How Does Current Flow in a Pure Inductor Connected to an AC Voltage Source?

[sahil_time]

if a pure inductor is connected to an AC voltage source!
The Back emf in the inductor=Thr instantaneous AC emf!
THEN
How can the current flow through the inductor if the NET VOLTAGE=0?
please explain

 

[sophiecentaur]

If the source is a true voltage source then the pd across the inductor will Have to the equal to the supply volts! If you want to explain this in terms of "back emf" then the "net volts" have to be zero. What else could they be? If there is series resistance, then the volts across the Inductor will be less than the supply emf. The amount of voltage 'drop' across the resistor is given by the ratio R/(R+XL) and the 'back emf' is what's left. This will be a complex quantity, of course.
Don't try to 'get this' from intuition. Just believe the Maths; it's right. Then come to terms with it. ;-)

 

[sahil_time]

But does a current flow when a purely inductive is placed on an AC source!..coz the net voltage=0?

 

[sophiecentaur]

Dealing with pure circuit elements and pure emfs is fraught and is not realistic. However:

The current that will flow will be V/jX where X = ωL
This involves a 90° phase difference.

The PD will be maintained at V, whatever. This must mean that the induced voltage (a further phase difference of 90°) will be equal and opposite to the supply V. But it doesn't have to mean that the current is zero. It's just in quadrature.

 

[sahil_time]

But doesn't the Net Voltage=0 imply something atleast?

 

[sophiecentaur]

What do you mean by "net voltage"?
You have to realize that ANY impedance other than Zero (for which there is no solution) applied across an emf of V will still measure V across the terminals because an emf is an emf - a constant voltage source. None of the above says that there will be no Current through the inductor, remember. The current will be V/jX, which can be an awful lot!

If anything is "implied" here, it is that you have not yet been prepared to believe the sums rather than your intuition.

I suppose one way of looking at this is that 'just enough current' is allowed through by the inductance to give a back emf of V. You would be happy enough, I am sure, if I said that, instead of an inductor, you placed a resistance R across an emf. The volts across the R would be IR (where the I was also 'just enough') and just the same value as the emf V. If there's nothing wrong with that then there should be nothing wrong with any impedance put across the emf.

 

[sahil_time]

Ohhh yeah that helped!:)But The voltage drop in resistance doesn't oppose the battery like in the inductor right?..

 

[sophiecentaur]

There's a voltage across it, isn't there? And it happens to be the same as the supply volts!

Would it help to look at Potential Difference in terms of Energy per Unit Charge? One Volt corresponds to one Joule per Coulomb. It isn't 'some kind of pushing force' but an energy difference.

Because an inductor is reactive, of course, there are no Joules expended as the current flows. I though i'd throw that one in and upset your evening. :-)

 

[sahil_time]

Well i just thought that the inductor is like this

 

[sophiecentaur]

But, if you are using batteries, this implies DC. There would be no resistance in that case. With AC, you need to think of the phases of currents and voltages. Remember that the AC behaviour of a circuit containing inductors or capacitors needs to be analysed by taking into account the rates of change of current or volts. It's usual to do this using complex numbers and to talk in terms of Impedance Z which consists of a real part R and a reactance X. So
Z = R +jX

There is no way of looking at it without a step up in the Maths, I'm afraid. If you're not familiar then you need to get reading!

 

[sahil_time]

How did u convince urself that the Mathematics is always right?..through experimentation or logic or intuition?

 

[sophiecentaur]

How did u convince urself that the Mathematics is always right?..through experimentation or logic or intuition?

We used to do experiments at School. Also I spent 24 years as an Engineer and another 20 as a Science Teacher. In a very precise branch of Engineering I found that, without exception, the Maths was right. If it ever seemed like being wrong, I found that it was down to me. You get a lot of confidence in the system from that sort of experience.

I also have found that the structure of Maths usually makes it hard to pull the wool over eyes with intuitive misconceptions. Maths is an amazing tool for figuring things out for myself (even with fairly limited calculus and knowledge of a few transforms).

I am always amazed at the number of people on this and other fora who cling madly to their own private notions about this and that with no more than a vague intuitive feeling to back them up. The numbers always count in Science and you just can't ignore them. You get some amazingly petulant and resentful reactions when you present a simple three lines of Maths to show them what's what.

When you say "the Maths is always right", it has to be the right Maths for the situation. You make a model and the Maths will take you so far and no further.
Phew - that was a bit of a rant, wasn't it? But Maths gets my vote nearly every time.

 

[davenn]

when I went to university for a BSc in Geology, I learned very quickly that maths is the language of science. If i was explaining some geophysics process to 2 guys with disciplines in fields other than geophysics, say a mathematician and an astrophysicist. Even tho thy did not know anything about geophysics if I explained it to them from a mathematical formula perspective, them would know and understand what I was talking about.
It truly is a universal language that needs to be learnt, at least to some extent.

Dave

 

[sahil_time]

Thank You!:)