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

• gracy
In summary: When you simplify component networks you lose analytical access to nodes that get absorbed into the replacement components. Sometimes you just have to leave some components unsimplified to retain access to nodes you need.
gracy

## Homework Statement

If 100 volts of potential difference is applied between a and b in the circuit .Find the potential difference between c and d.

2. Homework Equations

For Equivalent Capacitance in series

##\frac{1}{C}##=##\frac{1}{C_1}##+##\frac{1}{C_2}##For Equivalent Capacitance in parallel

##C##=##C_1##+##C_2##...

##Q##=##C####v##

## The Attempt at a Solution

[/B]
Equivalent capacitance between c and d (using series connection formula)
C1,C3 and C4 are in series
Ceq=2μF

this C eq and c2 are in parallel hence (using parallel connection formula)
Ceq'=8μc
This C'eq is in parallel with voltage hence it should have potential difference equal to 100 volts but it is wrong .I want to know why?

When you combined all the capacitors into one you "lost" the nodes c and d in the process. You simplified the circuit sown to a single capacitor that is situated between node a and b. In fact you lost c and d in your first simplification when you combined the three series capacitors into a single 2 uF capacitor. You mistakenly placed c and d node labels around it, but nodes c and d no longer exist in the simplified circuit. What you found is the potential across a and b, which you already knew since it's a given.

What you should do is first consider (as a slight but essential digression) a voltage divider that consists of capacitors in series. If you place a potential across a series-connect string of capacitors, how can you determine the potentials across the individual capacitors in the string?

gracy
gneill said:
When you combined all the capacitors into one you "lost" the nodes c and d in the process.
gneill said:
You mistakenly placed c and d node labels around it
No,I deliberately place c and d .
gneill said:
but nodes c and d no longer exist in the simplified circuit.
You mean nodes c and d should not exist in the simplified circuit?
But why ?As I have clarified this earlier https://www.physicsforums.com/threads/equivalent-capacitance.845489/page-2
post #22 and #23
that we can place equivalent capacitors wherever I want and should replace other component capacitor with wire,I did that.

gracy said:
No,I deliberately place c and d .

You mean nodes c and d should not exist in the simplified circuit?
But why ?As I have clarified this earlier https://www.physicsforums.com/threads/equivalent-capacitance.845489/page-2
post #22 and #23
that we can place equivalent capacitors wherever I want and should replace other component capacitor with wire,I did that.
A 'series' combination of 3 capacitors is connected across 100V supply. You don't need equivalent capacitance for calculating the voltage across the middle capacitor (Vcd).

gracy said:

## Homework Statement

If 100 volts of potential difference is applied between a and b in the circuit .Find the potential difference between c and d.

2. Homework Equations

For Equivalent Capacitance in series

##\frac{1}{C}##=##\frac{1}{C_1}##+##\frac{1}{C_2}##For Equivalent Capacitance in parallel

##C##=##C_1##+##C_2##...

##Q##=##C####v##

## The Attempt at a Solution

View attachment 92911 [/B]
Equivalent capacitance between c and d (using series connection formula)
C1,C3 and C4 are in series
Ceq=2μF

this C eq and c2 are in parallel hence (using parallel connection formula)
Ceq'=8μc
This C'eq is in parallel with voltage hence it should have potential difference equal to 100 volts but it is wrong .I want to know why?
C1 and C3 are not in between c and d. Also, if you want to calculate equivalent capacitance between C and D, your source should be between c and d. What you've calculated is the equivalent capacitance between a and b.

cnh1995 said:
your source should be between c and d.
you mean voltage source?

gracy said:
you mean voltage source?
Right.

gracy said:
No,I deliberately place c and d .
You created new labels c and d where they were not previously located. In the original circuit c and d surrounded the middle capacitor in the string of three. You collapsed the string and moved the labels c and d to the outside ends of the string where it connects to the rest of the circuit. In fact your new labels c and d are identical to the nodes identified by a and b.
You mean nodes c and d should not exist in the simplified circuit?
Right. They were abolished when you reduced the three capacitors. Effectively you cut off the branch containing the three capacitors and nodes c and d and replaced it with a single new capacitor. You lost c and d when you did that.

But why ?As I have clarified this earlier https://www.physicsforums.com/threads/equivalent-capacitance.845489/page-2
post #22 and #23
that we can place equivalent capacitors wherever I want and should replace other component capacitor with wire,I did that.
When you simplify component networks you lose analytical access to nodes that get absorbed into the replacement components. Sometimes you just have to leave some components unsimplified to retain access to nodes you need.

gracy and cnh1995
gneill said:
Sometimes you just have to leave some components unsimplified to retain access to nodes you need.
Here we need c and d so should I not simplify the series connection between C1,C3 and C4 ?

gracy said:
Here we need c and d so should I not simplify the series connection between C1,C3 and C4 ?
Yes. You needn't consider C2 too..

gracy said:
Here we need c and d so should I not simplify the series connection between C1,C3 and C4 ?
Correct.

gracy
gneill said:
Effectively you cut off the branch containing the three capacitors and nodes c and d and replaced it with a single new capacitor.
\
is blue node new node?It wasn't there previously?

The item in blue is new, yes, but there's no new node. The new capacitor connects to where the old string of three capacitors connected, and those two nodes existed before. In fact they are nodes a and b: just follow the wires to the labels near the battery.

gracy
in post #9
gneill said:
Remember, when you collapse a series of components down to one then you will be eliminating any nodes that were along the path, leaving just the two nodes at the ends where the "new" component attach.

Red nodes-eliminated nodes
Blue nodes-new introduced wires with equivalent capacitor
Black nodes-nodes which are left at the points of attachment of the new (equivalent )capacitor at both ends

#### Attachments

• eliminated.png
14.8 KB · Views: 489

I understood where I was wrong.All thanks to you guys.How shall I proceed from here?

gracy said:
How shall I proceed from here.

gneill said:
What you should do is first consider (as a slight but essential digression) a voltage divider that consists of capacitors in series. If you place a potential across a series-connect string of capacitors, how can you determine the potentials across the individual capacitors in the string?

gracy said:
I understood where I was wrong.All thanks to you guys.How shall I proceed from here.
Just use voltage divider for the series combination in the original diagram.

cnh1995 said:
Also, if you want to calculate equivalent capacitance between C and D, your source should be between c and d.
I did not understand this.Can not we calculate equivalent capacitance between two points if there is no battery between them?
Then how we did it here?There was no battery/voltage source in between A and B but we calculated equivalent capacitance between A and B.https://www.physicsforums.com/threads/equivalent-capacitance-circular-arrangement.845579/

gracy said:
I did not understand this.Can not we calculate equivalent capacitance between two points if there is no battery between them?
Then how we did it here?There was no battery/voltage source in between A and B but we calculated equivalent capacitance between A and B.https://www.physicsforums.com/threads/equivalent-capacitance-circular-arrangement.845579/

The equivalent capacitance between two points will be "seen" by a source placed between those points. That's the fundamental meaning of 'equivalent capacitance'. Here, if you place a source between A and B, it will "see" the capacitance of the network to be equal to the calculated equivalent capacitance between A and B, only for which the diagram has been modified. What you had calculated earlier was the capacitance "seen" by the voltage source between A and B which wouldn't be same as seen by a voltage source between C and D.

The presence of batteries or voltage sources does not change the effective capacitance of the circuit between nodes A and B. It depends only on the capacitors in the circuit and how they are interconnected.

cnh1995
gracy said:
But according to this applying voltage source matters
But in that problem, it was mentioned that there is a p.d. of 100V connected between a and b. Voltage between c and d was to be found out. So, there's no point in 'putting' a source between c and d.

gracy said:
But according to this applying voltage source matters
#post 51
It matters in that it would change the problem to a different problem. You can't determine the potential between c and d due to the 100 V at a and b if you place yet another source somewhere in the circuit. In fact, placing a source across c and d would force that potential to be whatever that new source supplies, and would be completely independent of the a-b source.

gracy
Thanks.

gneill said:
it implies that you need a voltage source in order to determine the equivalent capacitance.
But I have seen some problems in which I was asked to determine equivalent capacitance between two points and there was no voltage source.

Last edited:
A source won't change the capacitance between nodes?
Then what voltage source does?

gracy 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.
Then the potential difference must have been zero. Unless the voltage source was already charged capacitors and there were no paths for discharging them. Capacitors retain their charge if there are no closed paths for them to discharge through.

Perhaps you should post one of these problems as a new thread.

gracy said:
A source won't change the capacitance between nodes?
Then what voltage source does?
It provides a potential difference. In circuits where current can flow it provides the potential that drives current.

gracy
gracy said:
As in here
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.

gracy
gracy 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 said:
As in here
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.

SammyS said:
Equivalent capacitance is not equivalent potential.
I wrote that mistakenly.But my question is valid.

gracy said:
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..

SammyS 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.
Yes,what is wrong in here?

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