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AC Circuits confirmation

  1. Jan 19, 2012 #1
    For the circuit given in the power factor is 0.72 lagging and
    the power dissipated is 375 W.

    12.jpg

    Determine the:
    (1) apparent power
    (2) reactive power
    (3) the magnitude of the current flowing in the circuit
    (4) the value of the impedance Z and state whether circuit is inductive or
    capacitive.

    (1) Apparent power = True power / Power Factor (S=P/pf)

    = 375 / 0.72

    = 520.8333 VA

    (2) Reactive Power = SQRT (apparent power^2) - (True Power^2)

    = SQRT (520.8333^2)-(375^2)

    = SQRT (130642.3264)

    = 361.4448 VAR

    (3) Magnitude of Current

    Power= Voltage*Current *Power factor (P=V*I*pf)

    375 = 120*I*0.72
    I = P/(V*pf)
    I = 375/(120*0.72)
    I = 4.34 A

    (4) Total ohms = Voltage/Current
    = 120/4.34
    =27.6498 ohms (minus 10)
    z = 17.65 ohms

    The power factor is lagging within the circuit therefore the the virvuit is inductive



    Am i on the right lines here? or is there a better way of calculating the above results
     
  2. jcsd
  3. Jan 19, 2012 #2

    gneill

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    Staff: Mentor

    I find your calculation of total ohms (part 4) to be a bit dubious. Where you've made the assumption that you can just subtract the resistor value from the impedance magnitude to leave the impedance Z of the "mystery component" is also suspect. Impedance will have real and imaginary parts.

    I'd be more inclined to use the current magnitude and its known phase angle obtained from the power factor to construct the complex current: [itex] i = I (cos(\phi) - j\;sin(\phi)) [/itex], and then find the total (complex) impedance of the circuit:

    [itex] Z_{tot} = \frac{120 V}{i} = (19.91 + j 19.19) \Omega [/itex]

    Then subtract the given resistor from the real part of that to leave the impedance Z.
     
  4. Jan 20, 2012 #3
    Ok thanks gneill

    so ill use

    ϕ=acos(power factor)
    =acos (.72)
    = 0.76699

    then;

    i = 4.34(cos(0.76699)-jsin(0.76699))
    i = 4.339566
    Zt=120/4.339566
    =27.28723 (minus 10)
    =17.28723
     
  5. Jan 20, 2012 #4

    gneill

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    Staff: Mentor

    The complex current should retain real and imaginary components so that the impedance value obtained will also have real and imaginary components. These correspond to the resistance and reactance of the circuit.
     
  6. Jan 20, 2012 #5
    Think I've got it;

    i=4.34(cos(0.76699)-jsin(0.76699)

    i = 4.34(0.720003-0.693971)
    i = (3.12481-3.01183j)

    So

    Z=120/(3.12481-3.01183j)

    Z=19.9079 + 19.1882j

    Minus ten from real gives

    9.9079 ohms

    This better? Are my answers ok for the other parts of the question?
     
  7. Jan 20, 2012 #6

    gneill

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    Staff: Mentor

    Yes, that looks better. So the "mystery component" has both resistance and inductance.

    The other answers look fine to me; nicely calculated. :smile:
     
  8. Jan 20, 2012 #7
    Thanks for you help as always :)
     
  9. Jan 20, 2012 #8

    gneill

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    Staff: Mentor

    You're very welcome!
     
  10. Jan 24, 2012 #9
    gneill said - I'd be more inclined to use the current magnitude and its known phase angle obtained from the power factor to construct the complex current: , and then find the total (complex) impedance of the circuit:



    Charger said - ϕ=acos(power factor)
    =acos (.72)
    = 0.76699

    I work the phase angle out at 43.95, where does this 0.76699 come from? I'm having a real problem getting my head round this 'imaginary' thing.
     
  11. Jan 24, 2012 #10

    gneill

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    Staff: Mentor

    0.76699 is the angle in radians. You can work in degrees or radians as long as you're consistent.
     
  12. Jan 24, 2012 #11
    ok so i use cos-1 x power factor, gives me a phase angle of 43.95 degrees.

    i=I(cos(43.95)-jsin(43.95))
    =4.34(cos(43.95)-jsin(43.95))
    =4.34(0.7199-j0.694)
    =(3.1244-j3.012)

    is this right so far? this is where i'm getting a bit lost and not sure of the next step
     
  13. Jan 24, 2012 #12
    am i right in saying that the jvalue is a negative value?
    so it would actually be (3.1244 - (-3.012))
    giving 120/6.134 = 19.555 for the real value? then how do you get the second jvalue?
     
  14. Jan 24, 2012 #13

    gneill

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    Staff: Mentor

    You don't add the real and imaginary components of a complex number, they are separate entities. Take a look at post #5 in this thread. The impedance will have real and imaginary components, too.
     
  15. Jan 24, 2012 #14
    sorry gneill, i can only ask for patience here. i appreciate the time.
    i'm obviously missing something here, or just not grasping the concept at all!!
    if you don't add or subtract the jvalue then what purpose does it serve? how do you divide 120 by 3.124 and come up with 19.9 or whatever? my brain has packed up and gone home. :-)
     
  16. Jan 24, 2012 #15

    gneill

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    You should review the manipulation of complex numbers; how to add, subtract, multiply, and divide them. Here a link to get you started.

    If you enter "Complex number arithmetic" into a Google search you'll get more information, and even links to online complex number calculators.
     
  17. Feb 20, 2013 #16
    Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?
     
  18. Feb 20, 2013 #17

    gneill

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    Staff: Mentor

    27.63 Ω represents the magnitude of the impedance. Impedance is a complex value comprised of a real (resistive) and an imaginary (reactive) part. In this sense, yes, the magnitude will be greater than the size of the individual components: ##|Z| = \sqrt{R^2 + X^2}## . So for X nonzero, |Z| > R.
     
  19. Mar 11, 2013 #18
    i dont understand how get from Z=120/(3.12481-3.01183j) to Z=19.9079 + 19.1882j

    can anyone explain this please if possible followed the link at the end but can understand from that how it get to Z=19.9079 + 19.1882j
     
  20. Mar 11, 2013 #19

    gneill

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    Staff: Mentor

    It's basic complex arithmetic. You can convert the denominator to polar form (magnitude + angle) then do the division, convert the result back to rectangular form. Or, clear the imaginary term from the denominator by multiplying numerator and denominator by the complex conjugate of the denominator.

    A web search on "complex arithmetic" will likely turn up a tutorial.
     
  21. Mar 11, 2013 #20
    you see none of this is in the lesson book for this course the only way i can find is z=v/i

    there is polar form but nothing like whats on this page.. i still cant understand what your trying to tell me

    just tried a complex arithmetic calculator online and i dont get them answers
     
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