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Apparent Power, Real Power and Reactive Power

  1. Apr 1, 2014 #1
    A voltage source of v(t) = 169.7sin200t V is connected across a series combination of a resistance of 10 ohm, an inductance of 10mH, and a capacitance of 500 uF. Determine the current in the circuit, the apparent power, the real power, and the reactive power supplied by the source. What is the power factor of the circuit.


    I am having problems with this questions, as I don't know how to get the voltage since no time is given, and without the voltage, none of the equations are going to be any good.. Can someone point me in the right direction on how to get start with this question?
     
  2. jcsd
  3. Apr 1, 2014 #2
    Well, I figured out how to calculate the voltage = v = 169.7/sqr(2).. But I don't understand where the sqrt(2) comes from in this formula?
     
  4. Apr 2, 2014 #3

    gneill

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    Staff: Mentor

    That's the conversion from peak to RMS for the magnitude of the voltage source. If you're dealing with power in an AC circuit you almost always want to be working with RMS values.
     
  5. Apr 2, 2014 #4

    psparky

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    Gold Member

    If you are proficient in vectors it should be doable for you.

    impedance of cap: 1/jwC
    impedance of inductor: jwL

    w=200 in this case. (omega=200)

    J simply equals 1<90

    Add up your three impedances.
    10 + 1/jwC + jwL
    10+ (1/wC)<-90 + (wL)<90

    This will give you one vector at 0 degrees, one at -90 degrees and one at 90 degrees.


    Add them together and you will have one vector with an angle. Now divide your RMS voltage by this Vector with the angle. You now have the current....also a vector with an angle.

    Now multiply your RMS voltage by your current. This will give you your apparent power....also with an angle.

    Current and power should be complex conjugates.

    Now find the x and y axis of your apparent power from the angle. The x axis is your real power and the y axis is your reactive power.

    To find power factor, divide real power by apparent power.

    Notice how the "time" you mention above was not needed to find any of the answers.

    It takes new electrical guys quite a while to grasp all these concepts together. So if you don't get it the first time, or even 10th time, understandable.
     
    Last edited: Apr 2, 2014
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