Real power calculation

Discussion in 'General Electronics Chat' started by Justin Daubar, Aug 21, 2012.

  1. Justin Daubar

    Thread Starter New Member

    Aug 21, 2012
    1
    0
    I really need help with power calculations.

    Real power is the power into a resistive load.
    For my application, I know the Voltage (Vrms), the Impedance (Z), and the phase angle of the impedance.
    I am using the equation:
    P = (V * V) / (Z * Cos(angle)) Watts
    The denominator in the above equation is the resistance of the load.
    The above equation is equivalent to your:
    P = true power P = E*E/R

    Other people are saying that the equation should be:
    P = (V * V * Cos(angle) ) / Z Watts

    Using numbers from your inductance & resistance page, V=120 Vrms, Z=85.078 ohms at an angle of 45.152 deg.

    My equation gives the power to be: 240 Watts
    The other equation gives the power to be: 119.3 Watts. The same value that you get.

    So, here are my questions:
    1. Why is my equation (P = (V * V) / (Z * Cos(angle))) wrong?
    2. How was the other equation (P = (V * V * Cos(angle) ) / Z) derived?

    I am not too familiar with these forums. If someone can answer my questions, please reply to: <SNIP>

    Thanks,
    Justin Daubar
     
    Last edited by a moderator: Aug 21, 2012
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The correct equation is based on the relationship

    Real Power = Apparent Power * Power Factor

    The Apparent power is found from

    S=|V^2|/|Z|=120^2/85.078=169.26 VA

    The power factor is cos(45.152) = 0.705 [leading]

    So the Real Power = 169.26 * 0.705 =119.33 Watts

    In your case the term Z*cos(θ) is the resistive part of the total impedance.

    Your error lies in assuming the total source voltage is applied to the resistive part - which is incorrect. Only a portion of the source voltage appears across the resistive part. In this particular case the impedance may be thought of a series combination of a resistor and a capacitor connected to the source. In effect you have an AC voltage divider. If you consider that arrangement it's clear the full source voltage is not applied to the resistive part.
     
    Last edited: Aug 22, 2012
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  3. SVS

    Member

    Aug 16, 2012
    89
    4
    First of all Z * Cos(angle) is nothing but `R´!!
    Because Cos(angle)= R /Z.
    So, it is (P = (V * V) /R) which is the basic formulae!

    So, The Apparent power is found from

    S=|V*V|/|Z|=(120*120)/85.078=169.26 VA

    Then real Power is 169.26 * 0.705 =119.3 Watt

    2. How was the other equation (P = (V * V * Cos(angle) ) / Z) derived?
    How was this formulae derived? I am not sure about this.
     
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  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    There are formal proofs of the equation and its related equations.

    A simple approach is to assume that any complex load may be lumped as a series resistance R plus reactance ±jX having a magnitude Z Ω, where

    Z=\sqrt{R^2+X^2}

    If the source RMS voltage is V, then the load current [rms] magnitude is given by

    I=\frac{V}{Z}

    The real power dissipation is therefore simply

    P=I^2 \times R={\( \frac{V}{Z} \)}^2 \times R

    We also know that if the impedance angle is ±θ, where

    \theta=\arctan{\( \frac{X}{R} \)}

    and also that

    R=Z \times \cos{\( \theta \)}

    Then

    P=I^2 \times R={\( \frac{V}{Z} \)}^2 \times Z \times \cos(\theta)=\frac{V^2}{Z}\cos(\theta)

    If we note that cos(θ) is the power factor pf and S is the apparent power then

    P=\frac{V^2}{Z}\cos(\theta)=\frac{V^2}{Z}\times {pf}=S \times pf
     
    Last edited: Aug 22, 2012
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  5. SVS

    Member

    Aug 16, 2012
    89
    4
    If we note that cos(θ) is the power factor pf and S is the apparent power then

    [​IMG]


    Yeah!! Clear.
     
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