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Finding line impedance of an AC circuit

  1. Apr 21, 2012 #1
    So I've been trying to figure out what I'm doing wrong regarding a theoretical aspect of this lab. I've used both ideal analysis and experimental analysis and can't figure out if I'm doing this right.

    The circuit is shown produced in Multisim below:

    N5d7v.png

    The function generator produces a sine wave of 10 Vp, 1 kHz, 0 V DC offset

    The corresponding OSCOPE window is shown below with the peak values indicated and the time delay given:

    H1Q63.png

    My lab manual says that the line impedance is given by this equation:

    ELCcD.png

    It gives the derivation of this formula based on a similar circuit where Z_load is the load measured and Z_line is represented by the line impedance.

    So given:
    Z_L = 100
    V_S = 9.999 V
    V_L = 9.694 V
    dt = 30e-6 s
    ω = 2πf = 2 * pi * 1000

    Input into MATLAB:

    This does not seem to agree with what I know about the ideal line transmission! Why does this happen?

    From the answer:
    R_line = 1.3193 ohm =/= 1 ohm (but close enough?)
    X_line = 19.3277 = jwL --> 19.3277/1000 = L = 1.93 mH =/= 3.3 mH


    Keep in mind this is all done in Multisim, which I thought is supposed to be pretty close if not actually ideal measurements. I'm not sure what I'm doing wrong, or if indeed these are the correct values that I'm supposed to obtain. The lab manual mentions compensating reactance which is the negative X_line to correct the power factor to 1.

    Power tends to go over my head so I'd be most grateful if anyone could clear up the problem I'm having. Needless to say that since I'm confused in the ideal, theoretical case, I'm not getting anywhere in the experiments where I utilize these equations.



    I analyzed the circuit on pen and paper through a voltage divider:

    V_L = 10e^(j0) [ (100)/(100+j3.3) ] = 10e^(j0) * 100 / (101.05e^(j*1.87))
    V_L = 9.896e^(j*-1.87)

    Not getting anywhere with that either.
     
  2. jcsd
  3. Apr 22, 2012 #2

    gneill

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    Staff: Mentor

    It could be that your time measurements from the "scope" are not as accurate as they could be. I see that you're displaying about 5 cycles of the waveform. Why not adjust the timebase so that you see only a single cycle of Vs, and then take your Δt measurements between the zero crossings of each signal. This should be more accurate than trying to locate the peaks of the sinewaves.
    The power factor is the cosine of the angle that is the difference in phase between the current waveform and voltage waveform for the signal. If you "compensate" the line reactance by adding its negative in series with the line then the net reactance becomes zero and thus the phase difference between I and V becomes zero. When I and V are in phase you transfer maximum real power.
    What were you trying to determine?
     
  4. Apr 22, 2012 #3
    Well, Multisim allows me to set a cursor to the peak value--the values that I have in the oscope are not values that I have manually selected to be peaks. Therefore, they should be accurate. I've taken dt at the zero crossings and have noticed no difference, although during labs I'll take dt at zero crossings because it's easier to locate.

    There seems to be a sort of transient response in the first few cycles or so. Initially I was taking my peak values at the very beginning of my oscope reading and noticed I was getting slightly different values between the first few peaks... Off by a few mV but sometimes dt was also off and that was troubling.

    Now, I let the circuit run for a few seconds, letting it reach what I assume to be steady state before taking the measurements at the end of the reading. This seems to eliminate the inconsistent values I was getting.

    Yep! I noticed that when compensating the line reactance the pf goes to 1 and a maximum voltage load is achieved (and therefore maximum real power).

    I understand that. What I don't understand is that if the measurements I'm taking to find Z_line are accurate and ideal, shouldn't Z_line = R_line + jwL_line = 1 + j3.3 (for 1kHz for example)? Instead there seems to be some confounding factor that does not give that ideal value, unless I'm misunderstanding what I'm calculating in Z_line.

    I was trying to determine the power absorbed by the load at 1 kHz... Using a voltage divider I found V_L, V_S is given (as 10e^(j0)) and dt is a result of the difference of phase angles. It occurs to me, though, that I did not find the current phase angle and therefore could not calculate dt correctly.


    Update: My mistake has been trivial. I forgot to remember that, when determining X_L = jwL, w = 2*pi*f... Now, all of my values make sense and it would seem I've been doing the rest of this assignment correctly.
     
    Last edited: Apr 22, 2012
  5. Apr 22, 2012 #4

    gneill

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    Staff: Mentor

    It's possible that Multisim's timesteps for the curves are such that there will be some inaccuracies in the values ("discretization error"). This will be particularly true for horizontal or vertical parts of the curves where the "steps" are maximized along those directions. Does Multisim allow you to set a maximum timestep size?

    Also check that Multisim isn't adding a default value of series resistance to its inductors. Some simulators do this.

    I suspect that the Δt value should be closer to 32 μs.
     
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