# AC through an inductor

1. Mar 29, 2013

### asitiaf

While studying electrical science, I was told that when an alternating current passes through an inductor, current lags voltage by 90•. And when through a capacitor, current leads voltage by 90•.
But while studying transistors in electronics, I was told that AC passes through a capacitor but blocked by an inductor. And DC is blocked by a capacitor but allowed by an inductor.
So I am confused WHETHER AC PASSES THROUGH AN INDUCTOR.

2. Mar 29, 2013

### marcusl

The impedance of an inductor $X=j\omega L$ rises with frequency omega. At low frequencies, AC passes through--but we need to look at the load and other circuit elements. When the magnitude of the inductor's impedance becomes larger (at sufficiently high frequencies) than the characteristic impedance of other parts of the circuit, current flow diminishes and eventually becomes very small at "high" frequency. This is the sense in which an inductor "blocks" high frequencies.

3. Mar 30, 2013

### asitiaf

DC is blocked by a capacitor But AC passes by a capacitor. Why? If above mentioned logic is taken into account.
Taking the frequency consideration how the leading and lagging of current is interrelated.

4. Mar 30, 2013

### milesyoung

As marcusl wrote, the impedance of an inductor is given by:
$$\mathbf{Z} = j \omega L$$
and for a capacitor it's:
$$\mathbf{Z} = \frac{1}{j \omega C}$$
where ω is the angular frequency of the AC system, L and C are the inductance and capacitance of the inductor and capacitor, respectively.

The relationship between complex voltage, V, current, I, and impedance, Z, is given by:
$$\mathbf{V} = \mathbf{I}\mathbf{Z} \Rightarrow \mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}}$$
As ω tends towards infinity, the magnitude of Z tends towards infinity for an inductor and towards zero for a capacitor. This effectively makes an inductor an open circuit and a capacitor a short circuit as ω→∞. It's the other way around as omega tends towards zero (DC).

With regards to the current leading or lagging the applied voltage, with a bit of complex algebra, we have, for an inductor:
$$\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{|\mathbf{V}|\angle \mathbf{V}}{|\mathbf{Z}|\angle \mathbf{Z}} = \frac{|\mathbf{V}|}{|\mathbf{Z}|} \angle \left( \angle\mathbf{V} - \angle\mathbf{Z}\right) = \frac{|\mathbf{V}|}{|j \omega L|} \angle \left( \angle\mathbf{V} - \angle j \omega L \right) = \frac{|\mathbf{V}|}{\omega L} \angle \left( \angle\mathbf{V} - 90^\circ \right)$$
and for a capacitor:
$$\mathbf{I} = \frac{|\mathbf{V}|}{|\mathbf{Z}|} \angle \left( \angle\mathbf{V} - \angle\mathbf{Z}\right) = \frac{|\mathbf{V}|}{\left|\frac{1}{j \omega C}\right|} \angle \left( \angle\mathbf{V} - \angle \frac{1}{j \omega C} \right) = \frac{|\mathbf{V}|}{\frac{1}{\omega C}} \angle \left( \angle\mathbf{V} + 90^\circ \right)$$
The current will thus lag the voltage by 90° for an inductor and lead it by 90° for a capacitor.

Last edited: Mar 30, 2013
5. Mar 30, 2013

### technician

When giving help we should be sure to use correct terminology.
It is wrong to refer to the 'impedance' of an inductor. The relationship V/I for an inductor (and capacitor) is called REACTANCE (X).
IMPEDANCE (Z) is a combination of reactance and resistance.
The impressive equations above make no sense when referring to impedance, Z, because no RESISTANCE has been brought into the argument.
Z = √(R^2 + X^2)

Last edited: Mar 30, 2013
6. Mar 30, 2013

### milesyoung

No, it's impedance. Reactance is the imaginary part of the impedance.

Why would you bring resistance into it for a component which by definiton has none?

7. Mar 30, 2013

### technician

A component, which by definition has zero resistance, still has a relationship between V and I.... this is called REACTANCE.
When resistance is brought in the term to use for the relation between V and I for the combination is IMPEDANCE Z.
Do you recognise the expression Z = √ (R^2 + X^2) ???
Explain to the original post what the meaning of REACTANCE is.
Check STANDARD TEXT BOOKS (a condition of posts here) for the meaning of the terms REACTANCE and IMPEDANCE.

8. Mar 30, 2013

### milesyoung

No, the ratio of the complex voltage to current is the impedance, by definition (look it up). The impedance is thus complex in general. In Cartesian form:

Z = R + j*X

where R is the resistance and X is the reactance, both are real. Reactance can thus not, in general, represent the ratio between two complex numbers.

Yes, and? (edit: I assume you mean |Z|, the magnitude of Z).

Maybe you should have a look at "STANDARD TEXT BOOKS" because you're seriously misinformed.

Last edited: Mar 30, 2013
9. Mar 30, 2013

### technician

What is meant by REACTANCE
what is meant by RESISTANCE
How do these combine to produce IMPEDANCE
Is resistance an impedance
My text books:
Nelkon and Parker
Breithaupt
Duncan
All A level text books.
Which text books do you refer to
REACTANCE does not necessarily refer to the relationship between complex numbers. It is a relationship between measured quantities....volts and amps

10. Mar 30, 2013

### milesyoung

Let me quote my own post:
Well, take any of them and look up the definition of impedance. Go on.

11. Mar 30, 2013

### technician

I know what the definition of impedance is.
I also know that the imaginary components of impedance have special names....inductive and capacitative reactance.
The real part of impedance is resistance.
Impedance is the complex resultant of the combination of resistance and reactance.
Capacitors and inductors have reactance which, in combination with resistance produce an impedance.
For capacitors and inductors I = V/X....it is not I = V/Z
I cannot find any reference that talks about the 'impedance' of a capacitor or inductor, except of course here.
End of my contribution otherwise this will become a typical 30 post PF post that takes 1 page, at most, of a good text book.
Just use correct terms !!!!!!

12. Mar 30, 2013

### milesyoung

The impedance of an ideal inductor is Z = j*omega*L. This is the complex proportionality constant between the complex voltage across it and the complex current through it.

The reactance is X = omega*L. It's a real number and is not the complex proportionality constant between the complex voltage across the ideal inductor and the complex current through it, the impedance, Z, is.

I can't make it any more clear than that.

Edit:
This is simply not correct. X is a real number, if I = V/X was true then the current would always be in phase with the voltage.

Then I seriously question your knowledge of this subject.

Last edited: Mar 30, 2013
13. Mar 30, 2013

### technician

Here we go.....30 posts
"This is simply not correct. X is a real number, if I = V/X was true then the current would always be in phase with the voltage."
I can make a circuit to measureV and I which give a value I = V/X and the V and I are not in phase !!!!!
Goodnight
PS
In the same circuit there could be a V and I to give the same value as X and they would be in phase.

Last edited: Mar 30, 2013
14. Mar 30, 2013

### milesyoung

I challenge you to show me that calculation. Just take any example you'd like where V and I are measured quantities and show me how they're related by X.

Edit:
Also, I'm going to continue this because if your posts in this thread are allowed to remain unquestioned you'd potentially be misleading everyone else who come to this thread to learn something new.

15. Mar 30, 2013

### technician

Edit:
Also, I'm going to continue this because if your posts in this thread are allowed to remain unquestioned you'd potentially be misleading everyone else who come to this thread to learn something new.

Likewise

see attached and explain

Ps: I will have had enough by post 30 if not before !!

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Last edited: Mar 30, 2013
16. Mar 30, 2013

### milesyoung

The supply is 7 V but it drops 10 V across the impedances? (edit: It occurred to me that your diagram only include magnitudes instead of phasors).

Edit:
Let me preempt your next attempt and give an example for an ideal inductor, where I hope we can agree that the current lags the voltage by 90°:

V = 10∠0° V
I = 5∠-90° A

V/I = 10/5∠(0 - (-90°)) Ω = 2∠(0 + 90°) = j*2 Ω

which is the impedance of the inductor. The reactance is X = 2 Ω.

Last edited: Mar 30, 2013
17. Mar 30, 2013

### technician

Sorry....very wrong...how can 7V 'drop' to 10V across the impedances????
A (or perhaps it is B) is a capacitance (or is it an inductance?) with a reactance, X, of 5Ω.
B (or perhaps it is A) is a RESISTANCE of 5Ω.
The Impedance (Z) of the circuit is √(5^2 + 5^2) = 7.07Ω
7V supply gives a current of 1A through impedance of 7Ω (I have ignored sig figs)

I will make no further contributions until you can get this sorted out.
No need to go to 30 posts

18. Mar 30, 2013

### sophiecentaur

Try to ignore the technical bickering. It often happens on PF (I get involved myself, sometimes). It can really interfere with the answering of a simple question.
The reason why an inductor 'Impedes' AC is that the magnetic field around a coil takes time to build up and to decay. This causes a voltage to appear across the inductor which is in a direction such as to maintain the existing current through it. (There is a mechanical analogy of a flywheel on a shaft which takes some getting going but then is reluctant to slow down). This induced voltage is proportional to the rate of change of current (the time differential) so the voltage and current are 90°out of phase. For a very high frequency, this 'back emf' is so high that very little current can pass - hence the "AC Blocking".
A Capacitor is, of course, an open circuit to DC because, once charged to the value of the supply voltage, no more charge will flow, but charge can keep flowing in and out if AC is applied. The time derivative of the voltage across a capacitor is proportional to the current, which again, gives a 90° phase shift (in the other direction).

Ideal Inductors and Capacitors have a simple Reactance but real components always have some resistance so they will have an Impedance(Z), which is a (complex) combination of X and R.
So Z=R+jX
This applies to a particular frequency because X always depends upon the frequency in question. (This analysis only applies to single AC frequencies, of course).

19. Mar 30, 2013

### milesyoung

I should probably have picked up on this earlier, you're only dealing with the magnitude of the complex voltage, current and impedance.

If A is an ideal inductor with a reactance of 5 Ω (it must be an inductor since the reactance is positive), then its impedance is:
Z_A = (0 + j*5) Ω.

If B is an ideal resistor with a resistance of 5 Ω, then its impedance is:
Z_B = (5 + j*0) Ω.

The equivalent impedance of your circuit, Z, is then:
Z = Z_A + Z_B = (5 + j*5) Ω

For a current I = 1∠0° A, the supply voltage, V, must be:
V = I*Z = (1 + j*0)*(5 + j*5) V = (5 + j*5) V

The magnitude of V is:
|V| = sqrt(5^2 + 5^2) V = 7.071 V

The ratio between the magnitude of V and I is then:
|V|/|I| = 7.071/1 Ω = 7.071 Ω

and now comes the important part:
This is not the impedance of your circuit. It's the magnitude of the impedance.

And to reiterate my point:
The impedance is the complex ratio of the voltage to the current.

20. Mar 30, 2013

### The Electrician

Are you saying that if R is zero in the expression in red above, then it shouldn't be called impedance?

That would be like saying that if we have a 2 dimensional vector on the X-Y plane, for example (i,j), it is no longer a vector if i or j is zero, but it has become just a number, a scalar.

The concept of impedance admits of the possibility that either the real part or the imaginary part can be zero, and it's still an impedance. One can reasonably speak of a pure real impedance, or a pure imaginary impedance. What would be the benefit of disallowing this?

21. Mar 31, 2013

### Staff: Mentor

22. Mar 31, 2013

### Staff: Mentor

I think these are the key points, and hopefully the terminology discussion hasn't confused the OP too much.

I use the term impedance much more than the term reactance, because as Centaur points, out, real inductors and capacitors have some real resistance.

I'm going to leave this thread closed.