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Average power absorbed by resistor:

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img4/5259/wn5b.jpg [Broken]

    Find the average power absorbed by the resistor in the circuit
    if v1(t) = 10cos(377t + 60°)V
    and v2(t) = 20cos(377t + 120°)V


    2. Relevant equations

    average power = (1/2)Vmax*Imax*cosø

    x=rcosθ
    y=rsinθ

    mag = (r2 + j2)1/2
    θ = arctan(j/r)

    3. The attempt at a solution

    v1(t) = 10cos(377t + 60°)V; = (10∠60°)V; = (5 + 8.66j)V

    v2(t) = 20cos(377t + 120°)V; = (20∠120°)V; = (-10 + 17.32j)V


    Z = 1Ω; = (1 ∠ 0°)Ω


    Vtotal = v1 - v2 = 5 + 8.66j - (-10 + 17.32j)

    Vtotal = (15 - 8.66j) V

    Vtotal = (17.32 ∠-30°)V

    and I think these are all max values so:

    I = V/Z

    I = (17.32∠-30°)/(1∠0°)

    I = (17.32∠-30°)A


    Avg power = (1/2)(17.32)(17.32)(cos 0) = 150W
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 30, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Since you're dealing with power it would be prudent to convert the voltages to RMS values. Otherwise there will be a "missing" factor of 1/2 in your power calculation.

    EDIT: Never mind. I see that you're using the ##\frac{1}{2}V_m I_m cos(\phi)## formula which includes the 1/2 term. Your calculation is fine.
     
    Last edited: Dec 30, 2013
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