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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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    Quote Originally Posted by GavinK5 View Post
    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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    Quote Originally Posted by lester mcmanaway View Post
    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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    Quote Originally Posted by GavinK5 View Post
    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

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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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