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I'm taking an electronics course to get a physics minor (I'm an applied math major) and today, the professor shows a diagram of an ideal current source connected to an ideal inductor. Crazy stuff I know.

The problem I'm having is that the book and his lecture notes do not treat the independent current source as an actual independent current source. That is, the current is a function with respect to time t, rather than some numerical value like 1A. Hence it is a dependent current source. (Obviously this was done to make the problem easy and/or solvable.)

Despite the lack of care given for using the appropriate symbols, it made me wonder what the actual response would be for an ideal independent current source connected to an ideal inductor. I am a math major after all.

It's been a semester or two since I took ODEs and PDEs, so my differential equations are a bit fuzzy. But I have a feeling that a Heaviside step function and Laplace transforms are not going to work here.

I would argue that the voltage across the inductor would be indeterminate at the edge of the current step and zero after that. Essentially a discontinuous function that approaches infinity at t=0 and then 0 for t>0.

Obviously this isn't physically realizable, but I was curious how one would go about solving this even if the solution diverges.

So my question. How would you attempt to solve this problem? Heaviside step function and Laplace transform or some other method? Also, what change is needed to make this problem solvable? For instance, would a parallel resistor of some finite value do the trick (so you could convert to a Thevenin equivalent circuit)?

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  • \$\begingroup\$ related: electronics.stackexchange.com/questions/228093/… \$\endgroup\$ – Sredni Vashtar Mar 6 '18 at 5:29
  • \$\begingroup\$ Don't confuse the mathematical model with reality. Reality stays finite because it's more complicated than a pure inductor and a pure current source, neglect of residual capacitance being a significant difference between an ideal and a real inductor. With an ideal inductor, the voltage is a delta function, with a real inductor, it's just a 'large transient'. \$\endgroup\$ – Neil_UK Mar 6 '18 at 6:25
  • \$\begingroup\$ \$\frac{V_L}{L}=\frac{\text{d}I_L}{\text{d}t}\$. This inherently implies that it takes time for the current to change. So one solution that works is to assume that the initial current in the inductor is the exact same as the current source value. The other solution has \$\text{d}I\$ being a finite value rather than an infinitesimal value, but divided by the infinitesimal of time.. which is necessarily an infinite resulting ratio. Your intuition is right. See Dirac delta func. You will see Heaviside and Laplace while there. \$\endgroup\$ – jonk Mar 6 '18 at 6:51
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That is, the current is a function with respect to time t, rather than some numerical value like 1A. Hence it is a dependent current source.

You're misunderstanding how independent and dependent sources are usually defined.

A dependent current or voltage source is one who's value depends on some other circuit variable, such as a branch current or the difference between two node voltages.

An independent current or voltage source is one that doesn't depend on other circuit variables.

An independent current or voltage source that does vary over time is an entirely common device and used all the time in useful circuit models.

it made me wonder what the actual response would be for an ideal independent current source connected to an ideal inductor. ... I would argue that the voltage across the inductor would be indeterminate at the edge of the current step and zero after that.

Yes, this is correct.

Of course, real current sources never have perfectly discontinuous output functions (there is always a non-zero rise or fall time), and real inductors have interwinding capacitance, so this is never an issue in practice (although knowing the source rise-time and inductor parasitics might be an issue).

You could also approach this problem with a limit function: consider some well-behaved function for the current variation over time, and take the limit as the rise-time goes to zero.

How would you attempt to solve this problem? Heaviside step function and Laplace transform or some other method?

I've never had to use a Laplace transform to solve a circuit in the real world, though someone like you with a more math-oriented background might choose to do it that way.

I'd solve it by just knowing that the voltage must be a delta function to produce an instantaneous change in the current through an ideal inductor.

Also, what change is needed to make this problem solvable? For instance, would a parallel resistor of some finite value do the trick

As mentioned above, typically a real inductor has an interwinding capacitance. This is a capacitive parasitic that appears in parallel with the inductor. This is the first parasitic I'd add to make this model more realistic and avoid singularities in the circuit solution.

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