Self-Induction: Are the induced voltage and the applied voltage different voltages?

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Joined Jun 28, 2018
This may seem like a silly question, but I'm having trouble understanding and distinguishing between the applied voltage and the induced voltage in a simple inductor circuit when it comes to self-induction.

The images above are from the DC inductor chapter in the education section on AAC. To my understanding, if an AC voltage is applied to the simple inductor circuit above, the inductor will drop a voltage across its coils in a way that opposes the changing current and magnetic field that result from the applied voltage. This results in an induced voltage that leads the current and magnetic field of the applied voltage by 90 degrees. As the change in current and magnetic field increases, the induced voltage will oppose this change by orienting its polarity so that it is opposite that of current flow. When the change in current and magnetic field decreases, the induced voltage will flip its polarity, and try to aid the flow of current.

This 90 degree phase shift results in the induced voltage being in phase with the applied voltage. But the applied voltage and the induced voltage are two separate and distinct voltages, correct? Because of the reactionary effects of the inductor the two voltages are in phase and merely appear as one voltage (image above). I'm getting hung up on the idea that for at least part of the cycle the induced voltage opposes and reasoned that therefore it would have to be out of phase with the applied voltage. But this is incorrect. The induced voltage isn't opposing the applied voltage but rather the change in current and magnetic flux, which causes it to be in phase with the applied voltage.


Joined Feb 17, 2009
Well, form the ideal inductor definition we know that any voltage measured across an ideal inductor at any moment is the induced voltage.
And if you connect a voltage source directly to the inductor the applied voltage will be equal to the induced voltage.

And this 90-degree phase shift is a special case. Because the 90-degree phase shift is true only for pure sine waveform only.
Because by "accident" the derivative of the Sine is a Cosine (shifted by 90 degrees sine).
Do not forget that the true equation is

VL = L*dI/dt - the voltage across the inductor is proportional to the rate of change in current (proportional to how fast the current is changing)