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How to represent an inductive circuit 
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#1
Jan1912, 01:47 PM

P: 7

In all the tutorials I have seen on inductance and magnetism, I have never seen how to represent an inductor as a voltage or current source in a circuit. Please see figure. I am trying to determine V and I given the characteristics of the coil and the magnetic flux through the coil.
Note: Before you suggest an answer, would you please make sure your solution works for all values of resistance from zero to infinity? For example, Faraday's law for induced voltage only works into an open circuit. Thanks for suggestions. 


#2
Jan1912, 01:50 PM

Mentor
P: 40,964

Don't see an attachment yet, though... EDIT  Attachment added now. 


#3
Jan1912, 01:57 PM

P: 1,506

what do you mean when you say 'Faraday's law for induced voltage only works into an open circuit'?



#4
Jan1912, 02:02 PM

P: 7

How to represent an inductive circuit
I just added the schematic to my OP. To technician, if you have a solution that uses Faraday's law, by all means post it. I would like to focus on the solution to my problem. Thanks!



#5
Jan1912, 02:27 PM

P: 1,506

If you need an expression for the emf produced by the coil it comes from Faradays law....
induced emf = rate of change of magnetic flux linkage. (magnetic flux linkage = flux x number of turns on the coil = N∅) Therefore e = d(N∅)/dt Hope this helps with the rest of your analysis. 


#6
Jan1912, 03:39 PM

P: 7

Sorry, but it doesn't help and I hope there are other solutions out there. If you have a coil with two terminals and you short them together, clearly you can't have infinite current, so how does this help me? I tried to make this clear in my OP.



#7
Jan1912, 04:17 PM

P: 1,506

Sorry to hear that the responses you have had have been of little help.
Here is some more explanation: If you have a coil experiencing a changing magnetic flux linkage there will be an induced emf given by e = d(N∅)/dt (as previously stated) If you connect the ends of the coil together (assuming there is zero resistance) then a current will flow (it will not be infinite but....) the current will increase according to Faraday's law... e = LdI/dt (L is the inductance of the coil) so you have an expression for the rate of rise of the current. The current will increase uniformely (towards infinity) at a rate = e/L. If there is resistance in the circuit then there will be a max value for the current (= e/R) I hope that this information can be of some help in understanding what you need to know 


#8
Jan1912, 04:23 PM

P: 1,781

You don't see it because it's not part of circuit theory.
There's no such thing as a magnetic field in circuit theory. 


#9
Jan1912, 04:42 PM

P: 1,506

I agree, although magnetic field theory does give values for emf.
Without a changing field in this example there would be no emf and therefore no problem! 


#10
Jan1912, 04:53 PM

P: 7

My guess is the solution lies in this formula
I = V_{S} / (R + X_{L}) I wish I could give a dollar to the first person who can draw the circuit correctly and give me a complete solution. It may be that V_{S} is a voltage source that is a function of webers. 


#11
Jan1912, 05:04 PM

P: 1,506

I may not be able to draw the circuit you are after but it will not have I =Vs/(R+Xl)
The combination of R and Xl in series is called impedance....Z....and the relationship between R and Xl is Z^2 = R^2 + Xl^2 and I = V/Z I wish that I could grasp what you are after!! Your last post suggests that you are dealing with an AC circuit problem 


#12
Jan1912, 05:28 PM

P: 1,781




#13
Jan1912, 11:22 PM

P: 7

I really appreciate all of the comments I have received. At this point can we throw this open to other electronics fans out there? Just to review, we need a circuit that allows us to calculate current in the resistor and works for any value of resistance. We also know that the voltage or current generator will only produce AC signals and its amplitude will be related to flux. . 


#14
Jan2012, 07:52 AM

P: 1,506

I see what you mean now.... you mean Z = R + jωL I think
and Z = R jωC for capacitance. ???? 


#15
Jan2212, 01:12 PM

P: 9

The inductive coil by itself does not act as a voltage source, just a voltage drop, like any other impedence to current. It becomes a voltage source when it is the secondary coil in a transformer, and then of course only to an AC current. Calculations for current are according to Ohm's Law, for simple electronic AC circuits, which I presume is what you're dealing with.



#16
Jan2912, 02:23 PM

P: 7

I have asked the moderators to cancel this thread as it is 10 days old and we are nowhere getting an answer to my question and I don't think anybody is going to read through all of this. Just my opinion. 


#17
Jan2912, 06:29 PM

P: 5,462

An inductor is not, of itself, a voltage or current source. If you connect one into a circuit you do not magically get either source included in your circuit. So an inductor is an inductor, a voltage source is a voltage source and a current source is a current source. Each is different and thus awarded a different circuit symbol and obeys different circuit equations. So if you would like to try to explain your question again more clearly we could respond more accurately. 


#18
Jan2912, 09:54 PM

P: 1,781

Yes, studiot is correct. Your question is unclear.
Mr. Faraday knew nothing of circuit theory. He was doing physics. In circuit theory the inductors satisfy simple differential equations on their terminals. There are no fields and no generators. To model general coupling between multiple coils you have mutual inductance. You still don't have magnetic fields or sources in this case, just coupling between the windings. Your confusion seems to stem from a mixing of circuit and field concepts. 


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