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Calculating Power Distribution in Circuit

  1. Aug 19, 2016 #1
    1. The problem statement, all variables and given/known data
    I am trying to calculate the power of Rload in the circuit attached (SEE ATTACHED CIRCUIT). The coupler I am trying to model is the Agilent 778D. Can someone verify that I calculate this out correctly?

    2. Relevant equations
    y dB = 10 * log10(x)
    V = i * R
    P = v * i

    3. The attempt at a solution
    i. Calculating the percent of power the coupler uses:
    -0.6dB = 10 * log10( x )
    -0.06 = log10( x )
    10^( -0.06 ) = 10^( log10( x ) )
    x = 87.10%
    >> So the Coupler in the diagram consumes 12.90% of power seen at its input <<

    Power seen at couplers input = Power_Source - Power_R1
    So Power_Coupler = 0.1290 * ( Power_Source - Power_R1)

    Power_Source = Power_R1 + Power_Coupler + Power_Load
    Power_Source = Power_R1 + 0.1290 * ( Power_Source - Power_R1) + Power_Load
    Power_Source = Power_R1 + 0.1290 * Power_Source + 0.1290 * Power_R1 + Power_Load
    0.8710 * Power_Source = Power_R1 + 0.1290 * Power_R1 + Power_Load
    0.8710 * Power_Sourse = 1.1290 * Power_R1 + Power_Load
    0.8710 * V_Source = 1.1290 * V_R1 * i + V_Load * i
    0.8710 * V_Source = 1.1290 * V_R1 + V_Load
    0.8710 * V_Source = 1.1290 * i * R_1 + i * R_load
    0.8710 * V_Source = i * ( 1.1290 * R_1 + R_Load )
    V_Source = i * (1.1290 * R_1 + R_Load) / 0.8710

    i = (V_Load) / R_Load

    V_Source = V_Load / R_Load * (1.1290 * R1 + R_Load ) / 0.8710
    V_Source * R_Load * 0.8710 = V_Load * 1.1290 * R1 + R_Load
    V_Load = V_Source * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)
    V_Load * i = V_Source * i * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)
    P_Load = P_Source * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)

    So if R1 = R_Load = 50Ohms

    P_Load = P_Source * 0.4091

    Thusly,

    The power of the load = 40.91 % of the Power Source

    y dB = 10 * log10 (0.4091) = -3.8817 dB

    Conclusion:

    The power distribution is as follows:
    R1 = 46.19%
    Coupler = 12.90%
    RLoad = 40.91%

    and I should see a -3.8817 dB attenuation of the signal at the load
     

    Attached Files:

  2. jcsd
  3. Aug 20, 2016 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Did you include the power dissipated in the rf analyzer, which must also be a 50 ohm load?
     
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