Homework Help: Capacitors in series and Parallel

1. Apr 16, 2013

CAF123

1. The problem statement, all variables and given/known data
Calculate the capacitance between A and B in the attached diagram.
(Capacitor.png)

3. The attempt at a solution
I just want to clarify something in my book first: The definition for capacitors in parallel is given as 'the capacitors are directly wired together at one plate and directly wired together at the other plate..' So something like in the top picture of (Parallel.png). However, with the defintion given, I can also draw the bottom picture of Parallel.png which means (according to that defintion) that system is also in parallel. This does not seem right because there is only passage for charge through the circuit. Is the definition in the book a little vague or did I miss something?

Many thanks.

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• Parallel.png
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2. Apr 16, 2013

Staff: Mentor

If the whole circuit just has two capacitors in a circle, it is pointless to distinguish between parallel and serial capacitors.

3. Apr 16, 2013

Staff: Mentor

With just two components, it can be ambiguous as to whether they should be considered as being in series or in parallel. Both views are valid.

Ambiguity is removed when you specify how you are to "look" at the circuit for measurement purposes. Once you choose the "viewing port", the ambiguity is resolved:

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4. Apr 16, 2013

CAF123

the battery was implicitly there as well. When you say 'pointless', do you mean you cannot say whether it is a series or parallel or that you can, but there is no reason to?

Edit: gneills post cleared that up.

5. Apr 16, 2013

CAF123

What do you mean by 'viewing port'? By showing the junctions? Could you explain how the ambiguity is resolved? If I were to move the capacitors in the left picture up along the wire then I would have exactly the same diagram as the one on the right.

6. Apr 16, 2013

BBAI BBAI

See, for parallel condition, the voltage across each capacitor has to be same. In the second figure the capacitors are arranged in series or papallel can be decided form where the charge is flowing, the end points A and B as told by gneill matter. In his two diagrams both the capacitors have same voltage so they are parallel.

7. Apr 16, 2013

CAF123

So did you deduce they were both parallel circuits in this set up by saying that the current splits at point B and so the charge flow onto each of the capacitors is not the same, hence it can't be a series circuit thus it is parallel?

8. Apr 16, 2013

Staff: Mentor

The ambiguity is resolved by specifying how the combination is viewed externally. In the diagram, the added leads are where you'd measure voltage, inject current, etc., and between which you'd measure the capacitance.
Exactly! No ambiguity. They are the same when viewed from the external connection wires.

If there is no external framework by which to judge the connection, then you can choose to consider them to be either in series or in parallel.

9. Apr 16, 2013

CAF123

Thanks gneill.
In the attached circuit, I said that C3 and C4 are in parallel so the effective capacitance there is C3 + C4. Then C1 and C2 are also in parallel so the effective capacitance is C1+ C2 there. These two capacitances are in series, so I then add the reciprocals to obtain: $$\frac{1}{C_T} = \frac{1}{C1+C2} + \frac{1}{C3 + C4} = \frac{C3+ C4 + C1 + C2}{(C1+C2)(C3 + C4)}$$

The values of C4 = 150, C3 = 50, C2 = 100, C1 = 200 all in microfarads. Substitute in and I get $120\mu F$, which is 20μF over the suggested answer.

10. Apr 16, 2013

Staff: Mentor

If the attached circuit you're referring to is the first one from the initial post in this thread, then it appears to me that C3 is bypassed (shorted) by a wire, so it effectively disappears from the circuit.

11. Apr 16, 2013

CAF123

Shorted by the wire on the right (the wire parallel to C3)? If so, then surely the current would split at the junction right after C4?

12. Apr 16, 2013

Staff: Mentor

Yup.
Sure. Why don't you redraw the circuit without C3.

13. Apr 16, 2013

CAF123

It would be equivalent to this circuit:

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14. Apr 16, 2013

Staff: Mentor

Yes (although it's missing the top lead for the "B" connection).

Can you determine the net capacitance between the top and bottom nodes (B to A)?

15. Apr 16, 2013

CAF123

Yes and that gives me the required 100μF, however, I am still a little unsure about why exactly C3 is shorted out? Why is this the case? (I can see the wire parallel to C3 that shorts it but when the current splits at the junction ahead of C4, then some would surely accumulate at C3 ?

16. Apr 16, 2013

voko

Any capacitor has some equivalent resistance. When one is in parallel with a zero-resistance wire, how much current would flow through the capacitor?

17. Apr 16, 2013

CAF123

Hmm... I am thinking zero now since current flow always tends to flow towards regions of low resistance.
(If I am not mistaken, this is what can cause damage to wires due to overheating etc..)

18. Apr 16, 2013

Staff: Mentor

The shorting wire ties both leads of C3 together, making it a single node. By definition a single node is all at the same potential. That means there can be no potential difference across the capacitor, hence no charge can accumulate.

19. Apr 16, 2013

CAF123

Does that mean in the following sketch, the equivalent capacitance is C1?

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20. Apr 16, 2013

Staff: Mentor

Yes. Any component that's shorted becomes irrelevant and can be removed/ignored.

21. Apr 16, 2013

voko

One caveat: in a very high frequency circuit, "shorting" becomes tricky.

22. Apr 16, 2013

CAF123

Thanks. How in general would you identify a shorted component? Just by a wire that connects two leads of a component together?

@voko: Is my comment in post #17 correct?

23. Apr 16, 2013

voko

Yes. This is how "shorting" works. Most of the current flows through the lowest-resistance path, making the other path irrelevant.

24. Apr 16, 2013

Staff: Mentor

If both leads of a component connect to the same node, it is shorted.

25. Apr 16, 2013

Staff: Mentor

In a very high frequency circuit, MANY things become tricky A tale for another day (or another course!) methinks.