Polar capacitors and a lack of jargon

Question

Today I found myself wondering something about capacitors, and though my first reaction was to turn to Google, I also discovered that I don't know how to word my inquiry without getting ten thousand irrelevant results.

Assuming C1=C2, what would happen if polar capacitors were arranged in this configuration? What, if any, total capacitance would they have, or would they just pop? What if the values of C1 and C2 were different, say 100u:10u?

C1 will be damaged. What that damage entails depends on its chemistry. — Ignacio Vazquez-Abrams, Jun 13 '16 at 05:34
Crap, in my haste to put that diagram together I absent-mindedly used a battery for the voltage source. I meant to add an AC source, not DC. I'll change that. — Justin Pederson, Jun 13 '16 at 05:41
I assume that you mean that a load is connected at the rhs between C1 & C2 and NOT that they are joined.- not — Russell McMahon, Jun 13 '16 at 07:05
Yeah, some serious clarification about `whatever` is needed. If nothing is connected - remove it. If something is connected - draw it. We can't speculate about a circuit that anything can hook up to. — pipe, Jun 13 '16 at 08:35
Guys, by KCL, no current can flow into "whatever" so we really don't need to know what it is. — The Photon, Jun 13 '16 at 14:39
The circuit depicted is **evidently** worthless and both capacitors will instantly fail on the first ill-signed half period; see Russell McMahon’s answer. Even asked such a thing by mishap, one shouldn’t leave such bad jokes to linger at the site for months. Downvote. — Incnis Mrsi, Sep 04 '16 at 10:28

score 2 · Accepted Answer · edited Apr 13 '17 at 12:33

2

Assuming nothing is connected to the node labeled whatever, the expression you are looking for is series connected.

Identical polar capacitors connected in series is equivalent to a single capacitor which has half the capacitance of the individual capacitors, and double the voltage rating.

If their capacitances differ, the equation is the same as for parallel connected resistors: $$C_{tot} = \frac{1}{{1\over C_1}+{1\over C_2}+\dots}$$

Since you are connecting an AC source to a polar capacitor, even when constructed from several capacitors, the same bad thing will happen as if you used a single capacitor - pop.

If you flip one of the capacitors around, you have created a bipolar electrolytic capacitor. This is legitimate, and you can find more details in this question.

edited Apr 13 '17 at 12:33

Community

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answered Jun 13 '16 at 06:07

pipe

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Whoa. Not sure what happened there, but instead of an edit to my original answer, the edit was somehow posted as a new answer! I deleted my previous answer, but it may still be visible to users with a high enough rep. – pipe Jun 13 '16 at 06:10
Holy crap. Shows how little I actually understand about electronic engineering. What little I do know is self-taught; I got into modding and building stompboxes a while ago as a hobby, and I recently saw a schematic with one capacitor flipped as you described, which led me to ask this question. I didn't realize that that was what was going on. Thank you for taking the time to answer my dumb question. I obviously need to do a lot more reading... – Justin Pederson Jun 17 '16 at 12:21

score 2 · Answer 2 · answered Jun 13 '16 at 07:14

NB: If you like this answer and Pipe's answer and nobody else answers, then please accept his one as I am just adding to what he effectively said.

I assume that you mean that a load is connected at the rhs between C1 & C2 and NOT that they are joined. Thusly:

In this case the two capacitors have the correct polarity applied for half a cycle and the incorrect polarity for half a cycle. They die.

However, if, as pipe suggests, you reverse one capacitor (either one) then one is correctly biased on each half cycle and the other is reverse biased. The electrochemical magic of capacitor construction allows the reverse biased capacitor to act as a low value resistor and overall you get one or other capacitor acting regardless of polarity. As pipe says - this gives you a functional bipolar capacitor. If Ccap =C1 = C2 = C and Vcap = V1 = V2 = V then the resultant cap has V rating of ~= V and Capacitance of approximately C (I think :-).).

Polar capacitors and a lack of jargon

2 Answers2