Boundary conditions for a purely inductive load in an AC circuit

[FranzDiCoccio]

Hi all,
Kirchhoff's equation for this simple circuit is equivalent to
$$\dot I=\frac{V}{L}$$
Where $V=V_0 \sin(\omega t)$. Integrating both sides should give
$$I(t) = -\frac{V_0}{L\omega} \cos(\omega t)+c$$
where $c$ is an arbitrary constant (current).
Here, most of the derivations I've found simply drop the $c$, without so much as a word. Actually they would not even mention it in the first place.

Of course keeping that constant does not seem physically right. What could the value of $c$ possibly be? How would one determine it?
It should be ruled out by a boundary condition, but I am not really able to put my finger on it.
Somewhere I read: the current intensity must be maximum when the emf is zero... but that seems kind of a hindsight.

The only things I can think of are based on symmetry considerations. The "driving" potential is symmetric, so I expect the current to be symmetric too.
There is only one aspect that I can think of that could break the symmetry and introduce a "background" constant (DC) current. That would be the rotation direction of the coil in the AC generator.
But this makes me think that if the rotation of the coil is inverted, so would be the value of $c$.
So if the coil spins clockwise we get e.g. $c$, and if it spins counterclockwise we get $-c$.
On the other hand, the emfs of these two cases look exactly the same (especially if I do not know when the spinning has started). They only differ by a phase.
This makes me think that they should generate the same alternating current, which means $c=-c$, which of course means $c=0$.

Am I making any sense?

 

[kuruman]

The value of $c$ depends on the initial conditions, i.e. the value of the current at time t = 0. Once you specify that, you can solve the equation $I(0)=-\dfrac{V_0}{\omega L}+c$ to find $c$. Alternatively, use upper and lower limits when you integrate.

 

[FranzDiCoccio]

Dear kuruman,
I know that mathematically $c$ is determined by the initial conditions. But what's "initial" here? The emf is a periodic function. At $t=0$ (and every half period) the current intensity is maximum. But how does one justify that? As I say, it sounds like "hindsight" to me. I know that the current intensity is maximum there because I'm used to the standard expression for the current.
Not so intuitive, anyway.

 

[anorlunda]

Dear kuruman,
I know that mathematically $c$ is determined by the initial conditions. But what's "initial" here? The emf is a periodic function. At $t=0$ (and every half period) the current intensity is maximum. But how does one justify that? As I say, it sounds like "hindsight" to me. I know that the current intensity is maximum there because I'm used to the standard expression for the current.
Not so intuitive, anyway.

What was the current before applying the sinusoidal voltage?

 

[kuruman]

Dear kuruman,
I know that mathematically $c$ is determined by the initial conditions. But what's "initial" here? The emf is a periodic function. At $t=0$ (and every half period) the current intensity is maximum. But how does one justify that? As I say, it sounds like "hindsight" to me. I know that the current intensity is maximum there because I'm used to the standard expression for the current.
Not so intuitive, anyway.

I gather, although you have not explicitly said so, that you have a coil rotating in a magnetic field with angular frequency $\omega$. That's fine. You understand that mathematically $c$ is determined mathematically by the initial conditions. OK. When we do physics, $c$ is physically determined by the initial conditions. Let me explain. We have a physical entity here, the rotating coil, and $\omega t$ is the angle by which the coil rotates. Time parameter $t$ is measured with respect to a (usually imaginary) clock that starts clicking when the angle passes the line with respect to which the coil angle is measured. If you choose to draw the line so that the current is maximum when the clock starts clicking, then it will be a maximum at all subsequent time intervals separated by a period $T=\frac{2\pi}{\omega}$. You can also choose to draw that line at an angle of 90^o relative to the one I just mentioned. Then the current will be zero at t = 0 and at all all subsequent time intervals separated by a a period. And if you drew the line in between the previous two lines, the current will be between the maximum value and zero at t = 0.

 

[Dale]

Summary:: The mathematical solution to the problem would allow a constant "background" current. Of course that current is not there. What is a convincing reason for that?

Of course keeping that constant does not seem physically right. What could the value of ccc possibly be? How would one determine it?

Why would you drop it? It is perfectly valid to have a non-zero $c$. The value of $c$ is determined by your initial conditions. Nothing else.

 

[rude man]

Say you connect a battery E, resistor R and an inductor L in series. (battery + to R+, R- to L+, L- to battery -). Wait a long time. Then i ~ E/R. Now slap your voltage $V_0 sin(\omega t)$ directly across the inductor. Show that the inductor current is

$i = \frac {V_0} {\omega L}(1 - cos(\omega t)) + \frac {E} {R}$. Your initial condition is $i(0) = \frac { E} {R}$.

 

[FranzDiCoccio]

Hi ,
and thanks for taking the time for helping me.

What was the current before applying the sinusoidal voltage?

That is kind of my question. Here we are talking about a standard purely inductive load attached to a sinusoidale emf generator.
Perhaps naively I'd say that before applying the sinusoidal voltage there is *no current* in the system.
This does not feel like the correct way of thinking, though, because in the standard solution in general $I(t)\neq 0$ for $t= 0$ (and forever before the initial time, except at discrete times).

The initial condition seems to be that the current has to be maximum. Or better: there is no initial time and the current intensity (absolute value) is maximum when the voltage is zero.

This way, it works. Only, it feels a little "out of the blue"... How do we come up with that condition?

In the books and in the material I've found online the equation is solved by basically ignoring the constant, with no explanation whatsoever.

 

[FranzDiCoccio]

Why would you drop it? It is perfectly valid to have a non-zero $c$. The value of $c$ is determined by your initial conditions. Nothing else.

I do not think saying "let us keep it" solves the problem.
There should be a way of finding its value. Of course this has to do with a boundary condition.
This is precisely my question.
I'm considering a standard textbook circuit with a purely inductive load: a voltage generator and an inductor. What would be causing the constant current?
The solutions I found never keep the constant, and never mention why they drop it. No mention to the initial conditions either.
Perhaps the problem here is so super-obvious that it does not warrant the few drops of ink for an explanation...

 

[FranzDiCoccio]

I gather, although you have not explicitly said so, that you have a coil rotating in a magnetic field with angular frequency $\omega$. That's fine.

Well, that is just me trying to make sense of it. I'm assuming that any alternating voltage from a power plug ultimately comes from a coil rotating in a magnetic field.

You understand that mathematically $c$ is determined mathematically by the initial conditions. OK. When we do physics, $c$ is physically determined by the initial conditions. Let me explain. We have a physical entity here, the rotating coil, and $\omega t$ is the angle by which the coil rotates. Time parameter $t$ is measured with respect to a (usually imaginary) clock that starts clicking when the angle passes the line with respect to which the coil angle is measured. If you choose to draw the line so that the current is maximum when the clock starts clicking, then it will be a maximum at all subsequent time intervals separated by a period $T=\frac{2\pi}{\omega}$. You can also choose to draw that line at an angle of 90^o relative to the one I just mentioned. Then the current will be zero at t = 0 and at all all subsequent time intervals separated by a a period. And if you drew the line in between the previous two lines, the current will be between the maximum value and zero at t = 0.

Wait, it seems to me that here you are just changing the origin of the time axis.
This does not solve my problem, i.e. the value of the constant $c$. It seems to me that you are assuming $c=0$.
Indeed, the current still has the features you describe provided that $|c|$ be not too large.
Its maximum values are separated by $T$. It vanishes at times separated by $T$.
If $|c|$ is too large it never vanishes, though.

The boundary condition for the standard solution is, in my opinion, $I=0$ when $|V|$ is maximum and $|I|$ is maximum when $V=0$ (EDIT: this perhaps shows that the most significant time scale is $T/2$ rather than $T$).
But why do we choose that?
Is this based on a symmetry assumption? Is this based on real life examples? E.g. when I measure the current I see that it oscillates symmetrically about $I=0$?
By the way this is not as simple as saying that the average current or power vanishes. Indeed this is true also for $c\neq 0$.

 

[FranzDiCoccio]

Say you connect a battery E, resistor R and an inductor L in series. (battery + to R+, R- to L+, L- to battery -). Wait a long time. Then i ~ E/R. Now slap your voltage $V_0 sin(\omega t)$ directly across the inductor. Show that the inductor current is

$i = \frac {V_0} {\omega L}(1 - cos(\omega t)) + \frac {E} {R}$. Your initial condition is $i(0) = \frac { E} {R}$.

Uhm. Right, but the standard derivation of a purely inductive circuit never mentions any resistor, and
$$I(0) = \frac{V_0}{\omega L}$$
It seems hard to make contact with your answer.

Perhaps one has to study the transient, and in this sense you are perhaps right that one should include a resistive and perhaps a capacitive load as well.
Then the transient part of the solution will likely approach the "standard" one. Then, possibly, the boundary condition for the purely inductive load would become obvious.

But in the high-school texbooks I've found so far the approach is
1) describe the simple loads (resistive, capacitive, inductive) separately
2) put them together to describe a RLC circuit.

Also, in 1) the boundary conditions are pretty much never discussed.

So far the answer seems to be "$c=0$ because for some (possibly complicated) reason we know it".

 

[vanhees71]

As was several times pointed out before, you can determine $c$ only if you give initial conditions, e.g., assuming you switch on the AC at time $t=0$, so that $I(0)=0$.

Of course the main problem is that this example is somewhat artificial since most probably you have a coil with finite resistance. Then the initial condition plays a role only for a short time since the solution of the homogeneous equation is exponentially damped due to the Ohmic losses because of the resistor. Usually one is interested only in the quasistationary state after these transients have decayed, and that's why many textbooks just set $c=0$ to keep only the part of the solution which osciallates with the frequency of the applied AC. Of course a good textbook should discuss this!

 

[FranzDiCoccio]

As was several times pointed out before, you can determine $c$ only if you give initial conditions, e.g., assuming you switch on the AC at time $t=0$, so that $I(0)=0$.

I'm aware of that. I used "boundary conditions" instead of "initial conditions" in my post because in this specific, perhaps too simple circuit the initial condition did not feel enough for getting rid of $c$.

Of course the main problem is that this example is somewhat artificial since most probably you have a coil with finite resistance. Then the initial condition plays a role only for a short time since the solution of the homogeneous equation is exponentially damped due to the Ohmic losses because of the resistor. Usually one is interested only in the quasistationary state after these transients have decayed, and that's why many textbooks just set $c=0$ to keep only the part of the solution which osciallates with the frequency of the applied AC.

This sorts of agrees with my above reply to "rude man".
So the right answer seems to be "while are analysing the simple case of the purely inductive load, we need to choose a reasonable initial conditions which sets the correct value of the constant c. We are doing this based on a much more complex analysis which tells us that c should be 0".
I'm thinking high-school level here. Students know derivatives and perhaps have a grasp of indefinite integrals.

For sure a finicky student would react to the "standard approach" by asking "shouldn't we have a constant here? What is the reason for ignoring it?".

Of course a good textbook should discuss this!

The ones I found so far simply avoid addressing the problem altogether. At a first glance this seems to be true also for these course notes on the subject by MIT I've found online.
Can you think of a high-school text with a satisfactory answer to this problem?

 

[vanhees71]

For sure a finicky student would react to the "standard approach" by asking "shouldn't we have a constant here? What is the reason for ignoring it?".

In fact this is the kind of students everyone in the business of higher education looks for ;-)).

The ones I found so far simply avoid addressing the problem altogether. At a first glance this seems to be true also for these course notes on the subject by MIT I've found online.
Can you think of a high-school text with a satisfactory answer to this problem?
[/QUOTE]
That's an interesting question. Maybe one should browse the electrical-engineering literature, where you have of course plenty of books treating AC circuit theory in all details. There should be at least some that discuss "transient states".

 

[FranzDiCoccio]

In fact this is the kind of students everyone in the business of higher education looks for ;-)).

Yes, most of the students would not give much importance to the omission of the constant.
Some would, though, especially if they are taking a course in mathematics and the teacher insists that a given function has many primitive functions (or antiderivatives, or indefinite integrals), and they all differ by a constant.
Try discussing the purely inductive circuit and omitting the constant. I expect at least three or four students reminding you that you are forgetting the constant in the indefinite integral. :-)

 

[Dale]

There should be a way of finding its value.

The only way to find its value is with the initial condition as described by @kuruman in post 2. It must be given. It does not come from anywhere else.

So far the answer seems to be "c=0 because for some (possibly complicated) reason we know it".

Ugh, this is not correct. We do not know $c=0$. It could easily be non zero. It comes from the initial condition and only the initial condition.

Have you never used differential equations before? It seems like you don’t understand that initial conditions or boundary conditions must be given.

Perhaps the problem here is so super-obvious that it does not warrant the few drops of ink for an explanation..

It is super obvious but the ink has been provided repeatedly starting with post 2. Why are you ignoring it?

 

[hutchphd]

The general form of the formal solution has an undetermined constant. The specific solution to the problem requires one more constraint. It does not have to be an "initial" condition. You could, for instance, require the average current to be zero for long times.

That is in fact what we require in practical terms, and so the constant is zero. A solution is a solution.

 

[FranzDiCoccio]

The only way to find its value is with the initial condition as described by @kuruman in post 2. It must be given. It does not come from anywhere else.

Ugh, this is not correct. We do not know $c=0$. It could easily be non zero. It comes from the initial condition and only the initial condition.

So let me get this straight. You are saying that in a circuit with an AC generator and an inductor you can have a current that is alternating about a non zero value. I mean, physically?
Can you give me an example?

Have you never used differential equations before? It seems like you don’t understand that initial conditions or boundary conditions must be given.

It is super obvious but the ink has been provided repeatedly starting with post 2. Why are you ignoring it?

No, I'm not ignoring it. Perhaps I'm not making myself clear. Please bear with me, I'm not a native English speaker. I am perfectly aware that c is determined by the initial condition. The mathematical aspects are pretty clear to me. Here we are talking of a specific problem: an AC circuit with a purely inductive load.

In every discussion I've found (in high school textbooks, in the course notes by MIT I attached above), there is no mention of the initial conditions that allow us to conclude $c=0$.
It is more than obvious that choosing
$$I(0) = \frac{V_0}{\omega L}$$
sets $c=0$. This is mathematically trivial.
But what is the physical intuition for making this choice of the initial condition?
So far it seems to me that the answer is "because I do not want to see that $c=0$ around".

That is: it seems to me that the initial condition is (implicitly) chosen to get rid of $c\neq 0$ rather than from a physical consideration. And emphasis over "implicitly", because basically nowhere I find a discussion of this.
Do you see my problem?

 

[FranzDiCoccio]

The general form of the formal solution has an undetermined constant. The specific solution to the problem requires one more constraint. It does not have to be an "initial" condition. You could, for instance, require the average current to be zero for long times.

That is in fact what we require in practical terms, and so the constant is zero. A solution is a solution.

I do agree that it has not to be an "initial" condition. That is why I used the more generic term "boundary condition". But I'm all right with "constraint" too.

So your answer is "we want the average of the current to be zero, and hence c=0".
Ok, this is not very different from what I was proposing earlier: the current should oscillate symmetrically about 0.
Probably a high school student would be happy with this.

 

[Dale]

So let me get this straight. You are saying that in a circuit with an AC generator and an inductor you can have a current that is alternating about a non zero value. I mean, physically?
Can you give me an example?

Sure. The easiest example would be your same circuit but with a programmable generator. If you switch phase or frequency at t=0 then you can have it oscillate about a non zero voltage. Or, better yet, if you can switch your generator from a constant current source to the oscillating voltage source then it will oscillate about whatever constant current you had originally set.

But what is the physical intuition for making this choice of the initial condition?

I don't think there is one. Any condition is physically possible (within the limits of the material). I suspect it is a pedagogical choice. They didn’t want to spend class time on that so they just set it to zero (a perfectly valid solution) to move on with class. They thought it would be distracting to whatever point they were focused on.

You shouldn’t read too much into such things. For clarity it is often necessary to focus on one topic and to ignore others that are involved, but not the immediate focus of that specific section. Just because it is set to zero in the examples you have read does not imply that it must be zero always.

 

[vanhees71]

The general form of the formal solution has an undetermined constant. The specific solution to the problem requires one more constraint. It does not have to be an "initial" condition. You could, for instance, require the average current to be zero for long times.

That is in fact what we require in practical terms, and so the constant is zero. A solution is a solution.

Well, sometimes the question is whether it's the complete solution and often the application of the correct initial and/or boundary conditions are the most subtle point in finding a solution (e.g., it's a profound question why to choose the retarded rather than the advanced or anything else in between as a solution of equations, where the equation doesn't prefer a direction of time ;-)).

 

[hutchphd]

But of course such considerations in the real world will be superseded by when the coal runs out (or when the peasants ransack the power facility)...!..

 

[FranzDiCoccio]

Sure. The easiest example would be your same circuit but with a programmable generator. If you switch phase or frequency at t=0 then you can have it oscillate about a non zero voltage. Or, better yet, if you can switch your generator from a constant current source to the oscillating voltage source then it will oscillate about whatever constant current you had originally set.

So you're saying that you're starting with the "usual" current, oscillating about 0, and then you suddenly change the phase or the frequency when the current is nonzero.
That should force an oscillation about a nonzero background. I sort of see that.
And, assuming we do that, we should be able to measure a nonzero average current, right?

Also... What if we have the purely inductive load circuit and "close" it (if you like, "plug it" to the voltage source) at an arbitrary time? Before doing this, there is no current in the circuit. Does this mean that
$$c= \frac{V_0}{\omega L}$$
so that $I(0)=0$?
I am not sure about this. It does not feel right. If it's not, what would be?

I do not really understand the second part. I'm no expert, but I think that a current generator is a different thing.

I suspect it is a pedagogical choice. They didn’t want to spend class time on that so they just set it to zero (a perfectly valid solution) to move on with class. They thought it would be distracting to whatever point they were focused on.

You shouldn’t read too much into such things. For clarity it is often necessary to focus on one topic and to ignore others that are involved, but not the immediate focus of that specific section. Just because it is set to zero in the examples you have read does not imply that it must be zero always.

What if a curious student asks about the constant? Is the answer "you're right, this is just one of the many possible solutions. We choose this one because it is the simplest one, since it is 'as symmetric as the voltage' and has a zero average"?

 

[hutchphd]

We choose it because it matches our particular reality (constraints). We can do that!

 

[Dale]

So you're saying that you're starting with the "usual" current, oscillating about 0, and then you suddenly change the phase or the frequency when the current is nonzero.
That should force an oscillation about a nonzero background. I sort of see that.
And, assuming we do that, we should be able to measure a nonzero average current, right?

Yes, that is right.

Also... What if we have the purely inductive load circuit and "close" it (if you like, "plug it" to the voltage source) at an arbitrary time? Before doing this, there is no current in the circuit. Does this mean that
$$c= \frac{V_0}{\omega L}$$
so that $I(0)=0$?
I am not sure about this. It does not feel right. If it's not, what would be?

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.

Basically, you need some additional piece of information (initial condition, boundary condition, constraint, whatever) that you can use to determine $c$. Any value of $c$ is valid (for ideal components).

I'm no expert, but I think that a current generator is a different thing.

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.

Is the answer "you're right, this is just one of the many possible solutions. We choose this one because it is the simplest one, since it is 'as symmetric as the voltage' and has a zero average"?

That is what I would say. Or maybe just "... because it is convenient".

 

[Dale]

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.

 

[vanhees71]

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 $\varphi_0=\pi/2$.

 

[FranzDiCoccio]

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
$$c = \frac{V_0}{\omega L} \sin(\varphi)$$
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.

 

[FranzDiCoccio]

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).

 

[Dale]

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
$$c = \frac{V_0}{\omega L} \sin(\varphi)$$

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.

 

[rude man]

Uhm. Right, but the standard derivation of a purely inductive circuit never mentions any resistor, and
$$I(0) = \frac{V_0}{\omega L}$$
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.