Polarity of the voltage drop on inductance

We know from the Lenz's law that the induced voltage on an inductance has the polarity that determines an oppisition to the variation that has generated it.

But in circuit theory the inductance is seen not as a voltage source, but as an impedance. The convention is to say that the voltage drop on it is equal to VL = jwL I, with VL is opposed to the direction of I (if we run across the inductance in the current direction we see that the potential decreases). But, why is this true? I think it depends on if I is increasing (dI/dt >0) or decreasing (dI/dt<0). In the first case, VL will be opposed to the voltage source that provoked the increase of I; in the second one, VL will be opposed to the voltage source that provoked the decrease of I. How can the previous convention work?

For a resistance, saying that "if we run across the inductance in the current direction we see that the potential decreases" is ok for me, but for an inductance, how can it be ok? It should depend on the derivative of the current, not on the current.

• – G36 Aug 26 '19 at 18:47
• $V_L=L\cdot\dfrac{di}{dt}$ this is the starting point. Now, what's you short question? – Marko Buršič Aug 26 '19 at 20:02
• The idea with circuit theory is to maintain the appearances of Ohm's Law by expanding it to cover impedance, in general. So you want the complex-valued impedance equal to the complex-valued voltage divided by the complex-valued current. How familiar are you with complex analysis and Euler's as it applies here? – jonk Aug 26 '19 at 20:34
• Are you dealing with the time domain or frequency domain? – Chu Aug 26 '19 at 22:40