# Total XC for capacitors in parallel

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## Total XC for capacitors in parallel

Is it just me or is the answer to this is 44.78 ohms? Or am i crazy???

Three capacitors, a 12 µF, a 20 µF, and a 30 µF, are connected in parallel to a 60 Hz source. The total XC is

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Your answer seems to be correct but I got 42.78 ohms when I did the calculation...

Formula for capacitors in parallel is simply CT = C1 + C2 + C3 and then plug that into XC = 1 / (2 × π × f × C).

First, find total capacitance...
CT = 12 µF + 20 µF + 30 µF
CT = 62 µF

Then find capacitive reactance...
XC = 1 / (2 × π × f × C)
XC = 1 / (2 × 3.141... × 60 × 0.000062) = 42.78 ohms

Don't forget to use the decimal form for microfarads when calculating capacitive reactance (0.000062 not 62).

Edit: changed farads from 0.0062 to 0.000062
Last edited by GavinK5; October 24, 2016 at 01:18 PM.

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Originally Posted by GavinK5
Your answer seems to be correct but I got 42.78 ohms when I did the calculation...

Formula for capacitors in parallel is simply CT = C1 + C2 + C3 and then plug that into XC = 1 / (2 × π × f × C).

First, find total capacitance...
CT = 12 µF + 20 µF + 30 µF
CT = 62 µF

Then find capacitive reactance...
XC = 1 / (2 × π × f × C)
XC = 1 / (2 × 3.141... × 60 × 0.0062) = 42.78 ohms

Don't forget to use the decimal form for microfarads when calculating capacitive reactance (0.0062 not 62).
microfarad is to the millionth place. 1uF would be .000001. I believe it would be .000062

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Originally Posted by lester mcmanaway
microfarad is to the millionth place. 1uF would be .000001. I believe it would be .000062
ah, you are right, not enough zeros in my original equation. post updated. thanks, lester!

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Originally Posted by GavinK5
Your answer seems to be correct but I got 42.78 ohms when I did the calculation...

Formula for capacitors in parallel is simply CT = C1 + C2 + C3 and then plug that into XC = 1 / (2 × π × f × C).

First, find total capacitance...
CT = 12 µF + 20 µF + 30 µF
CT = 62 µF

Then find capacitive reactance...
XC = 1 / (2 × π × f × C)
XC = 1 / (2 × 3.141... × 60 × 0.000062) = 42.78 ohms

Don't forget to use the decimal form for microfarads when calculating capacitive reactance (0.000062 not 62).

Edit: changed farads from 0.0062 to 0.000062
I fat fingered that. 42.48 is what i got as well

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Hi, I have a question
What is the maximum voltage that can be applied between the terminals of the series combination of 0.1microF,250V and 0.2microF 500V capacitors

7. Originally Posted by GavinK5
ah, you are right, not enough zeros in my original equation. post updated. thanks, lester!
easily convert electrical units such as microhms to ohms, amps to milliamps, microfarads to farads, etc. -> http://testguy.net/content/210-Metric-Unit-Conversions

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## Agreed

Solving for total capacitance and finding reactance is the easiest and best way to prevent math errors. Simple math gives the answer as 42.78 ohms.

To double check: Find the reactance of each individual capacitor, then find the Rt for the three in parallel. I did it on a non scientific calculator with some quick rounding and came up with an answer of 42.9 ohm, which is close enough for me!

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