# Equivalent capacitance

1. Nov 28, 2015

### gracy

1. The problem statement, all variables and given/known data
I need to find equivalent capacitance between A and B.

2. Relevant equations
$C$=$C_1$+$C_2$.....
It was for parallel connection.

$\frac{1}{C}$=$\frac{1}{C_1}$+$\frac{1}{C_2}$...
It was for series connection.

3. The attempt at a solution
I tried to solve the problem as follows
First thing I did I used colored wire.Because i find it convenient .

Then moving forward I got the following

Am I correct till here?

Last edited by a moderator: Nov 28, 2015
2. Nov 28, 2015

### Staff: Mentor

Not quite. You've left gaps in the circuit wiring after replacing the series pairs with their equivalent values capacitance. When reducing a set of series capacitors you should replace all but one of them with a wire, so there are no gaps left in the wiring. You'll then want to revisit the coloring of the wiring to identify the "new" nodes.

3. Nov 28, 2015

### gracy

Right?

4. Nov 28, 2015

### Staff: Mentor

No, you've removed wiring, not filled in the gaps left by the capacitors you've eliminated.

When you combine series capacitors they become one capacitor. You can't just leave a "hole" in the circuit for the ones that get pulled into the combined capacitance.

Note the difference:
- When parallel capacitors are combined you erase all but one.
- When series capacitors are combined you replace all but one with wires.

5. Nov 28, 2015

### gracy

The reduction process left the capacitors in series with only one lead connected to the circuit.

Without both leads connected a component cannot carry current. Charge cannot move onto or off of a capacitor through one wire alone. The capacitor thus has no influence, no utility, as far as the circuit is concerned. Hence equivalent capacitance become C.

6. Nov 28, 2015

### Staff: Mentor

Read what I said above about the difference between combining series and parallel capacitors.

Combining series capacitors:

7. Nov 28, 2015

### gracy

But I am confused which wire to use?

8. Nov 28, 2015

### gracy

Right?We can see all three capacitors are in parallel,hence equivalent capacitance would be 2C.

9. Nov 28, 2015

### Staff: Mentor

Wire is wire... 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. So don't expect your previous coloring scheme to survive a series reduction. Replace the wiring for the "new" capacitor with the color of the end nodes.

10. Nov 28, 2015

### Staff: Mentor

Yes, that's good!

11. Nov 28, 2015

### SammyS

Staff Emeritus
This is the original figure . (No, I'm not clairvoyant.)

Last edited: Nov 28, 2015
12. Nov 28, 2015

### gracy

Yellow or blue?what if I had used yellow color instead of blue.It would be also correct,Right?It will also give 2C as equivalent capacitance.

13. Nov 28, 2015

### Staff: Mentor

The colors you use do not matter; they're merely an aid for identifying individual nodes when a circuit layout is complicated (or if it's purposely made tricky to interpret for a school exercise!). The colors don't change the circuit.

The circuit in this problem is simple and well laid out, so the coloring trick is not really necessary for identifying the nodes. The series connections are easy to spot right away and make a good first step in reducing the circuit.

14. Nov 28, 2015

### gracy

I am also thinking not to depend on colors but what to do I understood only when I used colors,I am afraid If I 'll not use them I'll not be able to understand but I am also afraid that I'll run out of time If I'll use colors.What to do now?

15. Nov 28, 2015

### Staff: Mentor

Practice. Recognizing parallel or series connections in a circuit gets easier with practice.
Only use colors when you can't find any parallel or series opportunities by normal inspection. It's usually a technique of "last resort" for circuits that are purposely made difficult to interpret for puzzles and quizzes.

16. Nov 28, 2015

### gracy

Ok.
@gneill I can't thank you enough for all the answer you have given me.
You are awesome!

17. Nov 28, 2015

### gracy

One last thing
Is my image(I mean image drawn by me)in post #5 a valid circuit?I know it is not correct for this problem but does the circuit like that exist?Because I can't decide it's ending point between two ending points.

18. Nov 28, 2015

### Staff: Mentor

If someone were to present it to you and ask if it was a valid circuit, the short answer would be no. A longer answer would require some context.

It is valid in that it is physically realizable. You certainly could take real components and connect them in that fashion. Whether or not it's useful is another matter. It has no voltage or current sources, and no closed paths for current to flow so it won't do anything as is. In that respect it is not a valid circuit. But you could connect it to other components, making it a sub-circuit of some other device. It's a common enough arrangement of capacitors that you will find it embedded in other circuits.

19. Dec 3, 2015

### gracy

Anyone ?

20. Dec 3, 2015

### Staff: Mentor

What is not clear?