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Homework Help: Convert between sinusoidal / phasor, find average power, impedance

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img834/8450/4y0t.jpg [Broken]

    Given V(s)t = 100cos100t volts, find the average power supplied by the source and the current i2t in the network.

    2. Relevant equations

    V = IR
    V = IZ
    Z = V/I

    inductor impedance: ZL = ωLj
    capacitor impedance: Zc = -1j/ωc

    j*j = -1

    Average power = (1/2)VmIm

    3. The attempt at a solution

    ω = 100

    L = 50 x 10-3

    ZL = ωLj

    ZL = (50 x 10-3)(100)

    ZL = 5jΩ

    C = (1 x 10-3)F

    capacitor impedance: Zc = -1j/ωc


    Zc = -1j/(1 x 10-3)(100)

    Zc = -10jΩ

    Now: 15Ω || 1mF

    =[ (1/15) + (-1/10j) ]-1; LCD:

    = (2j/30j - 3/30j)-1

    = (2j - 3/30j)-1

    = 30j/(2j - 3); multiply by conjugate:

    => 30j/(2j - 3) * (-2j-3)/(-2j-3)

    = -60j2 - 90j/(-4j2 - 6j + 6j + 9; now multiply out j^2s

    = 60 - 90j / 13

    = (4.615 - 6.92j)Ω; add to rest of impedance

    Z(total) = (14.615 - 1.92j)Ω

    I am really bad with the formula so not sure how to go about getting P(avg) now.

    TO get I of the whole circuit I think just do I = V/Z, and then do (1/2)VmIm ?

    but forgot how to get the sinusoidal function (ie 100cos100t) in phasor form
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 26, 2013 #2
    may I know why you didn't use scientific calculator? it simplifies the this problem..
  4. Dec 26, 2013 #3
    100cos100t should be converted to sine form.
    using trignometry

    if you want to convert to phasor form, you should get the rms value...
  5. Dec 26, 2013 #4
    but we will use maximum value itself since your formula uses maximum values of voltage and current..so the phasor form of voltage is
    V(s)t = 100|_(pi)/2(if your calculator uses degrees, then covert to degrees)
    if you don't have a scientific calculator, use euler's theorem and convert V(s)t to

  6. Dec 26, 2013 #5
    there are two ways..
    1.covert the impendence to phasor form..let it be Z(t). Divide the already found phasor voltage by this impedence.. you get the phasor representation of current.. You also get the maximum value, Im. You also know Vm. Thus you can get average power.
    2. Divide the two complex forms of voltage and impedence , use the properties of complex numbers, and get the complex value of current. Here the imaginary part is the representation of the current...Thus you can get average power
  7. Dec 26, 2013 #6
    I do have a scientific calculator, but just forgot / am not clear with the procedure.

    Ok so for that phasor form, 100 =/= ω, right? Just the value in front of cos (which is still 100 though)?

    Then V(s)t = ( 100 ∠ 90° )V ?

    Forgot to say the phasor form of the total impedance that I had was

    Z = (14.615 - 1.92j)Ω =>

    Z = (14.7 ∠ -7.484°)Ω

    so now I for the whole circuit would be

    I = V/Z

    I = ( 100 ∠ 90° )V/(14.7 ∠ -7.484°)Ω

    I(total) = ( 6.8 ∠ 97.484°)A ?

    Then I can just do current division for i2(t)... could I use this same value for
    average power = (1/2)VmIm ?

    Or are you saying the imaginary current is for the average power formula?
  8. Dec 26, 2013 #7
    Here in your formula,
    P= (1/2)VmIm,
    we are talking about average power dissipated in the resistive part of the circuit. To be specific, Vm is the maximum value of the applied voltage, Im is the maximum value of the current passing through the resistor.
    So, Vm = 100 V.
    Im= 6.8 A.

    You need not solve for i2(t)... sice you have simplified as you have found the eqivalent impedence...
  9. Dec 26, 2013 #8
    Ok, and you don't need to do anything with the angles?


    P =(1/2)VmIm

    => (1/2)( 6.8 ∠ 97.484°)(100 ∠ 90° )

    = (340 ∠ 187.5°)

    If you disregard the angle, is it just 340W?
  10. Dec 26, 2013 #9
    No.No.. What we are interested is in average value of power dissipated in the resistive part of the circuit...The average value of power is the sum of the all instantaneous powers dissipated in the circuit in time interval divided by the time interval. It is a constant. In the last expression, you got
    P=(340 ∠ 187.5°).

    This phasor expression represents that the average value changes with respect to time and in itself is meaningless. Average power dissipated(in W) is a constant when a sinusoidal voltage is applied to a circuit(composed of inductors, capacitors and resistors).
  11. Dec 26, 2013 #10
    Let me try this again.

    Sorry, did you mean:

    I = v/z

    V(t) = 100cos(100t + 0°)

    Z(t) = 14.7cos(100t - 7.484°)

    I = v/z

    I(total) = ( 6.8 ∠ 7.484°)A ? - (This should be the correct angle for it now, right?)

    x = rcosθ, y = rsinθ

    I = (6.742 + 0.8856954j)

    And the imaginary part (ie 0.8856954j) represents the current for Imax in the formula?
  12. Dec 26, 2013 #11
    you wrote Z(t) = 14.7cos(100t - 7.484°). This is wrong. For a given circuit, impedance is a constant(includeing the angle). There is no time component associated with impedence. From your expression, you say that impedence varies with time. It can't.
    I think you are getting confused because you didn't know how phasors simplify the problems. In-order to avoid that, read the proof of how complex number simplify the problem.
    You can also use differential equations to solve this problem. But the fact is that you should do more work to solve them.
    Also we take one angle as reference and so the problem. For example, I say I apply a voltage of 100sin(100t+(pi/2)). Here, 100t is the reference angle. From here I measure the angle of (pi/2). So my phasor representation is 70.7|_(pi/2). here 70.7 is the rms value. For this circuit, you say that the current is 60.6|_100(say), then the 100 rad which you are measuring is with respect to the angle of 100t.
    Impedance is constant and it is measured with respect to zero radians only.
    Please do read the proof.
  13. Dec 27, 2013 #12
    I am not understanding the proof very well. When doing P = (1/2)Vmax*Imax, you have to convert from the RMS values first (from only the magnitude)?

    I sort of understand, that you have a angle between voltage and current (in waveforms), but not so much the reference; still have trouble understanding how that works as well. (And sorry I forgot to put the "cos ø" in the formula of this phase angle between the voltage & current).

    Say for

    I = ( 6.8 ∠ (whatever angle)°)A

    then would the magnitude in waveform (the max value) be simply

    6.8 * √2?

    Does this mean the voltage source was already given in max form? Then it would be

    Imax = 9.61

    Vmax = 100

    so would the average power then be (at least):

    P = (1/2)(9.61)(100)*cosø ?

    What would then be ø?
  14. Dec 27, 2013 #13
    sorry, you do need power factor in your equation( cosø). this ø is the angle between the voltage phasor and the current phasor....or in other words, it is the impedance angle
  15. Dec 27, 2013 #14
    "Impedance angle" isn't the same as the angle from Z in the phasor right? Or it is? Is there another way to tell what the angle is?

    Should have also asked, what would the phasor for v(t) = 100cos100t look like, would it be

    v = 100 ∠ 0°? ( I don't think it is right though)

    And there does not seem to be enough information to calculate the power factor itself

    (I have that Pf = (power)/(V*I), but neither power nor the power factor is given).
  16. Dec 27, 2013 #15
    Yes the impedance angle is the same. It shoows the relationship between the current and voltage. For a circuit, the maximum value of voltage and current can appear at different times and the phase angle gives the relationship between these. Phase angle is determined by circuit components.

    The voltage phasor should be written in rms values only. I have read 2 to 3 textbooks and they use this convention. so v=70.7∠0. here i have converted to sinusoidal form and am using voltage as reference phasor. thus the angle is 0 rad in representation.

    And to your last question, the answer is that you need ø. for that, you should convert the complx impedence to phasor form. In that phasor form, you get the angle. from that angle, PF=cosø.
    Thus you got PF.
  17. Dec 27, 2013 #16
    Ok, so would average power then be

    P = (1/2)(9.61)(100)*cos(-7.484)

    = 476W?

    For the power in power factor it would be (using root mean square values again):

    cos(-7.484) = (power)/(6.8 * 70.7), it also ends up as

    power = 476.6W

    This is the same as the average power?
  18. Dec 27, 2013 #17
    the current phasor is 6.8 ∠ 97.484 A. So, Imax is 6.8 A and not 9.61 A
  19. Dec 28, 2013 #18
    So to get phasors from waveforms for all values (current, voltage, etc) you have to divide by √2 then?

    Or is Vmax still the same as 100? Still confused.
  20. Dec 28, 2013 #19
    Average power consumed is Vrms*Irms*PF or (Vmax*Imax*PF)/2
    these equations are same because
  21. Dec 28, 2013 #20
    Oh right sorry, 100 (voltage) was divided by the impedance (14.7), and they were the max values too, so 6.8 is max value too (but for current).

    So (6.8)(100)(cos(-7.484))

    Average power = 337W then ?
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