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Confusion about transformer 
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#19
Jun2414, 01:44 PM

P: 11

I worked as an engineer for an electric utility for over 30 years, specializing in transformer operation and maintenance, so I think I can answer this question.
Looking at the transformer at no load: The magnetizing current in a sense does lag the primary voltage by 90°, but the current isn't purely sinusoidal because there are a lot of odd harmonics (magnetic flux isn't a linear function of current). But you're right, there is a back emf that is essentially 180° out of phase with the primary voltage, so the back emf almost cancels the primary voltage. The back emf will have odd harmonics also, so it isn't purely sinusoidal. The residual emf  the primary voltage minus the back emf  is equal to the the emf that is required to force magnetizing current to flow through the primary circuit (the rest of the system connected to the primary winding). Since the magnetizing current is typically very small, the residual emf is small also. In an "ideal" transformer, you can assume the magnetizing current is zero. 


#20
Jun2414, 04:45 PM

P: 167




#21
Jun2414, 06:33 PM

Sci Advisor
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PF Gold
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#22
Jul2214, 03:23 PM

P: 70

@sophiecentaur. I think I understood the analogy for back emf: that, for the same back emf, the one with higher inductance will show a less change in current, compared to the one with low inductance showing a greater change in current. And that, for the limiting case of infinite inductance, the change in current will be zero. Right?? But, what I wanted to know was about the equation ø = L I, and L = ø / I. Here for L to be infinite, I has to be zero, irrespective of the value of ø. Which seems to mean to show that for infinite inductance, L , any magnetic flux, ø seems to be produced from zero current, I. Right? And, I think I understood that the back emf was the voltage drop across the inductor itself, i.e. L dI/dt, which is obviously equal to the supply voltage, V_{AC}, as seen from the picture. Am I understanding it right?? Would anyone kindly verify. 


#23
Jul3014, 11:30 PM

P: 70

Please. May I have some help here, in clearing this thing out. @sophiecentaur



#24
Jul3114, 12:30 AM

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If you have a loop with an ideal ac generator with emf E=E_{0}cos(ωt) and an ideal inductor with inductance L, ELdI/dt=0 according to Kirchhoffs Law. So dI/dt = E_{0}/L cos(ωt) and integrating, you get I=E_{0}/(ωL)sin(ωt) +C. Initially the current is zero, so C=0 and I=E_{0}/(ωL ) sin(ωt) The current is a sin function, lags the voltage by 90° and has the amplitude of E_{0}/(ωL ). It is not zero. ehild 


#25
Jul3114, 01:41 AM

P: 70




#26
Jul3114, 04:52 AM

Sci Advisor
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PF Gold
P: 12,186

Yes. That's why it is called 'back' emf; it subtracts from the supply voltage and, in a good transformer, it produces a very small net voltage. 


#27
Aug814, 11:59 AM

P: 70

Thank you very much. I think I get it now. But if there is any other problem I will just post it here. Thanks again.



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