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Calculating EMF in order for the powers to be equal

  1. Jan 16, 2017 #1
    1. The problem statement, all variables and given/known data
    In a circuit of constant current shown in the image, values are Ig=0.2 A AND R=750 ohm.
    What is the value of EMS of an ideal voltage source V for the powers of an ideal voltage and an ideal current source to be equal?



    circuit.PNG

    2. Relevant equations
    Pv=PIg

    3. The attempt at a solution
    I can calculate voltage on R in the middle and that is like it
     
  2. jcsd
  3. Jan 16, 2017 #2

    NascentOxygen

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    Staff: Mentor

    What is your expression for this voltage?

    Using the symbols already marked on the circuit, now write an expression for the current through the resistor on the right.
     
  4. Jan 16, 2017 #3

    NascentOxygen

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    I don't recognize the abbreviation. Is EMS correct spelling, or should it be EMF?
     
  5. Jan 16, 2017 #4
    Yes, sorry, EMS is EMF. Forgot to translate that (I translated whole problem into English )
    Expression for that voltage is V=IR =150V

    Current through the resistor on the right is Ir=V/R ( Ir being the current in that branch)
     
  6. Jan 16, 2017 #5

    NascentOxygen

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    Not 150 volts. If you are summing voltages, you need to add in the voltage across the current source, but we don't know it.
    Okay, so what is the expression for current that must be coming from the voltage source? Write it using only those symbols already marked on the circuit.
     
  7. Jan 16, 2017 #6
    I thought 150 V is on R in branch with current source
    About the current, could it be written as current divider or like I = (V/R) + (V/R)
     
  8. Jan 16, 2017 #7

    NascentOxygen

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    That was going to be my next question, but we can look at it now.

    The current source and its resistor are in series, so we have:
    voltage across ideal current sourcevoltage across series resistor = terminal voltage, V

    Rearranging this, you have the formula: voltage across the ideal current source = V + 150

    The sum of the currents is given by Kirchoff's Current Law, so we have:
    current from voltage source + current from current source = current through resistor on right side

    ⏩Rearrange this to get the formula for current from the voltage source, and write it using only symbols already shown on the circuit diagram.
     
  9. Jan 17, 2017 #8
    Isn't the voltage across the ideal current source = V - 150.
    If VR+VIg=V

    Are these equations any good?
    94806-19121bb49525af66c8a26666c83a75d2.jpg
    upload_2017-1-17_12-24-38.png
    I being the current from voltage source and Ir being the current in branch with only a resistor.
     

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  10. Jan 17, 2017 #9

    NascentOxygen

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    Is it +150 or –150? How to verify which sign is right? Try it this way:

    Draw just that branch, mark in the direction of the current Ig, and mark in the + and – signs to show the voltage across the current source. Now mark in the + and – signs to show the voltage across the series resistor due to branch current Ig. Finally, write the equation showing how the total voltage, V, is formed from these two voltages in series.

    No. Why do you think they could be?
     
  11. Jan 17, 2017 #10
    Okay, I didn't pay attention on the signs. About the equations, I thought that i could use current divider.
     
  12. Jan 17, 2017 #11

    NascentOxygen

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    What is the equation you'll be writing to indicate the specified condition that
    power from voltage source = power from current source ?
     
  13. Jan 17, 2017 #12
    V*Ig=V-R*Ig
    V2/R -VIg=VIg-RIg2
     
    Last edited: Jan 17, 2017
  14. Jan 17, 2017 #13

    NascentOxygen

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    I was expecting one equation.

    You need to explain these. What is u?
     
  15. Jan 17, 2017 #14
    u is V, We are here using U as voltage so I constantly make these kind of errors.
    Those two are the same equation.
     
  16. Jan 17, 2017 #15

    NascentOxygen

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    I don't recognize what you've done. The first equation looks wrong. Can you explain these equations?
     
  17. Jan 17, 2017 #16
    Second comes from this
    VI=VIgIg
    V( (V/R) -Ig)=(V-RIg)Ig
    (V2/R) -VIg=VIg-RIg2
    V2(1/R) - 2VIg+RIg2=0
    V1/2=...

    First one is wrong,

    I have found these equations, after posting here, written on page with the same problem posted here ( but that's it, no steps nor 1 calculation was done).
    My idea was to somehow calculate current through R in right branch, since voltage is the same on that R and V, if I'm not mistaken. But, I was unable to do so.
     
  18. Jan 17, 2017 #17

    NascentOxygen

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    So you have finished these calculations, and confirmed that the answers do give equal powers in the sources? If you haven't confirmed the answer, I think you should.

    Then come back and post here about what you've found.
     
  19. Jan 18, 2017 #18
    I'm unsure on what to. Result comes to 150V.
    The final result should be 362,13V (Looking at the solution)
     
  20. Jan 18, 2017 #19

    NascentOxygen

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    Perhaps you should go back to post #9 where I explained how to correctly sum a source and a resistive drop in series. In deriving your equations, you for some reason did not heed what I explained there.
    That is one of two possible solutions.
     
  21. Jan 18, 2017 #20
    Oh yes, my mistake. Well, with your help, now I got the answer right. Thanks
     
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