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If I take a closed circuit with two capacitors and a voltage difference, the circuit is apparently in parallel, but if I introduce a battery, the circuit is in series. Why does the presence of the battery make a difference?

Here's a diagram of what I mean:

enter image description here

According to my book, the left circuit is in parallel, but the right circuit is in series. I don't see why—after all, for any voltage to run through the circuit on the left, it has to pass through C1 before C2 (or vice versa).

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If you look at the structure of the circuit on the left, and where voltage must be, you will see that the wire connecting C1 and C2 ensures that the plates connected to point A are at one voltage, let's call it VA, and the other plates connected to voltage VB.

enter image description here

So this can be thought of as two batteries , each with voltage differential of VA-VB in parallel.

For the second circuit, the two capacitors are in series with respect to the battery (as mentioned by @FrankH)

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  • $\begingroup$ Just to complete the answer, mention that in the second circuit, with respect to the real battery, the two capacitors are in series... $\endgroup$
    – FrankH
    Commented Feb 15, 2013 at 10:32

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