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Calculate Real and Reactive Power and write equation for S = P + jQ

  1. Nov 30, 2017 #1
    1. The problem statement, all variables and given/known data
    v = 141.4sin(wt + 15)
    i = 14.14 cos(wt - 45)

    2. Relevant equations
    Draw Phasors.
    P = VI cos(theta)
    Q = VI sin(theta)
    S = P + jQ
    theta is angle between V and I
    3. The attempt at a solution

    V = 141.4sin(wt + 15)
    So Vrms = 141.4/1.414 sin(wt + 15)
    = 100 sin (wt + 15)

    I = 14.14 cos(wt - 45)
    Now cos (-theta) = cos(theta)
    So I = 14.14 cos (45-wt)
    So I = 14.14 sin(90-45 + wt)
    So I = 14.14 sin(45 + wt)
    I rms = 10sin(wt + 45)

    Now Vrms = 100sin(wt + 15)
    I rms = 10sin(wt + 45)
    I leads V by 30 degrees.
    theta = 30 degrees.
    P = VI cos(theta) = 100 * 10 * cos(30) = 866W
    Q = VI sin(theta) = 100 * 10 * sin(30) = 500VAR
    So S = P + jQ = 866 + j500

    But in answer it is given S = 866 - j500

    Where does the minus sign come from?
    I understand that it is leading power factor so it gives Q but mathematically sin(30) is positive value. So why should S have negative VAR?
     
  2. jcsd
  3. Nov 30, 2017 #2

    scottdave

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    Leading current (current leads voltage) gives phasors pointing down (negative angle). Lagging current (current lags voltage) give phasors pointing up (positive angle).
    That is the convention.
    phasor-diagram-of-RLC-Series-circuit-compressor.jpg
     
  4. Dec 1, 2017 #3

    cnh1995

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    By convention, leading VARs are negative and lagging VARs are positive, and S=VI*.
    (If it were the opposite, you'd have to use S=IV*.)
     
  5. Dec 1, 2017 #4
    So mathematically when we calculate Q then it is V multiplied by I multiplied by the angle by which the voltage leads the current.
    Thus here we'll get -theta.
     
  6. Dec 1, 2017 #5

    cnh1995

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    Take for example, V=20∠30° and I=4∠-30°.
    Here, the current is lagging and by the convention S=VI*, you'll get S=P+jQ i.e. positive reactive power.

    If you want lagging VARs to be negative (as your own convention:wink:), define S=IV*.

    https://www.physicsforums.com/threads/power-factor-convention.929532/
     
  7. Dec 2, 2017 #6
    I've also read that if P and Q are in opposite direction then motor/generator is working at leading pf. if P and Q are in same direction then motor/generator is working at lagging pf.
     
  8. Dec 2, 2017 #7

    cnh1995

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    There is no direction for Q. It is oscillating between the generator and the load.
     
  9. Dec 2, 2017 #8
    Then why do we say that capacitor acts as source of Q and inductor acts as load of Q?
     
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