Register to reply 
AC through an inductorby asitiaf
Tags: inductor 
Share this thread: 
#1
Mar2913, 10:40 PM

P: 19

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
Mar2913, 11:33 PM

Sci Advisor
PF Gold
P: 2,063

The impedance of an inductor [itex]X=j\omega L[/itex] rises with frequency omega. At low frequencies, AC passes throughbut 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
Mar3013, 03:53 AM

P: 19

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
Mar3013, 06:20 AM

P: 428

AC through an inductor
As marcusl wrote, the impedance of an inductor is given by:
[tex]\mathbf{Z} = j \omega L[/tex] and for a capacitor it's: [tex]\mathbf{Z} = \frac{1}{j \omega C}[/tex] 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: [tex]\mathbf{V} = \mathbf{I}\mathbf{Z} \Rightarrow \mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}}[/tex] 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: [tex] \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) [/tex] and for a capacitor: [tex] \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) [/tex] The current will thus lag the voltage by 90° for an inductor and lead it by 90° for a capacitor. 


#5
Mar3013, 01:57 PM

P: 1,506

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) 


#6
Mar3013, 02:22 PM

P: 428




#7
Mar3013, 02:32 PM

P: 1,506

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. PLEASE get these terms correct. 


#8
Mar3013, 02:44 PM

P: 428

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. 


#9
Mar3013, 02:51 PM

P: 1,506

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
Mar3013, 02:55 PM

P: 428




#11
Mar3013, 03:11 PM

P: 1,506

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
Mar3013, 03:16 PM

P: 428

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: Google: Impedance of an inductor Google: Impedance of a capacitor 


#13
Mar3013, 03:58 PM

P: 1,506

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. 


#14
Mar3013, 04:08 PM

P: 428

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
Mar3013, 04:39 PM

P: 1,506

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 !! 


#16
Mar3013, 04:49 PM

P: 428

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. 


#17
Mar3013, 05:01 PM

P: 1,506

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
Mar3013, 06:26 PM

Sci Advisor
PF Gold
P: 11,952

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). 


Register to reply 
Related Discussions  
Difference between boost inductor and flyback inductor?  Electrical Engineering  2  
The Inductor  General Physics  13  
AC inductor  Introductory Physics Homework  4  
Inductor in dc  Classical Physics  6  
How to get the Q of my inductor  Electrical Engineering  4 