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Electrical Engineering - circuits - Reactive Networks

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    There are multiple parts to this question, but I have only got through the first 2 before getting stuck. Please could somebody give me a hand as I am really struggling :frown:

    (a) Calculate the phasor voltage of the voltage source and the phasor current of the current source in both Cartesian and Polar coordinates. Use peak values when calculating the phasors.

    (b) What are the phasor voltage and impedance of the Thevenin equivalent circuit of the power supply? (Use peak values for the voltages and currents in your calculation)

    (c) What are the phasor current and impedance of the Norton equivalent circuit of the power supply? (Use peak values for the voltages and currents in your calculation)

    (d) At what frequency should the circuit operate for the current flowing through the load and the voltage across it to be in phase?

    (e) It is known that for a different load the voltage and current across the load are 240cos(ωot+45°) V and 4cos(ωot+60°) A, respectively. How much power is dissipated in that load?

    Screen_Shot_2016_01_29_at_14_52_10.png


    2. Relevant equations


    3. The attempt at a solution
    These are my attempts at part a, but I was unsure whether they're correct or not
    (a)
    Voltage source:

    |V| = 10 V; Φ = 75°

    V = 2.5882 + j 9.6593 V

    Current source:

    |I|=0.8 A; Φ = -30°

    I=-0.4 + j 0.6928 A

     
    Last edited by a moderator: Jan 29, 2016
  2. jcsd
  3. Jan 29, 2016 #2

    gneill

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    Your voltage source phasor looks okay, but your Cartesian form for the current source does not. You have correctly transformed the sine function to a cosine, but the components that you calculated aren't right. It's a good idea to make a quick sketch of the phasor to confirm that your components look reasonable.
     
  4. Jan 29, 2016 #3

    BvU

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    Out of curiosity: is that a short-circuit I see going down from between R and V ?
     
  5. Jan 29, 2016 #4
    I've re-calculated i(t) and for the cartesian my answer is 0.6929 -j0.4.

    As for the short circuit, yes I believe that is one
     
  6. Jan 29, 2016 #5

    gneill

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    That's better. So what's your next step?
     
  7. Jan 29, 2016 #6
    Part (b) requires the circuit to be reduced using Thevenin's theorem, so the overall circuit contains just a single voltage source and a series impedance. I'm not entirely sure how to go about it however
     
  8. Jan 29, 2016 #7

    gneill

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    What's the "algorithm" for determining a Thevenin equivalent?
     
  9. Jan 29, 2016 #8
    What do you mean exactly?
     
  10. Jan 29, 2016 #9

    gneill

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    What is the procedure to apply to determine a Thevenin equivalent?
     
  11. Jan 29, 2016 #10
    Firstly, remove the load and create an open-circuit. This renders both the current and voltage sources as 0.

    Secondly, I would draw up the remaining circuit which would only contain the remaining resistor, capacitor and inductor.

    Thirdly, I would work out the impedance for each of these components. I would then combine them into one impedance for the entire circuit.

    From there I am unsure what to do
     
  12. Jan 29, 2016 #11

    gneill

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    I think you should review your course notes or text for a description of the procedure and worked examples.
    No, that will only remove the load. The sources will still be active. You have to "manually" suppress the sources if want them suppressed. But removing the load is a correct first step.
    Yes, that will yield the Thevenin impedance. But you still need to find the Thevenin voltage.

    Start by creating a list of the component impedances from the given information. You should do this first thing when a problem specifies the component values and operating frequency. Then:

    1. Remove the load and find the voltage at the output terminals (all sources are still active). That's your Thevenin voltage.

    2. Suppress the sources: replace voltage sources with short circuits (wire), replace current sources with open circuits (remove them from the circuit). Then find the equivalent impedance of the resulting network looking into the load terminals. That's your Thevenin impedance.
     
  13. Jan 29, 2016 #12
    When calculating the Thevenin voltage, do I combine the capacitor with the inductor? Or would I use mesh analysis on each mesh?
     
  14. Jan 29, 2016 #13

    gneill

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    Use whatever technique you're comfortable with. To me it looks like a simple potential divider arrangement...

    Hint: Pay close attention to the effect of the short circuit that @BvU pointed out. It may make your task simpler. Can the current source affect the output voltage in any way? Where does its current flow?
     
  15. Jan 29, 2016 #14
    So, I'm planning on using the Voltage Divider Vo = V(Z2/Z1+Z2) equation, with Vo being the Thevenin Voltage.

    As for the Current source with resistor in series, to work out the voltage from that, what value for the current would I use? Would it be the modulus of I from part (a)? or just the real part of the cartesian for I?
     
  16. Jan 29, 2016 #15

    gneill

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    Good.
    Use whatever form of the source phasor is convenient for calculation. Since you're looking for voltages at this point, you must use the compete phasor information, not just real or imaginary parts. When you start looking at power dissipation, then you will need to be more careful with the individual components of the phasor (only "real" power counts for power dissipation).

    Have you determined that the current source can affect the output voltage in some way? What is the potential at the right end of the resistor with respect to the bottom rail (the common node running along the bottom of the circuit).
     
  17. Jan 29, 2016 #16
    Ok, I've used the value of the modulus of the current (0.8A) to work out the voltage from the current source and resistor in series.

    That worked out to be 8000V. As the voltage source is only producing 10V this seems a little off.
     
  18. Jan 29, 2016 #17

    gneill

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    Yes, but do you care at all about the voltage drop across the resistor? Where does this 8000 V potential occur in the circuit? Sketch it in on your circuit diagram. If you were to connect the negative lead of a voltmeter to the common reference node (the bottom rail), where would you place the positive lead to read this 8000 V potential?
     
  19. Jan 29, 2016 #18
    This is my circuit diagram for now. I've put the 8000V opposing that of the voltage produced by the voltage source. Is this correct? Or should it be flowing in the same direction. 20160129_190143.jpg
     
  20. Jan 29, 2016 #19

    gneill

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    What you've labeled as V1 and V2 are currents, not voltages. Voltages (potential changes) occur across components (between nodes), currents flow through wires and components.

    Your 8000 V potential change occurs across the resistor R. At which end of the resistor would you read 8000 V if you were to connect a voltmeter lead to it (with the other lead connected to the assumed reference node at the bottom).

    upload_2016-1-29_14-12-59.png

    What readings would you expect at the points labeled a, b, and c?
     
  21. Jan 29, 2016 #20
    Ah of course, my mistake!

    At a I would expect 8000V, but at b and c I would expect 0
     
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