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Reactive Power and voltage ratings

  1. Jun 9, 2007 #1

    Can someone please explain to me how you can get higher RMS voltage ratings across components than the supply voltage and how VAR ratings can become huge?

    Indeed what really is reactive power? I've been introduced to it for years but have never fully understood it and I could do with an explanation in Laymen's terms thanks.
  2. jcsd
  3. Jun 9, 2007 #2


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    You should use the term phasor here instead of RMS. Yes, its true than RMS is one component of a phasor quantity, but you're forgetting the angle. This is what makes the difference. If you add the phasors graphically, you'll find that you get the source voltage.

    I've read a bunch of definitions for reactive power, but they all simply state that it is the power which maintains operation of the system and is not for end-use i.e not converted to useful power. Reactive power is a function of the reactive elements in the circuit; true power is whats dissipated.
    http://www.pserc.wisc.edu/Sauer_Reactive Power_Sep 2003.pdf
  4. Jun 9, 2007 #3


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    Alternating current is just that - it moves backwards and forwards in a periodic fashion - sine curve. The current is induced by a time (sinusoidally) varying magnetic field (generator) and at the other end are often motors (which are inductive loads).

    Reactive power is that part of the power that gets stored in the inductances and capacitances of the AC electrical system. One can't get rid of inductances, but one can balance inductance with capacitance to minimize reactive power losses. It is real in the sense that it has to be transmitted with the 'real' or 'usable' power, and transmitting 'unusable' power still costs.

    Here is some discussion - http://en.wikipedia.org/wiki/AC_power. One might wish to avoid the math part.
  5. Jun 9, 2007 #4


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    Think of it this way adder: Suppose you have a very large coil wound on a very large iron core. Our hypothetical inductor is many Henries in value. This inductor is wired in series with an incandescent light bulb, a SPDT switch and a DC power supply. When the switch is thrown in one direction, the power supply is switched in and the circuit is complete. When the switch is thrown the other way, the power supply is switched OUT of circuit and the switch itself completes the circuit. Now we simply have the large inductor and the incandescent light bulb hooked together.
    Start out with the power supply switched out of circuit. Nothing unusual here. Now, throw the switch to switch in the power supply. The light will not come on instantly. Inductors oppose a change in current. It takes a specific amount of time for the magnetic field to build to the point where it is the full load current drawn by the light bulb. At the instant the switch was thrown, the FULL supply voltage was across the inductor and NO current flowed. As the field grew, the current gradually climbed and the voltage across the inductor gradually diminished to zero. Now think about this, while the current was increasing and the voltage accross the inductor was decreasing, some power had to be absorbed in the inductor because the inductor had a voltage across it in the same instant that it had current flowing through it. If you were to plot the voltages and currents in this circuit you would find that more power was delivered by the power supply than ever reached the light bulb. Some power was lost in this transition from power supply switched out to power supply switched in. But it wasn't actually lost.
    Here's why: When you switch the power supply out and replace it with a short circuit, the magnetic field in the inductor will fall and continue to supply the light bulb with power. Remember, inductors oppose a change in current. So it will try to continue the current flowing in the circuit. By the time the field has fallen all the way to nothing, you will have gotten back all of the power that was apparently absorbed in the inductor. This is why it's called apparent power. This explanation is very simplified and ignores losses. But I feel it explains it very well.
  6. Jun 10, 2007 #5
    Awesome replies, all three of them! I'm printing these off and they're all going in my learning folder which I only do for really good explanations!

    Nice one chaps that's much better in my mind now.

    There's only one concept I don't quite get which is, why if you get all the power returned to you with reactive/apparent power, is it considered a waste by the transmission companies?

    I assume only that with AC you get the apparent power returned on the falling part of the AC cycle where the inductor is returning current to the circuit when the voltage starts to drop on the second half on the sinusoid.

    Understand that bit but, again isn't that still obeying energy conservation laws? So why do the power companies consider it wasted energy?
  7. Jun 10, 2007 #6


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    The power companies consider it a waste of energy because more current is flowing in their transmission lines than necessary had the reactive part of the load been canceled out. This extra current in the transmission lines causes extra loss since the conductors in these lines are not perfect. They have resistance.
    Last edited: Jun 10, 2007
  8. Jun 11, 2007 #7
    Spot on thanks pal.
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