# Equivalent capacitance problem

1. Dec 4, 2015

### gracy

1. The problem statement, all variables and given/known data
If 100 volts of potential difference is applied between a and b in the circuit .Find the potential difference between c and d.

2. Relevant 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$
3. The attempt at a solution

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?

2. Dec 4, 2015

### Staff: Mentor

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?

3. Dec 4, 2015

### gracy

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.

4. Dec 4, 2015

### cnh1995

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).

5. Dec 4, 2015

### cnh1995

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.

6. Dec 4, 2015

### gracy

you mean voltage source?

7. Dec 4, 2015

Right.

8. Dec 4, 2015

### Staff: Mentor

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.
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.

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.

9. Dec 4, 2015

### gracy

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

10. Dec 4, 2015

### cnh1995

Yes. You needn't consider C2 too..

11. Dec 4, 2015

### Staff: Mentor

Correct.

12. Dec 4, 2015

### gracy

\
is blue node new node?It wasn't there previously?

13. Dec 4, 2015

### Staff: Mentor

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.

14. Dec 4, 2015

### gracy

in post #9

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

#### Attached Files:

• ###### eliminated.png
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15. Dec 4, 2015

### gracy

16. Dec 4, 2015

### gracy

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

17. Dec 4, 2015

### gracy

18. Dec 4, 2015

### cnh1995

Just use voltage divider for the series combination in the original diagram.

19. Dec 4, 2015

### gracy

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/

20. Dec 4, 2015

### cnh1995

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.