Voltage source starting with a phase difference?

Discussion in 'Homework Help' started by exidez, Sep 29, 2008.

  1. exidez

    Thread Starter Member

    Aug 22, 2008
    i have a simple circuit shown below

    however, in the question it is not know that it is an induction and the value is also unknown! All you know is that there is a 10 ohm resistor and capacitance or inductance in series

    You are given the waveforms from the oscilloscope for Ch1 (red) and Ch2 (green)


    Ch1 (input Voltage) has the wave form 20sin(523.59t + 48°) [note: voltage does not start at 0 when t= 0]
    Ch2 (the voltage drop across the resistor) is 5sin(523.59t)

    Does this mean the voltage source is 48 degrees out of phase to the current? I don't understand this since this is the source of the current, it has to start at the same time right?

    So if i wanted to work out "Is" it would simply be (5 < 0) / (10 < 0)
    The voltage drop of the resistor divided by the resistance.
    therefore being 0.5 A < 0 right? This agrees with the simulator when creating the circuit to suit these two waveforms.

    however, how can i find out the impedance of the Inductor (if it really is an inductor)?

    I can work out the voltage drop of the inductor this way:
    i know that Ch2 + Vxl = 20 < 48
    (5+0j)+(Vxl) = 13.38261213 + 14.86289651j
    Vxl = 3.3826 + 14.86289
    which equals 15.24295763 < 77.17°

    But this is suppose to be < 90 right ?
    whats wrong here, how do i go about this?
  2. mik3

    Senior Member

    Feb 4, 2008
    Yes, this means the voltage leads the current by 48 degrees.

    To calculate the inductor's impedance:

    divide the total voltage applied by the current flowing to find the total impedance of the circuit.


    then use the above formula to find X
    X=reactance of the inductor
    R=resistance (10 ohm)
    Z=total impedance

    you need to use complex number for all the calculations i mentioned above
  3. exidez

    Thread Starter Member

    Aug 22, 2008
    i thought it was that easy too.....
    (20 < 48 ) / (0.5 < 0) = 40 < 48
    4 < 48 = 26.76522425+29.72579302j

    Z = R + jX
    26.76522425+29.72579302j = 10 + jX

    R already just doesn't add up :confused:

    which would mean Is is wrong, well at least its angle. but how?


    I have thought about it a bit more. Can this be a trick question? The question also asks to state the type of reactance Xc or Xl, I knew it is an inductor due to the voltage phase. But As Ch2 is zero phase, it give the above impedance (26.76522425+29.72579302j) so it could also that it is a 29.72579302 ohm impedance with a 16.76522425 ohm resistor right? They never stated this in the question though, so somehow i think there is something incorrect in my understanding of this circuit.


    Ah ha! i get it now, it is not an ideal inductor! Such a simple concept causes me hours of thinking and research. Goes to show theory isn't everything.
    so how can i calculate the practical inductors impedance?? i know its going to be 16.76522425+29.72579302j impedance. And Its going to be around a 74mH Inductor but how do i calculate the real resistance side of the 74mH inductor at 83.333Hz ??
    Last edited: Sep 29, 2008
  4. mik3

    Senior Member

    Feb 4, 2008
    From my calculations i found the same results, so you may assume that the physical resistance of the coil is 16.76 ohm. Maybe that's the answer or something wrong is in the question.