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Impedance of parallel inductors in series with parallel capacitors

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    Impedance of parallel inductors in series with parallel capacitors

    11. Two 1mH inductors are connected in parallel and in series with two 45uF capacitors connected in parallel. What is the total impedance of the circuit?

    Quote Originally Posted by slts1991 View Post
    11. 0.41 miliohms

    How did you come up with 0.41 milliohms? I don't see how to solve this without frequency to calculate reactance, assuming 60hz this is what i got

    1mH = 2x 377 ohms in parallel= 188.50 ohms
    45uF = 2x 58.9 ohms in parallel = 29.45 ohms
    series 188.50 + 29.45 = 217.95 ohms
    help?

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    Quote Originally Posted by Inglefles897 View Post
    11. Two 1mH inductors are connected in parallel and in series with two 45uF capacitors connected in parallel. What is the total impedance of the circuit?




    How did you come up with 0.41 milliohms? I don't see how to solve this without frequency to calculate reactance, assuming 60hz this is what i got

    1mH = 2x 377 ohms in parallel= 188.50 ohms
    45uF = 2x 58.9 ohms in parallel = 29.45 ohms
    series 188.50 + 29.45 = 217.95 ohms
    help?
    It is 188.50 - 29.45 because they are 180 degrees apart in a 60 HZ system and when connected in series they subtract from each other. When they are in series the formula is Z = XL - XC. In parallel the formula is more complicated and is similar to R1xR2/(R1+R2). In this case it is (XL*-XC)/(XL-XC). This means that as XL-XC approach zero (resonance frequency) that the impedance of the parallel tank circuit will approach infinity at that resonant frequency, but since the reactor and capacitor have watt losses you will never get infinity and real world examples of a wave trap have been 2000 to 10000 ohms at resonant frequency. A wave trap blocks the resonant frequency from getting onto the operating bus and passes the 60 HZ, so a wave trap has to be rated to carry normal line loads. Your series resonant circuits are similar and we call those line tuners in the power line carrier world. As XL-XC approach resonant frequency the two cancel themselves out and it allows your resonant frequency to pass and it blocks your carrier frequency.

    Resonant Frequency equals 1/(2 PI (LC)^.5). I know that this formula is not needed for the above mentioned problem, but it helps you understand why a problem like this may be important and helps you figure out what size capacitor and reactor is needed for a wave trap and line tuner.

    So, the answer to this problem is actually 159.05.

    My old Navy instructors told me that the Smithsonian had a tank circuit ran for decades and only seemed to lose power when they took measurements, but I but that was just an electronic legend, lol.

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  5. mmacdonell is offline Junior Member Pro Subscriber
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    Quote Originally Posted by tonybeavers View Post
    It is 188.50 - 29.45 because they are 180 degrees apart in a 60 HZ system and when connected in series they subtract from each other. When they are in series the formula is Z = XL - XC. In parallel the formula is more complicated and is similar to R1xR2/(R1+R2). In this case it is (XL*-XC)/(XL-XC). This means that as XL-XC approach zero (resonance frequency) that the impedance of the parallel tank circuit will approach infinity at that resonant frequency, but since the reactor and capacitor have watt losses you will never get infinity and real world examples of a wave trap have been 2000 to 10000 ohms at resonant frequency. A wave trap blocks the resonant frequency from getting onto the operating bus and passes the 60 HZ, so a wave trap has to be rated to carry normal line loads. Your series resonant circuits are similar and we call those line tuners in the power line carrier world. As XL-XC approach resonant frequency the two cancel themselves out and it allows your resonant frequency to pass and it blocks your carrier frequency.

    Resonant Frequency equals 1/(2 PI (LC)^.5). I know that this formula is not needed for the above mentioned problem, but it helps you understand why a problem like this may be important and helps you figure out what size capacitor and reactor is needed for a wave trap and line tuner.

    So, the answer to this problem is actually 159.05.

    My old Navy instructors told me that the Smithsonian had a tank circuit ran for decades and only seemed to lose power when they took measurements, but I but that was just an electronic legend, lol.
    How is everyone getting 377 ohms for XL?

    2 x 3.14 x 60 x .001H = 0.377 ohms, not 377 ohms???

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    Quote Originally Posted by mmacdonell View Post
    How is everyone getting 377 ohms for XL?

    2 x 3.14 x 60 x .001H = 0.377 ohms, not 377 ohms???
    Not everyone! 0.377 is correct.

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    I came up with something completely different from everyone else, someone please check my work:

    1. Inductors in parallel are halved:
    (2) 1mH inductors in parallel = 0.5mH = 0.0005H
    Xl = 2*pi*60*0.0005 = 0.1885 ohms

    2. Capacitors in parallel are additive:
    (2) 45uF capacitors in parallel = 90 uF = 0.00009F
    Xc = 1 / (2*pi*60*0.00009) = 29.473 ohms

    3. Subtract in series to get impedance:
    Zlc = Xc Xl = 29.285 ohms

    I used this calculator to verify my answer: https://www.translatorscafe.com/unit...-lc-impedance/

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    Quote Originally Posted by SecondGen View Post
    I came up with something completely different from everyone else, someone please check my work:

    1. Inductors in parallel are halved:
    (2) 1mH inductors in parallel = 0.5mH = 0.0005H
    Xl = 2*pi*60*0.0005 = 0.1885 ohms

    2. Capacitors in parallel are additive:
    (2) 45uF capacitors in parallel = 90 uF = 0.00009F
    Xc = 1 / (2*pi*60*0.00009) = 29.473 ohms

    3. Subtract in series to get impedance:
    Zlc = Xc Xl = 29.285 ohms

    I used this calculator to verify my answer: https://www.translatorscafe.com/unit...-lc-impedance/
    Looks good. I think the issue I and others had was resolving the Xc of the caps and then mistakenly applying the 'caps in parallel' rules to the resultant reactance. Rather than finding the Xc of a single cap and then adding them together (wrong), you would then need to treat them as you would a resistor.

    As you did, finding the Xc of the resistors in parallel (adding capacitance together) is correct.

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