source transformation and how to decide which complex current is larger, IrwinExt8.15

Discussion in 'Homework Help' started by PG1995, Dec 26, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi

    Please have a look on the attachment; it also has my queries there. Please help me with them. Thank you.

    Regards
    PG
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    For Q1, I disagree with your conclusion that the specification of the polarity of Vo actually tells you anything about directions and whether the current due to one source is greater than that from another. We can arbitrarily define current direction arrow and voltage polarities before we solve the problem. If the answer for a given variable comes out negative, then we know that our starting assumption was backward in a certain sense. But, this is no problem. It is only after we solve the problem numerically that we really know the directions.

    For Q2, why would you not be able to handle the other source exactly the same way as you did the first one. Look at the original schematic. Both sources are in parallel, and there is no reason one you could not have transformed the other source first.
     
    Last edited: Dec 26, 2011
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, Steve.

    In the Q2, I wasn't able to transform the second voltage source into a current source because it had its positive terminal connected to the ground, 0 V, and it had an impedance connected in series with its negative terminal. Can we do the transformation to a current source even when a voltage source has a series resistance connected to its negative terminal? Please let me know. Thanks.

    Regards
    PG
     
    Last edited: Dec 27, 2011
  4. steveb

    Senior Member

    Jul 3, 2008
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    Yes, you can do that no problem. You can either change the direction of the current source, or you can redefine the voltage source to be positive up and negative down by adding 180 degrees to the phase.
     
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  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    I have solved it. And thanks for pointing out that I can also solve it by adding 180 degrees to the phase; I didn't know that.

    With many thanks
    PG
     
  6. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Please have a look on the attachment. Please let me know if my thinking is correct. Thank you.

    With best regards
    PG
     
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  7. steveb

    Senior Member

    Jul 3, 2008
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    I'm not sure if your thinking is correct. You are getting the correct answer, so it is possible you are correct. But, honestly I can't follow the logic of what you are trying to do. Whatever you are doing, it seems to be a roundabout way, so I don't recommend it, even if it is correct.

    There are a couple of ways to calculate the power. I've already seen that you know the power formula in terms of voltage and current and the cosine of the angle difference between them. Hence, just calculate the voltage on the resistor and calculate the current in the resistor and use that formula.

    Myself, I have my own roundabout way. I took the Vth and divided by 4 to get the current delivered by Thevenin voltage source. Why 4 ohms? Because I know the reactive components add to zero and I can just take the real part of the Thevenin resistance plus the real part of the load resistance. Also, I know the real parts are equal for maximum power transfer. When I use the power formula for the Thevinin voltage source, I get 90 W. And, I know that maximum power transfer implies half the power to the load, so the load power is 45 W.
     
  8. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
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    Thanks a lot, Steve.

    Please have a look on the attachment. I have tried to explain it what I was doing. This is important for me to know if my thinking was correct so that in future I can be careful. Thank you for the help and always being so kind to me.

    Best wishes
    PG
     
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  9. steveb

    Senior Member

    Jul 3, 2008
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    Ah, OK, I understand what you did now. I'm pretty sure that what you did is correct. I can't find any flaw in your logic and it gives the correct answer.
     
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  10. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
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    Hi

    Please have a look on the attachment and it would be very kind of if you could help me with the queries there. My knowledge about basic circuit analysis techniques is quite rusty right now so please excuse me if any of the queries is absurdly silly. Thanks.

    Regards
    PG
     
  11. steveb

    Senior Member

    Jul 3, 2008
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    Q1: Almost yes, but it's really -90 deg not +90 deg.

    Q2: Yes.

    Q3: You did not complete the question. I'm guessing you are asking what the two quantities are? One is the amplitude of the sine wave and the other is the DC offset of the sign wave.

    Note that it is the DC offset that will contribute to an average power once integrated over a complete cycle. This is why a resistor has real power loss, while the inductor and capacitor have only reactive power. Reactive power is power that flow to-and-fro, but averages to zero.
     
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  12. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you very much. I'm much obliged for your help.

    Yes, for Q3, I missed the end word "mean".

    Please have a look on the attachment. Ignore Q1, Q2, and Q3; please help me with Q4. Thanks.

    Best wishes
    PG
     
  13. steveb

    Senior Member

    Jul 3, 2008
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    For Q4, if there are no storage elements, then you have the case implied in Q2. The dotted line goes up to the point where the wave is never negative and the DC offset is equal to the wave amplitude.
     
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  14. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
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    Thank you very much, Steve.

    Best wishes
    PG
     
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