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A capacitor in my ceiling fan failed. I was hunting for this replacement:

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but at first I only found this:

enter image description here

I was wondering if the second one would work if I just capped off the red wire and didn't attach it to anything. With no potential difference across the third capacitor, it never stores any energy right?

Is my dim memory of idealized freshman physics correct? Even if it is, is there some real-world effect that would make doing this wrong in practice? (Don't worry! I found the exact replacement, so I'm not actually going to use the second one. I'm just curious.)

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You are totally correct, if you don't connect the private wire of the third capacitor (or short it to the common) it won't have any effect.

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  • \$\begingroup\$ Thanks, @Wouter. Also out of curiosity, is my question worded correctly? That is, would the third capacitor properly be called "disconnected"? \$\endgroup\$ – jtolle Jul 7 '13 at 20:03
  • \$\begingroup\$ A component that is connected by only one lead has no influence on the circuit. Period. But there are some components that cheat: an antenna for instance might seem to have only one lead, but in 'reality' it has the earth as other lead, or it might act as a kind of transformer all by itself. But at the frequencies we are talking about here (mains) such effects are not significant. \$\endgroup\$ – Wouter van Ooijen Jul 7 '13 at 21:00
  • \$\begingroup\$ Thanks. I edited the question title slightly. I'm really asking about a capacitor "connected by only one lead", not a "disconnected" one, although I understand the point that in this application it's pretty much the same thing. \$\endgroup\$ – jtolle Jul 7 '13 at 21:34

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