# Simple AC circuit

#### MrAl

Joined Jun 17, 2014
9,638
Hi,

Inductors can have no voltage across them but still have current in the same way that capacitors can have some voltage across them with no current flowing.

Here is a quick hint:

The inductor is the dual of the capacitor in that the roles of current and voltage are swapped as are the roles of the through and across variables.

The ideal capacitor where there is zero current through it gains or looses no voltage, so what does the ideal inductor do when there is zero voltage across it?

Another approach is to find the limit of:
i*e^(-t*R/L)

as R approaches zero Also note that the expression:
V/R*e^(-t*R/L)

would be invalid for this problem.

Last edited:
• cdummie

#### cdummie

Joined Feb 6, 2015
124
Hi,

Inductors can have no voltage across them but still have current in the same way that capacitors can have some voltage across them with no current flowing.

Here is a quick hint:

The inductor is the dual of the capacitor in that the roles of current and voltage are swapped as are the roles of the through and across variables.

The ideal capacitor where there is zero current through it gains or looses no voltage, so what does the ideal inductor do when there is zero voltage across it?

Another approach is to find the limit of:
i*e^(-t*R/L)

as R approaches zero Also note that the expression:
V/R*e^(-t*R/L)

would be invalid for this problem.
Let's see, voltage across inductor is zero, so it won't gain or lose any current so current i(t=0+)=i(t=0-)=IgRt/L assuming that approach and calculations in my previous post were correct.

#### MrAl

Joined Jun 17, 2014
9,638
Let's see, voltage across inductor is zero, so it won't gain or lose any current so current i(t=0+)=i(t=0-)=IgRt/L assuming that approach and calculations in my previous post were correct.

Hello,

Well, after a long time the inductor looks like a short circuit to DC current. So what's the current through the inductor after a long time with the switch open?

Note that if you have a short circuit in parallel with a resistance all the current flows through the short circuit.

Using the previous information, then what is the current after the switch is closed?

#### WBahn

Joined Mar 31, 2012
27,939
Let's see, i'll give it a shot

for i(t=0-) switch is still off so current through the inductor is:

IgR=Ldi/dt

di=IgRdt/L

i=IgR/L∫dt = IgRt/L is this correct?

for i(t=0+) switch is on and since there's no voltage on inductor, there no current either.
Where does IgR come from? You are taking Ohm's Law and just throwing the nearest I and the nearest R at it and hoping something sticks. Ohm's Law is only valid if you use the current that is actually flowing through THAT resistor!

Before the switch is changed, is anything changing in the system? If not, the what is di/dt in the inductor? What does that tell you about the voltage across the inductor? What does that tell you about the voltage across the resistor? What does that tell you about the current in the resistor? What does that tell you about the current in the inductor?

And the notion that because there is no voltage across the inductor means that there is no current in the inductor is completely bogus and shows that you don't understand the fundamentals of inductors. The defining relationship for inductance is

v = L di/dt

There is NOTHING there that implies that if v = 0 that i = 0. NOTHING!

#### cdummie

Joined Feb 6, 2015
124
Where does IgR come from? You are taking Ohm's Law and just throwing the nearest I and the nearest R at it and hoping something sticks. Ohm's Law is only valid if you use the current that is actually flowing through THAT resistor!

Before the switch is changed, is anything changing in the system? If not, the what is di/dt in the inductor? What does that tell you about the voltage across the inductor? What does that tell you about the voltage across the resistor? What does that tell you about the current in the resistor? What does that tell you about the current in the inductor?

And the notion that because there is no voltage across the inductor means that there is no current in the inductor is completely bogus and shows that you don't understand the fundamentals of inductors. The defining relationship for inductance is

v = L di/dt

There is NOTHING there that implies that if v = 0 that i = 0. NOTHING!
I am sorry, i forgot that i have R too, i mean, i did following:

i changed Ig with corresponding voltage generator and it's IgR with resistance R, only, i forgot to add that resistance to the equation, with that correction, i should have:

IgR = Ldi/dt + Ri

Ldi/dt = Ri - IgR

di/dt=Ri/L - IgR/L

di/i= Rdt/L - IgRdt/Li and i am stuck here, i don't know how to integrate this current by dt, since i don't know in which way it is dependent on t, anyway, is this any better?

I would really appreciate explanation on the rest of your questions since i am bit confused now.

#### cdummie

Joined Feb 6, 2015
124
Hello,

Well, after a long time the inductor looks like a short circuit to DC current. So what's the current through the inductor after a long time with the switch open?

Note that if you have a short circuit in parallel with a resistance all the current flows through the short circuit.

Using the previous information, then what is the current after the switch is closed?
Well, i am bit confused, since i don't understand what happens with Ig when switch closes, can you explain me that?

#### WBahn

Joined Mar 31, 2012
27,939
I am sorry, i forgot that i have R too, i mean, i did following:

i changed Ig with corresponding voltage generator and it's IgR with resistance R, only, i forgot to add that resistance to the equation, with that correction, i should have:

IgR = Ldi/dt + Ri

Ldi/dt = Ri - IgR

di/dt=Ri/L - IgR/L

di/i= Rdt/L - IgRdt/Li and i am stuck here, i don't know how to integrate this current by dt, since i don't know in which way it is dependent on t, anyway, is this any better?

I would really appreciate explanation on the rest of your questions since i am bit confused now.
Again, what is this Ig·R term. It would appear to be the voltage drop across R. But that is ONLY the case IF the current flowing THROUGH that R happens to be Ig. On what basis are you making that claim?

PLEASE, just answer my questions one by one. Since I'm making zero head way with asking you a serial chain of questions. Let's take if from the top and work through it one question at a time.

Q1) Just before the switch changes, the circuit is in stead state. So is there anything (any voltages or any currents) that are changing just before the switch is closed?

Forget about everything else and focus on JUST that ONE question.

#### MrAl

Joined Jun 17, 2014
9,638
Well, i am bit confused, since i don't understand what happens with Ig when switch closes, can you explain me that?
Hi,

Well dont feel too bad because this is a sort of a little tricky question, one that we usually just find on tests and rarely in real life, and that is because we are dealing with an ideal inductor. Had the inductor been a real inductor, there would be a second resistor in series with the inductor itself. That may help you understand this better because then you can first analyze it with a series resistor of some low value like 0.1 ohms and then let that get smaller like 0.01 ohms and see what happens as you do that.

When you say Ig do you mean the original source or the current in the inductor?
The original current source is short circuited as is the inductor, so both have zero ohms across them. The inductor had a previous current of lg amps because it too became a short circuit because of the fact that the switch was open for such a long time. So the current in the source is known, and the current in the inductor is known, so all you have to do is figure out what happens to the current in the inductor.

If it helps any, this problem is an initial value problem, and that means the inductor has an initial current through it at t=0 which is when the switch finally closes. An initial value in an inductor can be represented by an inductor with zero current flowing through it with an 'initial current generator' placed in parallel with it. The initial current generator has current equal to what the original inductor had through it, but now it appears outside of the inductor. So now you have a plain inductor with no current through it plus another constant current source in parallel with it. So when the switch closes, you are effectively shorting out an inductor with zero current through it plus shorting out a current source. You can write an equation with this in mind or just think about it. What happens when you short circuit a current source (the second current source now). Note the current source is also a constant current source just like the original one Ig.

Hope that helps a little anyway.

#### WBahn

Joined Mar 31, 2012
27,939
Hi,

Well dont feel too bad because this is a sort of a little tricky question, one that we usually just find on tests and rarely in real life, and that is because we are dealing with an ideal inductor. Had the inductor been a real inductor, there would be a second resistor in series with the inductor itself. That may help you understand this better because then you can first analyze it with a series resistor of some low value like 0.1 ohms and then let that get smaller like 0.01 ohms and see what happens as you do that.
And most hospitals have one of these circuits in real life -- they call it an MRI machine. The ideal inductor is usually called a superconducting magnet. The switch is called the persistent-mode switch, and the resistor (which is often several power diodes in series/parallel) is called the quench sump (which is where the energy gets dumped if the magnet should quench). For normal operation, the resistor (particularly if it's actually a diode bank) never has any appreciable current flowing in it. With the switch open you ramp of the current source until you get to the desired current. Then you close the switch (by turning off a heater so that it can go superconducting). Then you remove the current source and close up the unit.

If it helps any, this problem is an initial value problem, and that means the inductor has an initial current through it at t=0 which is when the switch finally closes. An initial value in an inductor can be represented by an inductor with zero current flowing through it with an 'initial current generator' placed in parallel with it. The initial current generator has current equal to what the original inductor had through it, but now it appears outside of the inductor. So now you have a plain inductor with no current through it plus another constant current source in parallel with it. So when the switch closes, you are effectively shorting out an inductor with zero current through it plus shorting out a current source. You can write an equation with this in mind or just think about it. What happens when you short circuit a current source (the second current source now). Note the current source is also a constant current source just like the original one Ig.
By the description you have given here, there is no reason for any current to flow in the inductor after the switch closes -- but that's not what is going to happen.

#### MrAl

Joined Jun 17, 2014
9,638
And most hospitals have one of these circuits in real life -- they call it an MRI machine. The ideal inductor is usually called a superconducting magnet. The switch is called the persistent-mode switch, and the resistor (which is often several power diodes in series/parallel) is called the quench sump (which is where the energy gets dumped if the magnet should quench). For normal operation, the resistor (particularly if it's actually a diode bank) never has any appreciable current flowing in it. With the switch open you ramp of the current source until you get to the desired current. Then you close the switch (by turning off a heater so that it can go superconducting). Then you remove the current source and close up the unit.

By the description you have given here, there is no reason for any current to flow in the inductor after the switch closes -- but that's not what is going to happen.
Hello there,

Thanks for the reply, but i dont understand your last sentence at all and here is why...

An initial value problem in electrical work involving inductors or capacitors is one where the component has some initial energy storage, or more simply put, an inductor has some current flowing before t=0 and a capacitor would have some initial voltage across it before t=0. I am almost sure you are well aware of this but just stating for clarity. This condition is often modeled by placing an initial condition generator in the circuit and reckoning the original component as having zero energy just before t=0.

That means that if there is current already flowing in an inductor before t=0 there's no question already about whether or not there is current in the inductor.

Where that current originally came from would have come from the analysis just before that point, and for this problem that is when the inductor looked like a short circuit in parallel with a current source. The short circuit trumps the resistor because after all it is zero ohms and the resistor value is finite so the inductor draws the entire current while the switch is still open, after some long time period where the period is determined by the values in the circuit (like the way the time constant is determined).

So at that point we would have had to know the current through the inductor already, so when the switch closes we know the initial current, and that is where the initial value problem idea comes from. Many times it's not just the initial value that drives the circuit but is also the forcing function, but in this case the forcing function goes to zero because the current source gets shorted out so the response (after that point in time) is due entirely to the initial value of the current in the inductor.

I hope that explains my point a bit better.

#### WBahn

Joined Mar 31, 2012
27,939
The source that is put in parallel with the inductor to account for the initial inductor current is not a constant current source, it is an impulse current source that is sized so that, at t=0, it "bangs" the inductor with an infinite voltage just enough to set the inductor current to the value it had at t=0-.

The problem here is that you have this impulse current source in parallel with both an inductor AND a short circuit. So the current can go either place and there is nothing mathematically to determine which. Conceptually the current should all go through the short and none of it into the inductor.

#### cdummie

Joined Feb 6, 2015
124
Again, what is this Ig·R term. It would appear to be the voltage drop across R. But that is ONLY the case IF the current flowing THROUGH that R happens to be Ig. On what basis are you making that claim?

PLEASE, just answer my questions one by one. Since I'm making zero head way with asking you a serial chain of questions. Let's take if from the top and work through it one question at a time.

Q1) Just before the switch changes, the circuit is in stead state. So is there anything (any voltages or any currents) that are changing just before the switch is closed?

Forget about everything else and focus on JUST that ONE question.
Well, i guess no, there's no AC generator, so there are no changes.

#### MrAl

Joined Jun 17, 2014
9,638
The source that is put in parallel with the inductor to account for the initial inductor current is not a constant current source, it is an impulse current source that is sized so that, at t=0, it "bangs" the inductor with an infinite voltage just enough to set the inductor current to the value it had at t=0-.

The problem here is that you have this impulse current source in parallel with both an inductor AND a short circuit. So the current can go either place and there is nothing mathematically to determine which. Conceptually the current should all go through the short and none of it into the inductor.
Hi again,

I am not seeing your reasoning here...
I am not sure why you would want to use an impulse source, because that's not necessary and it defeats the purpose of using an initial current generator in the first place. The purpose of this constant current generator is so that we can remove the current from the inductor and place it outside the inductor, conceptually, so that we can simplify the analysis. If we had to use an impulse source we would just be putting the initial current right back into the inductor at t=0 and that would not help because we already had that before we decided to use an initial current generator. This constant current generator would then be a constant current source which is just a normal current source with constant value of current like say 1 amp. It starts out at 1 amp at t=0 and stays at 1 amp forever, just like an ideal battery voltage source would have say 1v forever. The equations write out the same way as when we use any ideal source in a circuit.

Now reasoning out what happens with an inductor in parallel with this current source and also a short circuit in parallel with this (three things in parallel) is easy because any impedance like s*L in parallel with a short circuit yields a zero impedance, and since that zero impedance is in parallel with the current source, all the current flows through the short circuit. BUT, that current was really the intial current in the inductor, we just moved it outside conceptually not physically, and since the current does not change then the current through the inductor does not change. And since the current source is a constant current source, it must remain constant forever.

It is not uncommon at all to use a constant current generator in parallel with an inductor as the initial current generator for the initial current in the inductor. If the current at t=0 was 1 amp then we make it 1 amp, if it was 2 amps then we make it 2 amps, if 5.7 amps then we make it 5.7 amps etc., etc. Not much mystery left here What might seem odd is that the current NEVER goes away, even after t=infinity, because it is constant. The response can die out, but only due to the other components in the circuit like say an actual non zero resistor across the inductor (and current source). The math is then very very simple too, because we have just a resistor in parallel with an inductor with zero current though it at t=0 and in parallel with both of those we have a constant current source. At t=0 all the current flows through the resistor, so we see a voltage across the inductor at t=0, and over time the inductor starts to conduct current from the constant current source, and eventually it conducts ALL of the current from that source so the voltage goes to zero. The current through the fake inductor is the negative of the constant current source and the current through the physical inductor is the sum of the constant current source and the fake inductor current, which comes out to zero.
So for example, if the current in the inductor was 1 amp at t=0, we replace that with an inductor with zero amps through it and place a constant current source in parallel with it, and that means the resistor gets all the current at t=0. After much time has passed, the inductor gets all the current, but this is the fake inductor now not the real inductor. So the fake inductor has -1 amp through it. The constant current generator is 1 amp and the fake inductor has -1 amp, but the physical inductor is now made of the fake inductor and the constant current source, so the physical inductor current is the sum which comes out to zero.
So the thing to remember here is that the transformed model contains a new inductor plus a constant current source, and they BOTH make up the real physical inductor so the current through the fake inductor and the current source add together as the solution to the current in the real inductor. And in the end the fake inductor still has current flow, but it's only one part of the model that makes up the physical inductor so that doesnt mean the real inductor has current.

I think this helps clear up the view of the constant current initial current generator, but it may seem a little strange at first. A better view is using a constant voltage source as the initial voltage for a capacitor. If we write the equation for a resistor in series with a capacitor in series with a power supply in series with a battery as the initial condition generator, we end up with a circuit that simply has two constant voltage sources but keeping in mind that the initial voltage generator voltage and the cap voltage add together to show the true voltage across the true capacitor, because the model of the cap is then BOTH the cap AND a battery, so the voltage across the true cap would be the voltage across the two terminals of the model for the cap, which is a cap in series with a battery, not just the cap itself anymore.
For example, if the cap had 1v across it and the power supply was 10v, then the new cap has 0v and the battery is 1v. The new cap has zero volts across it but the physical cap has 1v across it, which is represented by the battery (ideal voltage source). In any solution using this model the true cap voltage is the Vfakecap+Vbatt.

As i was saying before this way of doing things is not that uncommon and we could probably find examples around the web for both the cap and inductor. There are other ways of doing it too, but the way i showed seems to be the simplest because we just have to introduce one more independent constant source, and independent constant sources are not that hard to deal with because they enter into the equations just like any other independent constant source like a power supply or battery (ideal source that is).

I hope i have explained this a little better this time. It might seem strange but it simplifies the analysis and intuitive view of circuits with inductors and caps that have initial conditions.

Here is an illustration: Last edited:

#### WBahn

Joined Mar 31, 2012
27,939
You say that the resistor initially (t = 0+) has all the current flowing in it and that, after a long time, the real inductor has no current flowing in it. Both of these are wrong. The resistor in that circuit never has any current flowing in it and the inductor has a constant current flowing in it for all time t > 0 that is equal to the current that was flowing in it before the switch was closed.

#### MrAl

Joined Jun 17, 2014
9,638
You say that the resistor initially (t = 0+) has all the current flowing in it and that, after a long time, the real inductor has no current flowing in it. Both of these are wrong. The resistor in that circuit never has any current flowing in it and the inductor has a constant current flowing in it for all time t > 0 that is equal to the current that was flowing in it before the switch was closed.

Hi there,

I am not sure what topology you are talking about here, but be aware that i was not only talking about the circuit in question in this thread but also for a circuit with a resistor and inductor only where the inductor has initial current at t=0. I did that because the single inductor single resistor circuit is easier to analyze than the one that introduces a zero ohm impedance into the circuit too.

So in other words, with a single resistor and single inductor where the inductor has 1 amp flow through it at t=0, the resistor must have voltage across it because there is a current flow already. So with 1 amp initial current and a 2 ohm resistor, the inductor current creates a voltage across both elements of 2 volts.

Now in terms of the inductor initial current generator and virtual inductor, at t=0 the virtual inductor has zero current through it, and the constant current generator has 1 amp through it, so the resistor has 1 amp through it at t=0 (the virtual inductor does not conduct current at t=0 because it can not respond instantaneously). So at t=0 the resistor and virtual inductor have 2 volts across both, as does the constant current generator.

After a little time has passed, the virtual inductor starts to draw current from the source. This causes the voltage across both to come down lower and lower as time goes on. Eventually the virtual inductor draws all the current of 1 amp and the device looks like a short circuit, so the resistor has no voltage and thus no current, and so all of the current from the CC source goes through the inductor. The CC source puts out 1 amp still, but now the virtual inductor conducts all of that current. The resistor now has no voltage across it.

If we look at the current flowing out of the combined leads of the virtual inductor and CC source however, we see zero current because the two currents sum to zero. Thus the physical inductor has zero current now.

Using the physical inductor view only, we see 1 amp initial current through it and also the resistor, so the voltage is 2v again. The current of 1 amp flows through the resistor and eventually the current gets lower and lower due to the normal response of an inductor with initial current in series with a resistor (no source):
I(t)=I0*e^(-a*t)

and that is the same exact response we see using the virtual inductor and CC source but remembering to treat the virtual inductor and CC source as ONE physical device when we want to know the real life current that we would measure if we measured the current in a real life inductor.

The circuit with the switch is more complicated so i thought starting with that simpler circuit would help show how the CC source and virtual inductor help simplify the analysis.

For the virtual inductor and CC source the analysis is then just like any other circuit:
V(s)=(I/s)*(sL*R)/(sL+R)

where as usual the impedance of the inductor is s*L abbreviated as just "sL".
The difference again though is that the real life inductor current is the sum of both the current in the inductor "L" in the above equation and the CC source current, taking current flow direction into account.

If you still dont agree i could set up a plot of the current and voltages using both methods to show that they both yield the same results for all time if you'd like to see some proof.

After all is said and done this makes things much simpler, as long as we keep in mind the way the currents have to sum to get the real life current. This is mostly because we can break the inductor into two distinct parts that are both known individual circuit elements (one inductor with zero current and one CC source) rather than trynig to deal with an inductor with initial current combined. Note that the direct use of "s*L" as the impedance says nothing about the initial current in the real inductor, but after breaking it into two parts "s*L" then only has to refer to the virtual inductor which has NO current flow through it so we no longer have that difficulty.
Of course there are other ways to handle the initial current, but initial current generators are one way to do it at least. I can promise that although we had to talk about this for some time, it does in the end help to simplify the analysis. We end up being able to draw the circuit with just zero current inductors and zero voltage capacitors and several initial condition generators that are either constant current sources or constant voltage sources. We can then proceed to analyze the circuit using the same methods we always use such as using Laplace transforms for all the elements.

The two circuits are shown in the attachment. If we look at the current in the resistor over time we see the same response in both circuits. Last edited:

#### WBahn

Joined Mar 31, 2012
27,939
I'd be interested in seeing the approach you describe applied to THIS thread's problem.

#### MrAl

Joined Jun 17, 2014
9,638
I'd be interested in seeing the approach you describe applied to THIS thread's problem.
Hello again,

Ok. I think i see why you persisted with this. It is not the ordinary kind of question because we have a short circuit which complicates things, but only a little. It means we really have to stick to the definitions to the letter to find out what happens. It's like being a lawyer for electronic circuits I used a non zero resistance with the real inductor first so the basic idea could be clearly illustrated. Of course i realize many people already know that, but i wanted to make sure anyone reading this despite their background could follow it too. I cant always do that but i wanted to this time. I will restate some things again too just for clarity and completeness.

What else i am hoping is that the OP made some progress by this time, regardless what method they choose to use. Now on to the analysis with the short circuit...

In the original circuit when the switch closes, the physical inductor (which i will call the real inductor) already has current flow through it. That is because of the final value we obtain from the previous topology (switch open). The final value of the current is thus applied as the initial value of the next topology (switch closed). So when the switch is closed at t=0 the initial value of current in the real inductor is the same as before i guess we can call it iL.

But to use the initial condition generator (which is supposedly a constant current generator) we have to make a change in topology again. We have to replace the real inductor with a model for an inductor with initial current using the following basic definitions:
1. We place an virtual inductor in parallel with a current source.
2. The current source is a constant current source with the same current amplitude and direction of current in the same direction as the original real inductor had, as if this current was measured in the leads of the real inductor.
3. The virtual inductor has ZERO current through it.

So let's look at what we have so far as viewed from the leads of the real inductor and the combined lead made from the connections of the virtual inductor and current source...
At t=0, the current out of the real inductor is iL (which could be 1 amp for example).
At t=0, the combined leads of the model has a current out that is also equal to iL.
So if the current was 1 amp before, it's still 1 amp. The only difference is we have to measure in the combined leads of the virtual inductor and current source, not just in the virtual inductor, and that is because the virtual inductor is just part of the model of the real inductor now.

Now very quickly we jump back to the circuit with a non zero resistance in it. What would happen at t=0+.
ALL the current from the current source would flow through that resistance, and a voltage would be immediately developed. That would place a voltage across the virtual inductor immediately, and so current would start to ramp up in the virtual inductor as it 'steals' some of the current little by little.
But we havent gotten that far in time yet, we are still at t=0. What is the virtual inductor current at t=0. Even with a non zero resistance it is zero amps. So the virtual inductor current is zero at t=0, and that is by definition of the model and by the fact that the current can not increase instantaneously in an inductor with a finite voltage across it...instead it must ramp up.

Now we jump back to the circuit with the short circuit. What happens at t=0.
The inductor has zero current through it (by definition).
The current source current must therefore be conducted ALL by the short circuit (no voltage, no time, equals zero current v=L*di/dt, or di=v*dt/L, so di=0*0/L=0).
So we now have a circuit with a constant current source in parallel with a short circuit. What is the current in the shorting switch. It is iL. What is the current in the real inductor. The real inductor current is the sum of the virtual inductor current plus the current source current, the total of which is iL. If iL was 1 amp, then it is 1 amp now also.

I can draw up a diagram if needed a bit later.

#### cdummie

Joined Feb 6, 2015
124
Again, what is this Ig·R term. It would appear to be the voltage drop across R. But that is ONLY the case IF the current flowing THROUGH that R happens to be Ig. On what basis are you making that claim?

PLEASE, just answer my questions one by one. Since I'm making zero head way with asking you a serial chain of questions. Let's take if from the top and work through it one question at a time.

Q1) Just before the switch changes, the circuit is in stead state. So is there anything (any voltages or any currents) that are changing just before the switch is closed?

Forget about everything else and focus on JUST that ONE question.

Sorry it took me so long to respond to this, but i think that i could give it a shot now, so:

since circuit is in stead stage there's no changes in the circuit just before switch closes.

In one of the previous questions you've asked me about current through inductor in the stead stage before closing switch and that current is equal to Ig, right?

Now, after switch is closed voltage on inductor is zero, and i know that u(t)=Ldi/dt so Ldi/dt=0 but how can i determine i(t) here, when t>0 ?

#### WBahn

Joined Mar 31, 2012
27,939
Sorry it took me so long to respond to this, but i think that i could give it a shot now, so:

since circuit is in stead stage there's no changes in the circuit just before switch closes.

In one of the previous questions you've asked me about current through inductor in the stead stage before closing switch and that current is equal to Ig, right?

Now, after switch is closed voltage on inductor is zero, and i know that u(t)=Ldi/dt so Ldi/dt=0 but how can i determine i(t) here, when t>0 ?

What is the basis for the claim that after the switch is closed that the voltage across the inductor is zero?

What is the continuity condition for an inductor.

#### cdummie

Joined Feb 6, 2015
124 