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Induced EMF within a part of a parallel circuit?

  1. Apr 5, 2015 #1
    A Power supply is connected to a parallel circuit with resistance R, there are three wires connected in parallel where the voltages would be the same,however, the current would be dived based on the number of wires and the resistance is the same.

    wVz7Qlu.jpg
    What would happen if wire (c) was placed in a changing magnetic field, and there is an induced EMF as diagram here:
    zd88Flt.jpg
    With two cases, one the polarity is opposing the power supply and and the other case is the induced EMF's polarity supporting the power supply as the diagram above.

    What is the resulting voltage? If the induced EMF is opposing the power supply then the EMF would be: ##V_ps - V_e##, and if the voltage is supporting the power supply(as the second diagram) the voltage is added?

    I could not predict the outcome when thinking about the theory.
     
  2. jcsd
  3. Apr 5, 2015 #2

    Svein

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    The way you have drawn it, there is no connection between the upper part of b and the upper part of a - c.
     
  4. Apr 5, 2015 #3

    NascentOxygen

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    That "hump" in your wiring should be a connection of all 3 wires.

    Perhaps it may be clearer were you to draw 3 resistors in those wires, then show the induced emf as though it is a battery added into that wire. See how that affects things.
     
  5. Apr 5, 2015 #4
    My apology for the terrible diagrams re-adjusted them based on your comments:
    Induced EMF opposing the power supply:
    fdwM0p5.jpg
    Induced EMF supporting the power supply:
    xWyWGtG.jpg

    I could not be certain of the outcome, for the first diagram where the induced-EMF is opposing the power supply I assumed that the voltage would be 15, and from the second diagram I assumed it to be 25V, however, not sure what is correct. Also, could not figure out the voltage at the blue points(a,c).
     
  6. Apr 5, 2015 #5

    Svein

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    In the upper diagram, the "induced E" will almost short-circuit the power supply. The voltage between a and c will be forced to 5V.

    The lower diagram isn't much better, since the PS will try to force +20V between a and c. As drawn, the "induced E" will definitely short-circuit the power supply.
     
  7. Apr 5, 2015 #6

    jim hardy

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    Svein is saying "doesn't that wire going from upper right 'c' down to 6 volt source have some resistance ? "
     
  8. Apr 5, 2015 #7
    Yes, it does.
     
  9. Apr 5, 2015 #8

    Svein

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    OK. Look at the upper diagram. Let us assume a very thin wire and assign it a resistance of 0.1Ω. The current through that wire will then be (20V - 5V)/0.1Ω = 150A. Something is definitely going to break!
     
  10. Apr 5, 2015 #9

    NascentOxygen

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    Don't bother posting your revised circuit. I intended for you to sketch it on paper and examine it by yourself. 3 wires, 3 resistances, with that extra battery in series with one of them. What is going to be different then?
     
  11. Apr 5, 2015 #10
    Its kind of frustrating that with all the time I've invested into understanding circuit theory I can't come up with a proper answer.
     
  12. Apr 5, 2015 #11

    jim hardy

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    Looks like a good time label some currents and write Kirchoff's laws equations around the loops....

    Nobody works these in their head until they've been at it for a semester or two.
     
  13. Apr 5, 2015 #12
    Sure I'll give it a try, but isn't this a short-circuit? As Svein stated it.
     
  14. Apr 5, 2015 #13

    jim hardy

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    looks to me like a 0.1 ohm resistor not a short circuit.

    In beginner courses a short circuit is zero ohms.
    When one moves on to real life "short circuit analysis", the resistance (and inductance) of the wiring comes into play.
    That's why you'll never get infinite amps out of your wall socket, but you might get a thousand for a millisecond or two.
     
  15. Apr 5, 2015 #14
    But wouldn't the existence of the induced-EMF on wire (c), cause a short circuit? Or would it change the overall voltage... not so sure I'll play around with KVL&KCL, and will share my results and see where i'm wrong.
     
  16. Apr 5, 2015 #15

    jim hardy

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    How ?

    KVL and KCL will not fail you. Admittedly it's hard work when we're new to it but
    what does not kill us , strengthens us.
     
  17. Apr 5, 2015 #16
    Not sure, just a gut feeling.

    Daily motto.
     
  18. Apr 6, 2015 #17
    I worked on a few attempts and here is my answer: Based on the wires resistance being the same(in fact for the whole circuit), the voltages for wire a/b would be 20V and 15V for wire c due to the induced-EMF opposing the power supply, and 25V when its non-opposing.

    Here is the most confusing part, the current. I think that the current would divide out based on each wire's voltage. PS is at 20V/100V = 0.2A/3 = 0.06A
    However, because the last wire has lower PD it should be getting much less then that value(Not sure how to find that even with KCL and KVL). The same assumption goes when the induced-emf is non-opposing 25V would require higher current than the other wires.
     
  19. Apr 6, 2015 #18
    Also, would the voltage still be 20V all around for both circuits. Im confused now with the induced-EMF changing the PD for the whole circuit? Or the single wire?
     
  20. Apr 6, 2015 #19

    jim hardy

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    Well, looks like you successfully applied KVL .... Hooray !

    Say what ?

    Focus on the three legs one at a time.
    What is voltage across R1 ? What does Ohm's law say the current must be through R1 ?
    Repeat for other two resistances.
    Then sum currents by KCL.

    This is how we work problems, one baby step at a time. Redraw between each step , gradually building up the number of known quantities and reducing the number of unknowns.

    Everybody wants to leap straight to the answer. But it is no more possible to do that than to climb stairs one whole flight at a time.
     
  21. Apr 6, 2015 #20
    Although you advice
     
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