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Alternating Current through pure inductors 
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#1
May1112, 05:09 AM

P: 35

I'm trying to understand the qualitative relationship of voltage and the a.c. through a pure inductor.
My books says that to maintain the current in the inductor, the voltage across the inductor provided by the a.c. supply must be equal to the back emf but in the opposite direction. I'm quite puzzled as to why the back emf must have the same magnitude as the ac provided, wouldnt that mean that the potential difference is 0 and there will be no current flow? and if anyone is willing, why is it that when the current is maximum, the voltage is zero? is there any significance physically?( i would prefer this to be explained physically not mathematically ) Thank you :) 


#2
May1212, 03:27 PM

P: 349

I hope the following will help !
The physics behind inductance is Faraday's laws of electromagnetic induction. When a conductor experiences a changing magnetic flux an emf is induced. If the flux is caused by current flowing through a coil and the current changes then an emf will be induced. In this case the induced emf is given by e = L(dI/dt). If a battery of emf E = 12V is connected to a pure inductance of 2H then (dI/dt) = 6A/s. There is zero resistance so you expect the current to be infinite !!! however inductance tells you how quickly the current rises.....6A/s in this case. The induced emf is the 'back emf' and it must equal the applied emf. Imagine what would happen if the back emf (e) was LESS than the applied emf (E).... the current would increase more rapidly and this would tend to make the back emf greater. Imagine what would happen if the back emf was GREATER than the applied emf.....the rate of current rise would decrease to make the back emf less. There is only one logical answer e = E In AC circuits the current is a Sine (or Cosine) function and the maximum rate of change (d/dt) of a sine (or Cosine) wave occurs when the Sine = 0 so you get maximum voltage when the current is passing through zero. You cannot easily explain everything without some maths but I hope this helps. 


#3
May1212, 04:34 PM

P: 4,663

It is useful also to look at the performance of a pure inductance using dc voltages and currents. If you apply a dc voltage V to a pure inductance L through a resistance R, you will build up a current I = V/R after some delay. This current through the inductance represents stored electrical energy in the form of magnetic field B. The stored energy E_{L} is proportional to the magnetic field B squared times the stored volume. This is exactly analagous to a capacitor storing electrical energy in an electric field (capacitor).
So in an inductance, to store energy in the magnetic field, the current must persist. The only way to do this is to short the inductance terminals together. Suprising but true. This is how the supercunducting magnets in MRI machines work. Read about persistent currents in http://en.wikipedia.org/wiki/Superconducting_magnet. When the terminals are shorted, the persistent current time constant is τ = L/R: [tex] I(t)=I_oe^{Rt/L} [/tex] In superconducting magnets, R is like nanoohms, so the persistence is very long. 


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