# To find the current through an inductance in an AC circuit.

1. Aug 31, 2014

### AakashPandita

1. The problem statement, all variables and given/known data
An inductance is connected across an AC current generator.

2. Relevant equations
v= L di/dt = Vsinωt

3. The attempt at a solution

i=∫di= V/L ∫sinωt dt= -V/ωL cosωt

This is what is written in my book. But wouldn't there be a constant at the end of the indefinite integral?

2. Sep 1, 2014

### Simon Bridge

Indefinite integrals should result in an arbitrary constant - so put it in and see if you can figure out what value it should have. (i.e. what should i(0) be?)

3. Sep 1, 2014

### AakashPandita

At t=0, v=0
so at t=0 i should be zero too

so if i= -V/ωL cosωt +c

c=V/wL

Is that right?

Last edited: Sep 1, 2014
4. Sep 1, 2014

### rude man

Why do you think that? It's incorrect.

Realize this: that if there is a constant term in addition to your cosine term, this implies a dc current through the inductor. Is this possible? (Hint: the answer is yes.)

Does a dc current generate a voltage across the inductor? So, can your inductor current have an ac term (cosine term) PLUS a dc term and still satisfy V = L di/dt?).

5. Sep 1, 2014

### ehild

No. Why should be the current zero at t=0?
The nonzero constant current would mean that the AC generator provides a constant current in addition to the AC one. It is not true. The average of the current in an AC circuit is also 0, just like the average voltage. The constant must be zero.

If you say that the voltage is v=Vsin(ωt) across the inductor it means that the generator has been connected to it for a very long time. So i=-V/(Lω)cosωt, and i(0)=-V/(Lω) at t=0.

You get different result if the generator is connected to the inductor at t=0. Then you have some transient phenomena, depending on the internal resistance of the generator and the small ohmic resistance of the inductor. The current will not be a simple cosine function. You will learn about it.

ehild

6. Sep 1, 2014

### rude man

A pure voltage source cannot have a dc component since then the current would build up to infinity.

But you could have a dc current together with the ac current if for example an ac sine voltage and a dc current from a dc constant-current supply are simultaneously applied to the inductor.

7. Sep 1, 2014

### AakashPandita

No but I only have an ac source connected to an inductance.

8. Sep 1, 2014

### rude man

OK, so then you know there is no dc current and therefore your constant of integration has to be zero.

9. Sep 1, 2014

### AakashPandita

Thanks.

10. Sep 7, 2014

### Vibhor

Hello,

Could you explain in simple language how and why i(0)=-V/(Lω) at t=0 i.e i≠0 at t=0? I am finding it difficult to understand how current already exists in the circuit at t=0 even without applying the voltage source.AC voltage source is applied across the inductor at t=0.

Many thanks .

11. Sep 7, 2014

### ehild

Choosing zero for the time is arbitrary. You can start the stopwatch at any instant.

When you say that the voltage is V0sin(ωt), it is a periodic function, the same pattern repeated from infinite time of past to infinite time of future... In this case, the current is I=-V0/(Lω)cos(ωt).

If you mean to switch the AC source to the inductor at t=0, it is a different problem then. The generator voltage is the following function: V=0 if t<0, V=V0sin(ωt) if t≥0.
The solution for the current will not be the simple cosine function as before, and the voltage across the inductor is not the same as the generator voltage. You can not ignore the resistances both that of the coil and the internal resistance of the source. The current has a transient term which decays with time. The solution is more complicated, but the current is indeed zero at t=0.

ehild

12. Sep 7, 2014

### vela

Staff Emeritus
It would help if you two would clarify exactly what the situation is. Is it a voltage source, as one of you says, or a current source, as the other says? Are you two looking the same problem? Is the source turned on at t=0?

13. Sep 7, 2014

### rude man

AC voltage source only. This is a steady-state problem; no transients.

My point was that there could be no dc component in the source voltage, but I added that one could add a dc constant-current source if one wanted to. That would be one way to have a non-zero integration constant.

14. Sep 7, 2014

### vela

Staff Emeritus
How do you know it's a voltage source? The OP said AC current generator whereas Vibhor says it's an AC voltage source.

15. Sep 7, 2014

### rude man

I assumed this was a steady-state problem (ac souce on for a long time). If the ac source is turned on only at t=0 then that is an entirely different problem and a more involved one.

16. Sep 7, 2014

### ehild

The OP assumed that the voltage is given and calculated the current, by integrating V/L.

If it was an ideal current generator supplying current I0sin(ωt) the voltage across the inductor is U=LdI/dt and no integration constant is involved. But the whole text in the OP is about given voltage, and finding the current.

ehild

17. Sep 7, 2014

### Vibhor

Oh ! I was erroneously thinking about the latter situation you mentioned . Thanks for clearing that up .

What does negative in i(0)=-V/(Lω) at t=0 signify ? Is it that the current is flowing in the positive terminal of AC voltage source at t=0 ?

18. Sep 7, 2014

### ehild

Yes, the current and voltage are shifted in phase. At t=0, the voltage is zero, and the current is negative. But you can not speak about the positive terminal of the source as the polarity changes in every half period.
See the picture.

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• ###### currentvoltage.jpg
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19. Sep 7, 2014

### Vibhor

But at t=0 ,we can assign positive to one side of the terminal of AC voltage source and negative to other side .In that case current seem to enter from the positive terminal .

If that is not the case then what is the significance of current having a minus sign at t=0 ?

20. Sep 7, 2014

### ehild

You can write '+' at one terminal of the source, and still, the voltage is positive for half period and negative in the other half. And it is zero at t=0. The current is positive if it flows out from the + terminal. The voltage leads the current on an ideal inductor. When the voltage is maximum, the current is zero. When the voltage is zero at t=0, the current is negative, so it flows into the generator at the positive terminal. At a little time later, the voltage is positive, and the current is still negative.
When you write the function representing the time dependence of the current, the sign belongs to the function. I=-V0/(wL) cos (wt).

ehild