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AC analysis, multiple frequencies.

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data

    http://www.aijaa.com/img/b/00322/3721717.jpg [Broken]
    http://www.aijaa.com/img/b/00092/3721734.jpg [Broken]

    Both supplies are basic sine waves and the frequency of the supply 2 is 100Hz. The assignment is to calculate the power that is transformed to heat (i.e the power the resistor is using).

    2. Relevant equations

    3. The attempt at a solution

    I know there are many ways to approach this one, but I calculated the current in the circuit and then using [tex]P=|U||I|cos( \phi _u- \phi_i)[/tex] for both supply independently. Then I simply calculated P=P1+P2. Is this one of the correct methods?

    One more problem I had was with the total impedance of the circuit. [tex]Z=R+j \omega L+ \frac{1}{j \omega C}= \frac{j \omega CR- \omega ^{2}LC+1}{j \omega C}=R+ \frac{j \omega ^3LC^2-j \omega C}{ \omega ^2 C^2}=R+ \frac{j( \omega ^2LC-1)}{ \omega C}[/tex]. If I plug [tex]Z=R+j \omega L+ \frac{1}{j \omega C}[/tex] into my calculator, I get a different answer if I plug in [tex]Z=R+ \frac{j( \omega ^2LC-1)}{ \omega C}[/tex]. I found no errors in my calculations, but if some one finds an error, please let me know. I blindly trusted my calculator and used the former value (Z=10-j3.308 for f1) for getting the current. Was it the correct one?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 6, 2009 #2
    You need to use superposition for this circuit, solving it in each of its separate frequency domains. You cannot use P = P'+P'' while doing superposition, it is a linear analysis technique, and power is nonlinear.

    Short Source 2 and calculate the impedance using Source 1's frequency of operation, 50 Hz. Find the voltage across R and the current through R. These values should be labeled: V'R and I'R.

    Next, you will have to short Source 1 and calculate the impedance again using Source 2's frequency of operation, 100 Hz. Again, find the voltage across R and the current through R. These values should be labeled: V''R and I''R.

    ...and back to the original circuit:
    Current through resistor R: IR = I'R + I''R
    Voltage across resistor R: VR = V'R + V''R

    That should be more than enough information to solve for the resistive power dissipation in the original circuit. Let me know if I need to be more clear.
  4. Mar 6, 2009 #3
    I just checked your impedance calculation for f = 50 Hz, it was correct.
  5. Mar 6, 2009 #4
    Thanks for checking my impedance calculations, mplayer, I must have done a mistake with calculator input then. I've used the superposition method to calculate the current in the circuit. But then my method differs a little bit from yours. The only component in the circuit with resistive power is the resistor. So I should get the right answer by simply solving the real part of the power in the whole circuit, right? So it would simply be [tex]S=Re[UI^*] [/tex],where * is conjugate. Now, I will re-check this one with your method and see if I get the same or different answer. I sure hope it's the same 'cause otherwise my solution is probably wrong. :biggrin:

    But I got to disagree with you on P=P'+P''. Since power in the resistor is [tex]P= \frac{U_{rms}^{2}}{R}[/tex]. And now that we have to frequencies, [tex]u_r(t)=u_{r1}(t) + u_{r2}(t)[/tex], right? So [tex]U_{rms}[/tex] (rms = root, mean, square) for [tex]u(t)[/tex] would be [tex]U_{rms}= \sqrt{ \frac{1}{T} \int_{0}^{T} (u_{r1}(t)^2+u_{r2}(t)^2)dt}= \sqrt{|U_{r1}|^2+|U_{r2}|}[/tex]. I left quite a bit off of the calculation, but I'm just lazy. :biggrin:
  6. Mar 6, 2009 #5

    The Electrician

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    This is not true for the problem at hand. Since the two voltage sources are sinusoids of different frequencies, the voltages are orthogonal functions. This means that the power due to each source can be calculated individually, and the two results just added (look up Parseval's theorem).

    You already got the 50 Hz impedance, 10 - j3.308. This gives a 50 Hz current of magnitude .7595 amps (the angle doesn't matter). The 60 Hz impedance is 10 + j.6994, giving a 60 Hz current of magnitude .399 amps.

    The effective total current is Ieff = SQRT(I1^2+I2^2) = .85796 amps. Then the power in the resistor is Ieff^2*R, or 7.361 watts.
  7. Mar 7, 2009 #6
    That's how I did it, basically. But you must have read the value for f2 wrong. Instead of 60Hz it's 100Hz and it would give P=3.197W. I hope it's right, I need to post this in about 12 hours.

    One adittional question about power. When I need to use the whole value (including angle) if I'm calculating power? Is it only when I want to know the complex power (i.e S=UI*=P+jQ).
  8. Mar 7, 2009 #7

    The Electrician

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    You're quite right. I didn't notice the 100 Hz in your text, and because of my North America-centric bias , I used 60 Hz!!

    At 100 Hz, I get an impedance of 10 + j12.483, giving a current of magnitude .25008 amps.

    So, Ieff is .79964 amps, and the power is .6394*10 = 6.394 watts.

    If you calculate the powers individually, the 50 Hz power is .7595^2 * 10 = .5769 * 10 = 5.769 watts, and the 100 Hz power is .25008^2 * 10 = .06254 * 10 = .6254 watts. Adding the two powers I get 6.394 watts.

    In your very first post, you have the correct value for the 50 Hz impedance. What do you get for the current? You have to set e2 to zero when you calculate the current due to e1, and vice versa. The power due to the 50 Hz current alone is greater than 3.197 watts.

    Maybe you should show the details of how you got 3.197 watts, so I can check for errors.

    I would be tempted to say yes, but I'm not even sure without thinking about it some more how complex power would be defined when more than one frequency is present.
    Last edited: Mar 7, 2009
  9. Mar 7, 2009 #8
    Okay, here we go:

    Impedance for f1 is Z2=10-j3.308 ohms and for f2 is Z2=10+j12.48 ohms. Then I1=E1/Z1 giving me the magnitude .537 A and I2=E2/Z2 giving me .177 A. Where E1=5.66/60, Z1=10.53/-18.30, E2=2.83/0 and Z2=15.99/51.30. Using degrees in the phasors.

    If you can't read that just reply and I'll post the whole thing with Latex.
    Last edited: Mar 7, 2009
  10. Mar 7, 2009 #9

    The Electrician

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    OK. I see the problem. Your class materials are using a notation I'm not familiar with.

    The symbol e1 has a circumflex accent over the e, apparently denoting peak voltage.

    If this is so, then you have the calculations exactly right.

    Taking that into account, I have a final power in the resistor as 3.197 watts.

    Sorry for the confusion.
  11. Mar 7, 2009 #10
    Yep, that's the case. So you used the ê's as RMS values?

    It's okay. The main thing is I had a confirmation I had done this right. Thanks for the help!
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