Electrical – How is the phase of the applied voltage related to the phase of the induced emf in an inductor

circuit analysisimpedanceinductanceinductioninductor

As I understand it the induced emf in an inductor is generated directly in opposition to the applied voltage, so would that mean that it is 180∘ out of phase with it?

Also, what then is the phase relationship between the source voltage and the current through the circuit?

In both cases I am simply talking about an inductor connected to an ac voltage source.

If anyone could draw a diagram that would be great. There are plenty on the relationship between the current and phase across the inductor but none on the relationship between the source voltage and the above parameters.

Best Answer

A sinusoidal AC voltage will produce a sinusoidal-shaped current in the inductor. However, that current lags the voltage by 90 degrees because: -

\$v = L\dfrac{di}{dt}\$ or

\$i = \dfrac{1}{L} \int v\cdot dt\$ and this produces the well-known 90 degrees lag seen below: -

enter image description here

Current flow produces magnetic flux proportional to that current and, a changing magnetic flux produces a back-emf in accordance with to Faraday's Law of induction: -

enter image description here

So, from the originating sinusoidal voltage, you get a current that is 90 degrees lagging and this produces a flux whose rate of change is in phase with the original voltage hence, the back emf is 180 degrees out of phase to the original voltage due to the "minus" sign in Faraday's law.