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Hoffman Amps Forum image Author Topic: capacitance in series  (Read 2492 times)

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Offline tubenit

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capacitance in series
« on: August 18, 2011, 07:23:48 pm »
Is this correct?

(C1 x C2)  divided by (C1 + C2) = C

(.002 x .00039)  divided by  (.00239) =  .0002928  or  293pf ?


With respect, Tubenit

Offline Rich

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Re: capacitance in series
« Reply #1 on: August 18, 2011, 08:08:40 pm »
Just cranking the numbers I get 3.263598

Offline Fresh_Start

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Re: capacitance in series
« Reply #2 on: August 18, 2011, 08:09:39 pm »
AFAICT formula is correct, calculation is not.

Chip
« Last Edit: August 18, 2011, 08:13:05 pm by Fresh_Start »
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Offline tubeswell

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Re: capacitance in series
« Reply #3 on: August 18, 2011, 08:12:58 pm »
Is this correct?

(C1 x C2)  divided by (C1 + C2) = C

(.002 x .00039)  divided by  (.00239) =  .0002928  or  293pf ?


With respect, Tubenit

Yes the formula is correct (except you made a mistake in your calculation)

CT =  (C1 × C2)/(C1 + C2)

I prefer to do it as:

1/CT = 1/C1 + 1/C2 ...

which in your example is 1/.002 + 1/.00039 = 1/.00032 (which if you think about it is how the answer will always be, the sum will be smaller than the smallest contributing value, same as with resistors in parallel)

Incidentally, for the purpose of popping these into a freq roll-off formula, a microfarad is 1 millionth of a farad, so 1uF would be 0.000001F (and  0.002F would be 2,000uF and 0.00039F would be 390uF). But me wonders whether you meant to mean 0.000002F (2uF) in series with 0.00000039F (390nF)?, or 0.0000002F (200nF) in series with 0.0000000039F (3n9F)?, or 0.00000002F (20nF) in series with 0.00000000039F (390pF)?
« Last Edit: August 19, 2011, 04:22:08 am by tubeswell »
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