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Homework Help: Find power of resistor and source in AC circuit

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    If an average power of 500W is dissipated in the 20Ω resistor, find Vrms, I S RMS, the power factor seen by the source, and the magnitude of VS
    (Based on circuit in attached diagram)

    2. Relevant equations

    Pave= Irms*Vrms*pf*[itex]\frac{1}{2}[/itex]
    Imaginary number referred to as "j", not "i".

    3. The attempt at a solution
    By a node equation at node A, we see that [itex]\frac{V}{20}[/itex]= [itex]\frac{VS}{-j*20}[/itex], so V = VS [itex]\angle[/itex]-90

    Loop 1: (j*20-j*20)IS -j*20I= VS
    By observation, I=V/20, so V=VS[itex]\angle[/itex]90.
    Loop 2: (20+j*20)I-j*20IS=0, so IS= 1.41[itex]\angle[/itex]135 *I
    = 1.41[itex]\angle[/itex]135 *[itex]\frac{V}{20}[/itex]= 0.0705 V [itex]\angle[/itex]135

    Since I and V are in phase, the power across the resistor is 1
    Solve for V:
    500=V*[itex]\frac{V}{20}[/itex][itex]\frac{1}{2}[/itex], so V=[itex]\sqrt{20,000}[/itex]=

    Vrms=[itex]\frac{V}{\sqrt{2}}[/itex] =100,

    Since V = 100[itex]\sqrt{2}[/itex] IS =7.05 *[itex]\sqrt{2}[/itex] [itex]\angle[/itex]135, so I S RMS = 7.05

    V=VS[itex]\angle[/itex]90, so VS=V[itex]\angle[/itex]-90
    And IS= 0.0705 V [itex]\angle[/itex]135,
    So power factor pf = cos(135-90)= 0.707

    VS=V[itex]\angle[/itex]-90 = -141.4*j, so magnitude given by 141.4.

    I think I got this right, but I just want to make sure.

    Attached Files:

  2. jcsd
  3. Apr 6, 2013 #2
    Just realized the diagram I attached was small, hopefully this one's easier to read.

    Attached Files:

  4. Apr 6, 2013 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Bad start.
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