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Finding current in parallel RLC Circuit

  1. May 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A parallel RLC circuit with every component in parallel.
    The resistance is 120 Ohm, the reactanse over the capacitor is 150 Ohm and the reactanse over the inductor is 70 Ohm. The total current is equal to 100mA.

    Calculate the current passing through the inductance.



    2. Relevant equations
    Look below...


    3. The attempt at a solution
    I have been searching the web for equations to calculate the total impedance of a parallel RLC.
    I am not familiar with the names of inverse resistance and inverse reactance so they will be written as 1/R.

    I found one equation saying...
    (1/Z)^2=(1/R)^2+(1/XL-1/XC)^2

    By using that equation , I calculated the total impedance. Then by using the total current calculated the Power source voltage. Which equals 8.9 V, then via Ohms law came to the conclusion that the current through the inductance is equal to 127mA.

    This result works wery well with the right angle triangle, using the Pyth. Teorem.

    The paper we got says that the current should equal 9.5mA, have I done something wrong?

    Some places I just see the equation
    1/Z=1/R+1/XL-1/XC
    Are both correct?

    Many thanks in advance!
     
  2. jcsd
  3. May 20, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Mrhu! Welcome to PF! :smile:
    But what is the frequency? :confused:
     
  4. May 20, 2012 #3
    Re: Welcome to PF!

    Hi Tim! Thank you

    The frequency is not given, would it be to any use?
    We know the resistance and reactances plus the total current, is that not sufficient?
     
    Last edited: May 20, 2012
  5. May 20, 2012 #4

    ehild

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    No, your solution looks completely correct.

    No, 1/Z=1/R+1/XL-1/XC is not correct. When people calculate with complex impedances, (Z a complex number) they can write 1/Z=1/R+i(-1/XL+1/XC) where XL=ωL,XC=1/(ωC) and i is the imaginary unit. I guess you have not learned about complex impedances yet, but it would yield the same result.

    ehild
     
  6. May 20, 2012 #5
    Glad to hear that, we have used complex numbers in math but have not applied them to Physics yet.

    But how does one come to the equation: (1/Z)^2=(1/R)^2+(1/XL-1/XC)^2?
    Considering how the equation looks I would say that they relate to each other by a right angle triangle.
    How come?

    Is it because the voltage across the components are equal, so the proportions of the currents are equal to the reactances? We have learned how the currents are related.
     
  7. May 20, 2012 #6

    ehild

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    If you studied complex numbers, you know how to find the magnitude from the real and imaginary components. If z=u+iv, |z|^2=u^2+v^2. Complex numbers are similar to two-dimensional vectors.
    The impedance of a resistor is 1/R, those of an inductor and a capacitor are iωL and -j/(ωC), respectively.
    In the parallel RLC circuit, the reciprocal impedances ("admittance " is the name) add up. You are right, it is because the voltage is the same across each component and the currents add. Current is voltage/impedance. I=V/R+V/(iωL)+V/(-i/ωC)=V(1/R+i(ωC-1/(ωL)).

    Sometimes the complex admittance is denoted by Y, so I=YV
    .
    The admittance of a resistor is 1/R, those for the inductor and capacitor are -i/(ωL) and iωC.

    The total admittance is Y=1/R+i(ωC-1/(ωL)). It is a complex number with real part 1/R and imaginary part ωC-1/(ωL). You find the magnitude by adding the squares of the real and imaginary parts and taking the square root.

    ehild
     
    Last edited: May 20, 2012
  8. May 20, 2012 #7
    Many, many thanks!
    It really makes sense when you put it like that.

    Grateful for the quick and understandable answers!
     
  9. May 20, 2012 #8

    ehild

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    You are welcome:smile:

    ehild
     
  10. May 20, 2012 #9

    tiny-tim

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    ah, i didn't read the question properly! :redface:

    although the impedance of a capacitor or inductor always depends on ω (the frequency),

    the question gave you the reactance (in ohms), which includes ω, so (as ehild :smile: has showed) you didn't need ω
     
  11. May 20, 2012 #10
    No worries, if it makes you feel any better I felt that the frequency and peak voltage was missing as well.

    As you're saying omega is important.
     
  12. May 20, 2012 #11

    ehild

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    When the value of voltage or current is given for an AC circuit, it is usually understood as the rms voltage or current (The peak value divided by √2.n 100 mA total current means 100√2 mA peak current supplied by the AC generator.
    The value of reactance refers to a certain ω (which was not specified in this problem).

    ehild
     
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