Boundary conditions for a purely inductive load in an AC circuit

  • #26
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We choose it because it matches our particular reality (constraints). We can do that!
Right, but we need some additional information about our particular reality than simply the circuit diagram and the voltage of the source in order to determine which of all of the possible values of ##c## corresponds to ours.
 
  • #27
vanhees71
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This seems to be a perfect example for the often observed fact that apparently simple unrealistic examples are either not really simple (my favorite is the question about the momentum of a particle in a infinite-wall potential pot in quantum mechanics) or more confusing than the realistic case.

So let's to the more realistic case of a coil made of real-world wire with a finite resistance. Then the equation for the current reads (using Faraday's induction law in integral form as usual for AC circuit theory)
$$L \dot{I}+R I=U_0 \cos(\omega t).$$
This is a linear, inhomogeneous differential equation of first order.

Math is your friend and tells you that the general solution is given by the superposition of a special solution of the inhomogeneous equation (i.e., the full equation written above) and the general solution of the homogeneous equation, which is given by
$$L \dot{I}+R I=0.$$
So let's start with this equation. It is easily solved by "separation of variables", i.e., you bring it into the form
$$\mathrm{d} I \frac{1}{I}=-\frac{R}{L} \mathrm{d} t.$$
Formally integrating on both sides of the equation you get
$$\ln \left (\frac{I}{c} \right) = -\frac{R}{L} t,$$
where ##c## is an integration constant. Solving for ##I## you get the general solution of the homogeneous equation
$$I_{\text{hom}}(t)=c \exp \left (-\frac{R}{L} t \right).$$
Now you need only one special solution of the full inhomogeneous equation. Physics intuition tells you that an "ansatz of the type of the right-hand side" may be promosing.

Here it's simpler to make a little detour using complex functions, writing ##\cos(\omega t)=\mathrm{Re} \exp(\mathrm{i} \omega t)##. So we first look for a solution of the complexified equation
$$L \dot{I}_c + R I_c = U_0 \exp(\mathrm{i} \omega t).$$
It's obvious that the ansatz
$$I_c(t)=A \exp(\mathrm{i} \omega t)$$
must leads to a solution. Indeed pluggin the ansatz in and cancelling the common exp-factor leads to
$$(\mathrm{i} \omega L + R) A=U_0 \; \Rightarrow \; A=U_0 \frac{1}{R+\mathrm{i} \omega L} = U_0 \frac{R-\mathrm{i} \omega L}{R^2 + \omega^2 L^2}.$$
From this you get
$$I_c(t)=U_0 \frac{R- \mathrm{i} \omega L}{R^2 + \omega^2 L^2} \exp(\mathrm{i} \omega t).$$
You get a special solution of the original real equation by taking the real part of this, which yields
$$I_{\text{inh}}(t)=\frac{U_0}{R^2+\omega^2 L^2} [R \cos(\omega t)+\omega L \sin(\omega t)].$$
So the general solution of the ODE is
$$I(t)=I_{\text{hom}}(t)+I_{\text{inh}}(t) = c \exp \left (-\frac{R}{L} t \right) + \frac{U_0}{R^2+\omega^2 L^2} [R \cos(\omega t)+\omega L \sin(\omega t)].$$
The indetermined integration constant ##c## is physically determined by the initial condition. In this case you must give ##I(0)##. There's only 1 initial condition, because the ODE is of 1st order.

Now in electrical engineering one very often is not interested in any kind of specific initial condition, but only on the long-time behavior. The reason is that the solution of the homogeneous equation exponentially decays, i.e., after a time scale of the "relaxation time" given by ##\tau=L/R##, the specific initial condition doesn't play much of a role any more. That the solution exponentially decays is of course due to dissipation of energy in the resistance of the coil. Thus in the artificial limit, where you have set ##R=0## the relaxation time goes to infinity, and the circuit keeps "a memory" of the initial state forever. That's of course unrealistic (except if you have a superconducting coil, but in that case you have to use different equations to begin with, so we'll not discuss this case any further).

Now it's understandable, why many books simply leave out the general homogeneous solution: It's because they are only intersted in the "steady-state solution", which is the special solution of the inhomogeneous equation which vanishes whenever you set the inhomogeneity to zero (i.e., in your case the external AC voltage, ##U_0=0##).

Then you are left with
$$I_{\text{steady-state}}(t)= \frac{U_0}{R^2+\omega^2 L^2} [R \cos(\omega t)+\omega L \sin(\omega t)].$$
Now in the limit ##R \rightarrow 0## you get
$$I_{\text{steady-state}}(t)=\frac{U_0}{\omega L} \sin(\omega t)=\frac{U_0}{\omega L} \cos(\omega t-\pi/2).$$
As you, see in this limit the current's phase stays behind the voltage's phase by ##\varphi_0=\pi/2##.
 
  • #28
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Yes, that is right.

Yes, this is correct. If it was an open circuit previously then, by definition, the current is zero at ##t<0##. So then the initial condition would be ##I(0)=0## and the circuit would oscillate around a non-zero average current. I am not sure why it doesn't feel right to you, but that is the solution.
No wait, perhaps I see where I'm mistaken. If we are switching the circuit on at an arbitrary time we cannot say that ##V=V_0 \sin(\omega t)##. We should write ##V=V_0 \sin(\omega t + \varphi)## in order to take into account the instantaneous value of the potential.
So the constant should be
[tex]
c = \frac{V_0}{\omega L} \sin(\varphi)
[/tex]
Earlier I was not convinced because I expected a relation between the initial value of the potential and the constant in the current, which was not there because of my mistake.

So we could say that the "standard" solution corresponds to switching the circuit on exactly when the voltage is at its maximum value.


In my circuits lab back in college we had programmable sources. We could set it to provide either a current source or a voltage source, as needed. It also had a built in oscilloscope so that we could provide input and measure output to our circuits.
Ok but the AC circuit with a purely inductive load is usually attached to a voltage generator, right? I mean, in the discussions you find in textbooks.
Perhaps naively, I think that providing a sinusoidally alternating current makes the problem easier. The voltage is proportional to the derivative of the current. The "problem" with the constant appears when you provide a voltage and look for the current, right?

That is what I would say. Or maybe just "... because it is convenient".
Uhm... I like the first one better. The second one feels a little dismissive of the student's curiosity.
 
  • #29
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The reason is that the solution of the homogeneous equation exponentially decays, i.e., after a time scale of the "relaxation time" given by τ=L/Rτ=L/R, the specific initial condition doesn't play much of a role any more. That the solution exponentially decays is of course due to dissipation of energy in the resistance of the coil. Thus in the artificial limit, where you have set R=0R=0 the relaxation time goes to infinity, and the circuit keeps "a memory" of the initial state forever. That's of course unrealistic (except if you have a superconducting coil, but in that case you have to use different equations to begin with, so we'll not discuss this case any further).
I like this very much. Thanks. If I could I would add more "likes" to your reply.

So this seems to prove that, whatever the initial state is, the long time solution would be the same.
This is what I suspected from the beginning. It does not seem to be consistent with other replies, though (or with my understanding of those replies). I'll look more carefully into the maths of both versions.

One nice thing of "your" solution is that one can make ##R## small, but non zero, and ramp up ##\omega##. That should be almost like having ##R## vanish altogether.
After a reasonable time one should find the expected behaviour (pretty much the one in the textbooks).
 
  • #30
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If we are switching the circuit on at an arbitrary time we cannot say that ##V=V_0 \sin(\omega t)##. We should write ##V=V_0 \sin(\omega t + \varphi)## in order to take into account the instantaneous value of the potential.
So the constant should be
[tex]
c = \frac{V_0}{\omega L} \sin(\varphi)
[/tex]
Well, across an inductor the voltage can change instantaneously no problem. It is just the current that cannot jump instantaneously. So it is perfectly fine to start with an open circuit (##I(t<0)=0##) and jump immediately to ##V(t>0)=V_0\sin(\omega t)##. The voltage will be discontinuous but finite and the current will be both continuous and finite.
The voltage is proportional to the derivative of the current. The "problem" with the constant appears when you provide a voltage and look for the current, right?
Yes.
 
  • #31
rude man
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Uhm. Right, but the standard derivation of a purely inductive circuit never mentions any resistor, and
[tex]
I(0) = \frac{V_0}{\omega L}
[/tex]
It seems hard to make contact with your answer.
The resistor is there just to set up the initial condition. When you apply your voltage ##V_0 sin(\omega t) ## it no longer forms any part of the inductor current.

You have set up an initial current and have then applied your voltage directly across the inductor.
 

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