
#1
Nov2012, 06:08 PM

P: 9

Hi everyone,
I encountered a problem as a part of the solution of which I needed to get the iV relation across a solenoid to gain some intution in the other parts of the problem, which is the well known expression V_{L}=Ldi_{L}/dt, where V and i are referenced with respect to the passive sign convention. Nothing tricky here, just basic stuff  however I wanted to quickly verify this before I moved on in the problem. The following is a simple MSpaint sketch that I have just created for illustrative purposes The rest of the details are in the above picture. The question I had in mind is why I am not able to derive the simple relation V_{L}=Ldi_{L}/dt for an inductor (solenoid in this case), even though I referenced everything in line with the passive sign convention? Thanks! 



#2
Nov2012, 06:16 PM

P: 9

The following link is the closest match I could find on the Web that describes a similar problem,however the answer given is not satisfactory in that case:
http://answers.yahoo.com/question/in...0194553AAVDdIw 



#3
Nov2112, 10:58 AM

P: 9

Any ideas? Something I may be overlooking?




#4
Nov2512, 10:40 AM

Admin
P: 21,628

Inductance of a coil (negative inductance?)
This may help.
http://hyperphysics.phyastr.gsu.edu...farlaw.html#c2 http://hyperphysics.phyastr.gsu.edu...indcur.html#c1 http://faculty.wwu.edu/vawter/Physic.../Solenoid.html Note in the last link: 



#5
Nov2612, 01:04 PM

P: 9

I will give this some more thought and follow it up thereafter.
Many thanks for the response! Note: I know how an inductor should behave under a timevarying excitation current and in this context I know why it makes sense for an inductor to have the particular iV relation that it does have, V = Ldi/dt. I just have a problem showing this through Maxwell's eqns, but I think I'll hopefully manage to see the error in my approach if I give it some more thought. 



#6
Nov2712, 06:13 AM

P: 217

The voltage drop across an inductor is L dI/dt and not L dI/dt.
Your first drawing is O.K. if dI/dt>0 and you consider the initial point is where the current enters the inductor and the output where the current exits the inductor. From that drawing you can easily see the voltage drop is negative (look at the plus and minus signs). That expalins everything. 


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