Say I wanted 2 uf for example, and I had (2) 1 uf caps. Which of these arrangements would equal 2 uf?
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Caps clarification
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jfrey
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Caps clarification
Another noob question for the ILF diy wizards. I read that you can add caps together to replace higher value caps.
Say I wanted 2 uf for example, and I had (2) 1 uf caps. Which of these arrangements would equal 2 uf?
Say I wanted 2 uf for example, and I had (2) 1 uf caps. Which of these arrangements would equal 2 uf?
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Scruffie
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Re: Caps clarification
The Bottom Picture.
Caps Add In Parallel, Resistors Divide.
Resistors Add in Series so that means Caps...
Caps Add In Parallel, Resistors Divide.
Resistors Add in Series so that means Caps...
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jfrey
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Re: Caps clarification
...Divide?
So the top picture would have what value? 1 uf? Would there ever be a point then to having caps in series?
So the top picture would have what value? 1 uf? Would there ever be a point then to having caps in series?
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Scruffie
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Re: Caps clarification
Correct, well it's actually a little different that, it's to do with Conductances, you'd need to know ohms law, but that's it simplified, you can use this sum to find out the value of Parallel Resistors - R1*R2/(R1+R2).
Caps in Series share the same Sum - C1*C2/(C1+C2) so you'd actually have 500nF or thereabouts.
So yes if you need a value you don't have it can be useful or say you have a 3 way toggle, it's handy in that situation to know how to get the resistance/capactiance you need for each throw of the switch with different values.
Caps in Series share the same Sum - C1*C2/(C1+C2) so you'd actually have 500nF or thereabouts.
So yes if you need a value you don't have it can be useful or say you have a 3 way toggle, it's handy in that situation to know how to get the resistance/capactiance you need for each throw of the switch with different values.
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jfrey
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Re: Caps clarification
Ok, I get it now. Thanks!
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randel_07
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Re: Caps clarification
or if you want
1/Ceq = 1/C1 + 1/C2 + ...... 1/Cn
1/Ceq = 1/C1 + 1/C2 + ...... 1/Cn
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