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Homework Help: Basic AC power calculations

  1. May 16, 2010 #1

    I have the following circuit:
    [PLAIN]http://img152.imageshack.us/img152/1848/powertransfer.png [Broken]

    My task is to calculate the power dissipated in the resistor, and also the power delivered by the V1 source.

    Here's what i tried.

    First i /dev/null'ed the second source and calculated the impedance
    [PLAIN]http://img293.imageshack.us/img293/7569/powertransferzero2.png [Broken]
    [tex]\frac{1}{(1/j6)+(1/-j4)} = -J12[/tex]

    This gives [tex]|Z| = 12-J12 \approx 17\ohm[/tex]

    [tex]I_2 = \frac{48}{17} = 2.82A[/tex]

    Then for the other source shorted:
    [tex]\frac{1}{(1/12)+(1/6)} = 4[/tex]

    This gives [tex]|Z| = 4-J4 \ohm \approx 5.65\ohm[/tex]

    [tex]I_1 = \frac{8}{5.65} = 1.41A[/tex]

    Now, do you add the currents or do you calculate the power dissipation in the resistance for
    each of the voltage sources?

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 16, 2010 #2
    Sorry for not answering your questions, but I'd just do mesh analysis with complex numbers. Also a warning, remember equations for power are different for peak and rms values.
  4. May 16, 2010 #3
    No worries..

    I would really like to solve this problem by using superposition, just for practice.
  5. May 16, 2010 #4
    You add the currents. Superposition in circuits only works due to the linearity of current and voltage. Differently, power is proportional to the square of current or voltage.
  6. May 16, 2010 #5
    So, in other words you can't calculate the power dissipation caused by the individual sources and add them together?
  7. May 16, 2010 #6
    Yes, you cannot. You must add the currents flowing through the component due to each source to calculate the power dissipated by that component.
  8. May 17, 2010 #7
    So my answer to this question would be:

    [tex]1.41 + 2.82 = 4.23A[/tex]

    [tex]P = I_{rms}*R \rightarrow \bigg(\frac{4.23}{\sqrt{2}}\bigg)^2 * 12= 36W[/tex]
  9. May 17, 2010 #8
    The currents could have different phase shifts, so you cannot do regular addition of the two current magnitudes. You must do vector addition (meaning you keep the angle).

    Also, when you found the equivalent impedance for the parallel combination after shorting out the "other source," you didn't properly use vector math. You have to keep that j component in there.
    [tex] (\frac{1}{12} + \frac{1}{j6})^{-1}=2.4 + 4.8j[/tex]

    Also the total current delivered by v2 is not the same as the current delivered by v2 to the resistor. You need to do current division between the 12 and the 6j impedances of that total current (and don't drop the angles!)

    Seriously, though, I'd recommend doing mesh analysis:
    [tex]i_1(12 + j6) + i_2(-j6) = \frac{48}{\sqrt{2}}[/tex]
    [tex]i_1(-j6) + i_2(j6 - j4) = -\frac{8}{\sqrt{2}}[/tex]
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