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Self-inductance electromagnet circuit

  1. Aug 22, 2014 #1
    1. The problem statement, all variables and given/known data
    The wire in an electromagnet has the resistance 125 ohms and a parallel resistor has the resistance 4 kiloohms. The magnet is kept magnetized by a DC voltage 40V. What is the voltage over the electromagnet soon after switch K is opened? Also comment on the polarity of the voltage.

    2. Relevant equations
    emf=-L(delta I/delta t)
    V=IR

    3. The attempt at a solution
    I have no idea because I'm not able to extract enough data from that text. The polarity should be the same according to Lenz's law. Answer is 2,3 kV.
    Please don't use calculus btw, appreciate any help :) ImageUploadedByPhysics Forums1408696012.272072.jpg ImageUploadedByPhysics Forums1408696012.272072.jpg
     
  2. jcsd
  3. Aug 22, 2014 #2

    NascentOxygen

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    After K has been closed for an adequate length of time, the circuit conditions reach steady conditions. What is this steady DC current through the inductor?
     
  4. Aug 22, 2014 #3
    Self-inductance electromagnet, circuit


    KVL: 40V-12.5I=0 I=0.32 A. I got that far. But now I realized what to do. :) Very soon after K is opened, I is still 0.32 A through the inductor, which now forms a closed circuit with the resistor. The voltage drop over the resistor is V=IR=1.3 kV, which equals the induced emf (according to KVL). :)
     
    Last edited: Aug 22, 2014
  5. Aug 22, 2014 #4

    NascentOxygen

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    What is the value of R here? :wink:
     
  6. Aug 22, 2014 #5

    (4000+125) of course! I actually thought of that after I did the above, but I didn't reflect over that they give the same value after rounding off, so assumed the first one was right haha. Thanks!
     
  7. Aug 22, 2014 #6

    NascentOxygen

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    There is still this part to be answered. :cool:
     
  8. Aug 22, 2014 #7

    gneill

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    Actually, using the 4kΩ value is correct. The resistance associated with the electromagnet itself is distributed throughout the wire comprising the wiring. You cannot separate this resistance from the electromagnet in any way, so if you were to measure the voltage across the electromagnet (say with a voltmeter), then you would necessarily have to include this resistance between the meter's probes.

    Fortunately the 4k resistor is in parallel with the electromagnet and must share this net voltage.
     
  9. Aug 22, 2014 #8
    Self-inductance electromagnet, circuit


    So when the switch is opened, you shouldn't apply KVL so that emf-125i-4000i=0? Then it would work like a battery with internal resistance. Or is that voltage drop not taken into account here?
     
    Last edited: Aug 22, 2014
  10. Aug 22, 2014 #9

    The induced current should try to counter-act the decrease, so the induced current has the same direction. The emf would then have "the same" direction as the initial emf from the battery. Does the upper part of the inductor have positive potential then? Then the current goes through the resistor from positive to negative potential.
    But through the coil it's the other way around which I don't really understand, is that analogous to batteries in any way? Do positive charges (in conventional current terms) go from negative to positive inside the battery?
     
  11. Aug 22, 2014 #10

    gneill

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    The voltage drop due to the internal resistance of the coil needs to be accounted for, sure. When you consider the voltage across the electromagnet it includes the internal resistance since it is inseparable from the electromagnet.

    attachment.php?attachmentid=72371&stc=1&d=1408733478.gif

    Drawn as "lumped components" the electromagnet consists of an inductor and resistor (at least in this approximation of a "real" electromagnet). The voltage V in the diagram is measured across the whole electromagnet, and the resistor R happens to be in parallel with the whole electromagnet.
     

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  12. Aug 22, 2014 #11

    NascentOxygen

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    I see, the wording was a bit ambiguous. For finding the induced voltage, it's the product of current x resistance, and OP's working is correct. Had the wording "resistor" been "resistance" it wouldn't have misled you (and led on to the confusion that then had to be sorted out).
     
  13. Aug 22, 2014 #12

    NascentOxygen

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    Correct. Current in the coil continues on much as it had been.

    You can't say that. The voltage across an inductor can change instantly, it's the current that is difficult to change rapidly.

    Since you know the direction of current through the 4k you know the polarity of the voltage across it. That voltage is also the coil's voltage.

    Having decided that, compare it with what you know of the L. di/dt formula to resolve any uncertainties you may have.
     
  14. Aug 23, 2014 #13

    The formula gives a positive value for the emf which makes sense. But does the upper part of the coil have higher potential?
     
  15. Aug 23, 2014 #14
    Thanks for helping to clear this up! :) In the original post it says the answer is 2.3 kV but it should be 1.3 kV.
     
  16. Aug 23, 2014 #15

    NascentOxygen

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    After the switch opens, what is the direction of current through the 4000 Ω resistor?

    Current in a resistor always flows from the positive end towards the negative end. So which end of the resistor is positive?

    In conclusion, which end of the inductor is positive?
     
  17. Aug 23, 2014 #16

    The upper end of the resistor is positive since current flows downward. Then it would make more sense to me that the upper end of the inductor is positive because the emf increases the potential. Can I compare it to the motional emf of a moving rod? In the picture, current flows counter-clockwise and the upper end of the rod is positive. ImageUploadedByPhysics Forums1408791617.396674.jpg
     
  18. Aug 23, 2014 #17

    NascentOxygen

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    [STRIKE]The battery causes inductor current flow downwards. After the switch opens, inductor current continues to flow downwards through the inductor. Where path does that current follow after it exits the inductor?[/STRIKE]

    EDIT: I overlooked the battery polarity being - upwards, so my current directions were wrong. You are correct in saying the inductor current causes the top of the resistor to be positive with respect to the lower end. Sorry for the confusion. :redface:
     
    Last edited: Aug 23, 2014
  19. Aug 23, 2014 #18

    NascentOxygen

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    BTW, to dispel any uncertainty, when a switch goes "open" a gap opens up between its contacts and it becomes an open-circuit and ceases to conduct current.
     
  20. Aug 23, 2014 #19

    I'm just lost at this point. How can current go downward through the inductor when the circuit is in steady conditions, does conventional current not go from positive to negative? The current then goes upwards through the resistor as well, and when the switch opens the current's direction through the conductor is the same but the current through the resistor is now downwards, it's then in series with the inductor. That's how I understand it, maybe I should ask a teacher to explain face-to-face/in Swedish haha.
     
  21. Aug 23, 2014 #20

    NascentOxygen

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    Sorry, I failed to notice the battery polarity, so you are correct. After the battery is disconected, the top of the resistor becomes positive, likewise the top of the inductor. You do have it correct.

    (I have edited my earlier post to fix my mistake.)

    No need for a translator!
     
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