Frequency does not matter for the "lag" ?

[Yuri B.]

AC, as is known, "lags" voltage in a coil and we are familiar with the drawings of the two sine waves with the 90 degree angular "shift" between them. For 50 Hz the shift translates into 5 µs "lagging" but, for instance, for 0.05 Hz AC it would indeed be 5 s ?

What I mean : frequency does not matter for the "lag" ?

 

[cepheid]

AC, as is known, "lags" voltage in a coil and we are familiar with the drawings of the two sine waves with the 90 degree angular "shift" between them. For 50 Hz the shift translates into 5 µs "lagging" but, for instance, for 0.05 Hz AC it would indeed be 5 s ?

Well, for a 90 deg. phase shift, you'd have π/2 = ωt_lag

t_lag = π/(2ω) = π/(2*2π*(0.05 Hz)) = 1/(4*0.05 Hz) = 1/(0.2) s = 5 s.

So, sure.

 

[cepheid]

What I mean : frequency does not matter for the "lag" ?

No, it does not matter. Impedance of an inductor is Z_L = iωL = ωLe^i(π/2) or ωL∠π/2 in phasor notation.

So you can see that, for a fixed sinusoidal AC voltage applied to an inductor, the the amplitude of the inductor current changes with frequency, but not the phase, which is fixed at -π/2 relative to the voltage, since I = V/Z, these being complex (phasor) currents and voltages.

 

[Yuri B.]

Roughly speaking, a generator's rotor pole passing the stator pole will indeed cause no current in a (load) coil so far the poles have not passed over their midpoint ? No matter how slow moves the rotor ?

 

[Yuri B.]

Thank you !

 

[technician]

No, it does not matter. Impedance of an inductor is Z_L = iωL = ωLe^i(π/2) or ωL∠π/2 in phasor notation.

So you can see that, for a fixed sinusoidal AC voltage applied to an inductor, the the amplitude of the inductor current changes with frequency, but not the phase, which is fixed at -π/2 relative to the voltage, since I = V/Z, these being complex (phasor) currents and voltages.

what is the reactance of this inductor?
Are you assuming the inductor has zero resistance and has zero associated capacitance

 

[Yuri B.]

what is the reactance of this inductor?
Are you assuming the inductor has zero resistance and has zero associated capacitance

You are correct. It is always assumed so as you say, but in reality the "loss in copper" current starts immediately the emf is applied.
What is form of the current on oscilloscope ? (I have never had an access to one) A slope of the resistive current, then the sine of the magnetizing current (shifted 90 degrees to the emf)?