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AC circuit: is the voltage source absorbing or delivering?

  1. Jun 28, 2013 #1
    I'm looking at an AC circuit. There is a voltage source, a resistor, an inductor, then another voltage source. One loop. I am supposed to find the power in each element and tell weather it is absorbing or delivering. I think I remember that when a current goes into the negative terminal of a voltage source, the source is delivering, and in an AC circuit the curent goes back and forth. I found the current to be 21 ∠93° Amps. The voltage on the left is 240√2∠60°, and the one on the right is 280√2∠30°. Both have negative terminals at the ground. The resistor is 5Ω, and the inductor is J8Ω. Do I need to use superposition. Can I find the power with the total current that I found using mesh????? What's happening here?
     
  2. jcsd
  3. Jun 28, 2013 #2

    gneill

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    Staff: Mentor

    Why the √2's in the voltage descriptions? Are you converting from RMS to peak? If you're dealing with power you should leave things as RMS, or convert to RMS if peak values are given.

    You've remembered correctly regarding how to determine if a source is delivering (sourcing) or consuming (sinking) power.

    You can use superposition if you wish. But it looks like you've already determined the net current. What is the the assumed direction direction of your current? Do you know how to determine the complex power associated with a voltage source given the current out of the + terminal (same as into the - terminal)? The real component of the complex power is the real power.
     
  4. Jun 28, 2013 #3
    I assume the book used square root of 2's to make for easy transfer to RMS. I am given Peak, and plan to convert to RMS. To find the current I assumed clockwise around the mesh.
    Complex power = 1/2 VI*. Never considered direction of current when looking at it. Pretty new concept to me. So how do I use the assumed direction of current to know if the voltage source is absorbing or delivering?
     
  5. Jun 28, 2013 #4

    gneill

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    Okay, so drop the square roots from the peak values at the outset. Then the complex power for the source will be just VI*. With the assumed clockwise direction for the current, the complex power produced by the first source is V1I*. Do the math and take a look at the sign of real component. If it's positive, the source is producing power. If it's negative, it's consuming power. Same goes for the second source, only here the assumed current direction is flowing INTO the source's + terminal. So V2I* yields the power consumed. If the sign of the real term is negative, it's "consuming" a negative amount of power, i.e., it's producing power.
     
  6. Jun 29, 2013 #5

    rude man

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    I would proceed as follows:
    Without loss of generality, make V1 angle = pi/6, then V2 angle = 0. V1 on left, V2 on right.

    Compute complex I. Using superposition is a good idea.
    Compute I*, the complex conjugate of I. Also compute complex conj. of V1. You already have complex conj. of V2 of course.

    Then compute P1 = 1/2 (V1 I* + V1* I) and P2 = 1/2 (V2 I* + V2* I).

    Either source will be generating real power if its P is +, and will be absorbing real power if its P is -.

    Hints:
    1. Use exponential notation thruout, including Z.

    2. [exp(j theta)]* = exp(-j theta).
     
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