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Ac circuit

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data
    the power factor of a load is 0.866 lagging.the voltage is 200v and the current is 5A.caculate the equivalent series reactance and resistance of the load.

    2. Relevant equations

    cos.theta = power factor = resistance / impedence= 0.866
    volt = ampere / impendence

    3. The attempt at a solution

    trying but still stuck at this question..can someone help me
    btw i am new here if i did anything wrong please guide me...thanks guys
  2. jcsd
  3. Feb 12, 2010 #2


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    First of all: Voltage = Current * Impedance (not /)
    You know the voltage and the current. What is the total impedance?
    The resistance and the inductive impedance form the sides of a right triangle, whose hypotenuse is the total impedance.
  4. Feb 12, 2010 #3
    sorry about the wrong equiation given but i tried
    200v=5A * impendence
    so impendence = 200/5
    = 40 ohm
    but the answer given by my teacher is 20 ohm
    is he wrong?
    does equivalent series reactance means impendence??
  5. Feb 12, 2010 #4
    [tex]\tilde{}[/tex]I think you are mixing terms.

    Impedance is the name given to the whole complex number, defined by [tex]Z=\frac{\tilde{V}}{\tilde{I}}=R+jX[/tex] (tilde denotes phasors, hence both amplitudes and phases)

    While reactance is defined as only the imaginary part of Z: X.

    Also the power factor will be [tex]cos(\theta}=\frac{R}{|Z|}=\frac{R}{\sqrt{R^{2}+X^{2}}[/tex]
    (You got it almost right actually, I just want you to distinguish the impedance, which encodes amplitude and phase, from only the amplitude of the impedance)
  6. Feb 12, 2010 #5
    hmm so how do i get the 20 ohm that my teacher did???
  7. Feb 12, 2010 #6


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    The reactance is one of the sides of the triangle. You already know the hypotenuse (40 ohm) and the cosine of the angle between the hypotenuse and the side representing the resistance (0.866).
    You can calculate both sides.
  8. Feb 12, 2010 #7
    so i take sine30*40 to get the 20 ohms???
  9. Feb 12, 2010 #8


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