was that was being referred to, and in post #26, when you edit the post, it helps to point out what change you make if the change is significant.

 

[epenguin]

Related to what I have advised before you need to be less plodding.
If I tell you the answer to this question (the one you started with) is obvious and not even needing to write down any calculations, does that make it obvious?

Relevant equations.

1. The potential drop (voltage as you and I naturally say, but I correct myself - pedantically) across a capacitor depends on some way or other that doesn't matter now on just the charge and capacitance, nothing else.

2. About the charges on capacitors in series we know something. Well we have gone into it recently so hopefully we do.

3. The capacitances are given in the problem. Is one, by any chance, related to another?

 

[cnh1995]

Then what voltage source does?

Suppose V=100V. In this problem, if you were asked to calculate the total amount of charge supplied by the source, you'll first calculate the equivalent capacitance between a and b (what you calculated earlier as 8μF ) and then use Q=C(eq)*V.. This is where adding a voltage source and calculating the equivalent capacitance matter. Because again, equivalent capacitance is always the one which is "seen" by a source connected between those points.

 

[gracy]

voltage divider

I want to know is voltage divider rule applicable only when two resistance are there in series,I never found R3 in the below formula.I know that it is applicable for capacitors also what I am asking is it applicable only when two components are in series I know also that it is applicable in series connection only ,I am trying to stress on two

 

[epenguin]

I can probably guess what you're after with this question but shouldn't have to.

If you are asking whether there is any analogy between logic and equations of capacitors in series and resistors in series, yes there is a kind of analogy.

With resistors in series what is the physical quantity that is the same in each of them?

With capacitors in series what physical quantity is the same in each of them? - you have recently discussed this in detail.

If several equal resistors are in series how is the potential difference over each one related?

May be suggestive for your problem.

 

[cnh1995]

I want to know is voltage divider rule applicable only when two resistance are there in series,I never found R3 in the below formula.I know that it is applicable for capacitors also what I am asking is it applicable only when two components are in series I know also that it is applicable in series connection only ,I am trying to stress on two

Do you know how the formula for voltage divider is derived? Do you know Ohm's law?

 

[gracy]

Do you know Ohm's law?

$V$=$I$R

 

[cnh1995]

$V$=$I$R

Right. Now can you apply it to derive the voltage divider formula?

 

[gracy]

derive the voltage divider formula?

Vs = I × Rtot

However, we know that Rtot = R1 + R2 so we substitute this into the equation above to give the following expression.

Vs = I × (R1 + R2)

The above expression transposes to give the following expression for current.

I = Vs / (R1 + R2)

We know that the following expression for the output voltage is true.

Vo = I × R2

Substituting the previous expression for current into the above equation, we get the following.

Vo = Vs × R2 / (R1 + R2)

 

[gracy]

I think we can use it for any number of resistance.

 

[cnh1995]

Vs = I × Rtot

However, we know that Rtot = R1 + R2 so we substitute this into the equation above to give the following expression.

Vs = I × (R1 + R2)

The above expression transposes to give the following expression for current.

I = Vs / (R1 + R2)

We know that the following expression for the output voltage is true.

Vo = I × R2

Substituting the previous expression for current into the above equation, we get the following.

Vo = Vs × R2 / (R1 + R2)[/QUOT

I think we can use it for any number of resistance.

Exactly..

 

[gracy]

But it is derivation for resistance as we know ohm's law can't be applied for capacitors,how to derive this for capacitors?

 

[cnh1995]

But it is derivation for resistance as we know ohm's law can't be applied for capacitors,how to derive this for capacitors?

Use Q=CV. In series circuit, Q is same on each capacitor. Can you proceed from here?

 

[cnh1995]

But it is derivation for resistance as we know ohm's law can't be applied for capacitors,how to derive this for capacitors?

It will be not as straightforward as it is for resistors. Because resistors is series add directly while capacitors in series don't.. You can derive it but can't memorize it because it will change with number of capacitors.

 

[gracy]

Vo

What is Vo?

 

[cnh1995]

What is Vo?

Vo = I × R2

Voltage across R2..

 

[gracy]

In general?

 

[cnh1995]

In general?

In general it is the output voltage..

 

[gracy]

it is the output voltage..

Voltage across last resistance/capacitor?

 

[gracy]

Vs

Emf of battery?

 

[cnh1995]

Voltage across last resistance/capacitor?

Not last. It can be across any component. That depends on the problem. Here in this capacitance problem, Vo is voltage between c and d.

 

[cnh1995]

Emf of battery?

Right.