1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Questions on Ac power.

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data

    I have some questions about powers in AC circuit. What is the maximum and minimum current power on resistor and inductance.

    2. Relevant equations

    Um = U / V2
    Im = I / V2

    3. The attempt at a solution
    prmax = U*I=(Um/V2)*(Im/V2)=(Um*Im)/2. But answer seems to actually be that prmax = Um*Im, how can that be? And for plmax = 1/2 Um * Im plmin=- 1/2 Um *Im
  2. jcsd
  3. May 9, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    hi 0fibonacci1! :smile:

    your notation is difficult to understand :redface:

    here's an extract from the pf library on impedance

    Power = work per time = voltage times charge per time = voltage times current:

    [tex]P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)[/tex]
    [tex]=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2[/tex]​
    (because [itex]2cosAcosB = cos(A-B) + cos(A+B))[/itex])
    [tex]=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)[/tex]
    [tex]=\ \left(V_{rms}^2\left/|Z|\right.\right)(\cos\phi + \cos2\omega t)[/tex]​

    So (instantaneous) power is the constant part, [itex]P_{av} = V_{rms}I_{rms}\cos\phi[/itex] (the average power), plus a component varying with double the circuit frequency, [itex]V_{rms}I_{rms}\cos2\omega t[/itex] (so a graph of the whole power is a sine wave shifted by a ratio [itex]\cos\phi[/itex] above the x-axis).​
  4. May 10, 2013 #3


    User Avatar

    Staff: Mentor

    It almost seems that you might be using V2 as denoting 2 ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted