Self-inductance electromagnet circuit

In summary, when switch K is opened, the voltage over the electromagnet is equal to the induced emf and is equal to the voltage drop over the parallel resistor. The polarity of the voltage is the same as the initial emf from the battery, and the current in the coil continues in the same direction. The voltage across the inductor can change instantly, but the current is difficult to change rapidly.
  • #1
Daltohn
30
0

Homework Statement


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.

Homework Equations


emf=-L(delta I/delta t)
V=IR

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
 
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  • #2
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?
 
  • #3
Self-inductance electromagnet, circuit

NascentOxygen said:
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?
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). :)
 
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  • #4
Daltohn said:
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). :)
What is the value of R here? :wink:
 
  • #5
NascentOxygen said:
What is the value of R here? :wink:
(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!
 
  • #6
Also comment on the polarity of the voltage.
There is still this part to be answered. :cool:
 
  • #7
Daltohn said:
(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!

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.
 
  • #8
Self-inductance electromagnet, circuit

gneill said:
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.
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?
 
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  • #9
NascentOxygen said:
There is still this part to be answered. :cool:
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?
 
  • #10
Daltohn said:
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?

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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  • #11
gneill said:
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.
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).
 
  • #12
Daltohn said:
The induced current should try to counter-act the decrease, so the induced current has the same direction.
Correct. Current in the coil continues on much as it had been.

The emf would then have "the same" direction as the initial emf from the battery.
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.
 
  • #13
NascentOxygen said:
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.
The formula gives a positive value for the emf which makes sense. But does the upper part of the coil have higher potential?
 
  • #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.
 
  • #15
Daltohn said:
The formula gives a positive value for the emf which makes sense. But does the upper part of the coil have higher potential?
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?
 
  • #16
NascentOxygen said:
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?
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
 
  • #17
[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:
 
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  • #18
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.
 
  • #19
NascentOxygen said:
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?
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.
 
  • #20
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!
 
  • #21
Ok haha, good! Thanks!
 

1. What is self-inductance in an electromagnet circuit?

Self-inductance is the ability of a circuit to generate an electromotive force (EMF) in itself due to changes in the current flow. It is a property of the circuit's inductor and is measured in henrys. It causes the circuit to resist changes in the current flow and can be both beneficial and detrimental in different applications.

2. How does self-inductance affect the performance of an electromagnet circuit?

Self-inductance can have both positive and negative effects on the performance of an electromagnet circuit. On one hand, it can help stabilize the current flow and prevent sudden changes, making the circuit more efficient. On the other hand, it can also cause voltage spikes and limit the circuit's ability to respond quickly to changes in the input current.

3. What factors can affect the self-inductance of an electromagnet circuit?

The self-inductance of an electromagnet circuit can be affected by several factors, including the number of turns in the inductor, the shape and material of the core, and the presence of nearby conductors. Changes in these factors can alter the circuit's inductance and therefore its overall performance.

4. How is self-inductance different from mutual inductance?

Self-inductance refers to the EMF generated in a circuit due to changes in its own current flow, while mutual inductance refers to the EMF generated in a circuit due to changes in the current flow of a nearby circuit. Both are caused by inductors, but self-inductance only affects the circuit itself, while mutual inductance can affect other circuits as well.

5. How can self-inductance be controlled in an electromagnet circuit?

Self-inductance can be controlled by using different types of inductors with varying properties, such as different core materials or shapes. It can also be mitigated by using other components, such as capacitors or resistors, to balance out the effects of self-inductance. Additionally, careful circuit design and layout can also help minimize the impact of self-inductance on the overall performance of the circuit.

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