Simple AC circuit

WBahn

Joined Mar 31, 2012
27,880
Ah, it is I that looked too quickly. You are correct, the voltage across the inductor is 0 V once the switch is closed. I now recall going through this previously earlier in the thread.

The continuity condition for an inductor and a capacitor is a consequence of the conservation of energy, which says that since energy can neither be created nor destroyed but only changed to other forms, that the magnetic energy stored in an inductor (and the electric energy stored in a capacitor) cannot change instantaneously. Since the magnetic energy stored in an inductor is a function of current, this says that the current in an inductor cannot change instantaneously. Thus, the current in the inductor immediately after the switch is closed is the same as it was immediately before. Similarly, the voltage across a capacitor must be continuous.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Ah, it is I that looked too quickly. You are correct, the voltage across the inductor is 0 V once the switch is closed. I now recall going through this previously earlier in the thread.

The continuity condition for an inductor and a capacitor is a consequence of the conservation of energy, which says that since energy can neither be created nor destroyed but only changed to other forms, that the magnetic energy stored in an inductor (and the electric energy stored in a capacitor) cannot change instantaneously. Since the magnetic energy stored in an inductor is a function of current, this says that the current in an inductor cannot change instantaneously. Thus, the current in the inductor immediately after the switch is closed is the same as it was immediately before. Similarly, the voltage across a capacitor must be continuous.

Ok, so in the moment right after switch closes (t=0+) current in the inductor is still Ig. But what happens after that? I mean, usually, when i have this kind of circuits i use kirchoff's laws in order to write a differential equation so i could find current as the function of time. But, now switch is closed so it confuses me. How can i use Kirchoffs law here and how can i find i(t) t>0?
 

MrAl

Joined Jun 17, 2014
9,633
Hello again,

Did you read over the previous posts? I am starting to think that maybe you are not ready for the level of detail required in order to understand how this works. But let's try one more time.

Forget about the circuit. Just take a circuit with inductor with a short across it, and some initial current say 1 amp through the inductor. What happens to the current as time goes on?

One of the basic inductor equations is:
v=L*di/dt

and solving for di we get;
di=(v/L)*dt

Now for our problem v is also a constant so let's write:
di=(V/L)*dt

With a short across the inductor, the voltage is zero, so we have:
di=(0/L)*dt

and from this we can see that it doesnt matter what the value of L is as long as it is above zero, so we get:
di=0*dt

and simpler:
di=0

So the CHANGE in the current through the inductor is ZERO.

The change in the current would be something that is added to the current that is already in the inductor to find the new current:
Inew=Iold+di

but in this case di is zero because the voltage is zero. So what is the new current Inew here?
 
Last edited:

WBahn

Joined Mar 31, 2012
27,880
Ok, so in the moment right after switch closes (t=0+) current in the inductor is still Ig. But what happens after that? I mean, usually, when i have this kind of circuits i use kirchoff's laws in order to write a differential equation so i could find current as the function of time. But, now switch is closed so it confuses me. How can i use Kirchoffs law here and how can i find i(t) t>0?
We've tried to lead you to the answer many times before. Let's try again. First and foremost, answer the following question:

What is the rate of change of the current in an inductor if the voltage across it is zero volts?
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Hello again,

Did you read over the previous posts? I am starting to think that maybe you are not ready for the level of detail required in order to understand how this works. But let's try one more time.

Forget about the circuit. Just take a circuit with inductor with a short across it, and some initial current say 1 amp through the inductor. What happens to the current as time goes on?

One of the basic inductor equations is:
v=L*di/dt

and solving for di we get;
di=(v/L)*dt

Now for our problem v is also a constant so let's write:
di=(V/L)*dt

With a short across the inductor, the voltage is zero, so we have:
di=(0/L)*dt

and from this we can see that it doesnt matter what the value of L is as long as it is above zero, so we get:
di=0*dt

and simpler:
di=0

So the CHANGE in the current through the inductor is ZERO.

The change in the current would be something that is added to the current that is already in the inductor to find the new current:
Inew=Iold+di

but in this case di is zero because the voltage is zero. So what is the new current Inew here?

Well, if there's no change of current then the new current should be Ig, does this means that i=Ig for every t>0 ?
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
We've tried to lead you to the answer many times before. Let's try again. First and foremost, answer the following question:

What is the rate of change of the current in an inductor if the voltage across it is zero volts?

EWell, as MrAl in his previous post stated u=Ldi/dt since u is zero, it means that di/dt is zero, derivative of constant is zero, so current is not dependent on time and it is equal to Ig, right?
 

MrAl

Joined Jun 17, 2014
9,633
Hello again,

In a word, "yes" :)

I guess it takes time to become accustomed to thinking about these components in this way. Initial values do make the analysis a little more difficult, but after you do a few of these you'll see the way this works.

Inductors are basically constant current genrators for very short periods of time, and in this case since there was no voltage across it that means it's going to be a constant current generator for all time.

It's also not easy to think intuitively about the constant current generator in parallel with the inductor so that the current generator completely takes over the role of the initial current, but again once you do a few it becomes very apparent why this works so well. It's such a general method and it allows separating the initial current from the inductor itself, and breaking elements into their basic constituents is always a good idea because then we can follow simpler rules, even though there may be more of them.

There are other ways to handle these circuits too but i dont know what you have studied so far. it's good to stick with the course work that's been given so far.

The capacitor is the 'dual' of the inductor, and so for that element it acts like a constant voltage source for a very short period of time.

For the inductor with a short across it, it acts like a constant current source forever, and the capacitor with an open circuit acts like a constant voltage source forever. These are ideal elements however, as in practice we usually see at least some resistance in series with the inductor because it is made out of wire that has resistance, and for the capacitor we usually see at least some resistance in parallel with it because the dielectric conducts a small but measurable current. Over time, either of these devices would discharge in the real world, but the ideal elements are taken to be completely free of any energy dissipators.
 

WBahn

Joined Mar 31, 2012
27,880
Well, if there's no change of current then the new current should be Ig, does this means that i=Ig for every t>0 ?
YES!

Now go back and look at Post #17

http://forum.allaboutcircuits.com/threads/simple-ac-circuit.120808/#post-966228

And some of the other posts in the vicinity and try to figure out why you were stumbling with this notion back then. If you can identify it, then you can cement a significant insight into your reasoning that was previously missing.
 

WBahn

Joined Mar 31, 2012
27,880
For the inductor with a short across it, it acts like a constant current source forever, and the capacitor with an open circuit acts like a constant voltage source forever. These are ideal elements however, as in practice we usually see at least some resistance in series with the inductor because it is made out of wire that has resistance, and for the capacitor we usually see at least some resistance in parallel with it because the dielectric conducts a small but measurable current. Over time, either of these devices would discharge in the real world, but the ideal elements are taken to be completely free of any energy dissipators.
Though, as pointed out previously in the thread, this circuit is essentially the same circuit used by persistent-mode superconducting magnet systems where you charge them up with a current source and then close a switch (which is actually just a section of the superconducting coil with a heater wire wrapped around it. The current source is then removed and the coil maintains the current for months before dropping a percent or so and needing to be recharged.
 

MrAl

Joined Jun 17, 2014
9,633
Though, as pointed out previously in the thread, this circuit is essentially the same circuit used by persistent-mode superconducting magnet systems where you charge them up with a current source and then close a switch (which is actually just a section of the superconducting coil with a heater wire wrapped around it. The current source is then removed and the coil maintains the current for months before dropping a percent or so and needing to be recharged.
Hi again,

Yes thanks for reiterating. Pretty cool if you ask me :)
What is interesting is the first thought view we take sometimes, that the current can not continue, yet we have no problem with understanding that a voltage can continue if there is no load. If there is nothing to dissipate the energy nothing changes. The common experience is more with voltage that's all.

We take a lot of things for granted until we look at them in more detail. Energy rules the universe :)
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Hello again,

In a word, "yes" :)

I guess it takes time to become accustomed to thinking about these components in this way. Initial values do make the analysis a little more difficult, but after you do a few of these you'll see the way this works.

Inductors are basically constant current genrators for very short periods of time, and in this case since there was no voltage across it that means it's going to be a constant current generator for all time.

It's also not easy to think intuitively about the constant current generator in parallel with the inductor so that the current generator completely takes over the role of the initial current, but again once you do a few it becomes very apparent why this works so well. It's such a general method and it allows separating the initial current from the inductor itself, and breaking elements into their basic constituents is always a good idea because then we can follow simpler rules, even though there may be more of them.

There are other ways to handle these circuits too but i dont know what you have studied so far. it's good to stick with the course work that's been given so far.

The capacitor is the 'dual' of the inductor, and so for that element it acts like a constant voltage source for a very short period of time.

For the inductor with a short across it, it acts like a constant current source forever, and the capacitor with an open circuit acts like a constant voltage source forever. These are ideal elements however, as in practice we usually see at least some resistance in series with the inductor because it is made out of wire that has resistance, and for the capacitor we usually see at least some resistance in parallel with it because the dielectric conducts a small but measurable current. Over time, either of these devices would discharge in the real world, but the ideal elements are taken to be completely free of any energy dissipators.

YES!

Now go back and look at Post #17

http://forum.allaboutcircuits.com/threads/simple-ac-circuit.120808/#post-966228

And some of the other posts in the vicinity and try to figure out why you were stumbling with this notion back then. If you can identify it, then you can cement a significant insight into your reasoning that was previously missing.

Thanks a lot guys, you really helped me. That short circuit was very confusing, but i think i finally understood whats going on, but i need to practice these kind of examples a lot more, this doesn't seem very complicated, understanding what is going on is the most important thing here, and you guys helped me a lot to understand it. Thanks a lot again!
 
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