# why current lags 90 deg behind the voltage when inductor is present?

When we supply a voltage to an inductor then the phase difference between voltage and current is 90 deg (ideally).

Mathematically I have seen how it becomes 90 deg but can anyone help me to understand it in an intuitive way?

• Lenz's Law, in opposing the change of currents in inductors, causes that lag. – analogsystemsrf May 1 '18 at 3:58
• why 90 deg?? why not any other deg? – pankaj prasad May 1 '18 at 3:59
• Because d/dt sin(t) = cos(t) = sin(t + 90°) – Hearth May 1 '18 at 4:17
• Or maybe it was sin(t - 90°). Too sleepy to remember right now. – Hearth May 1 '18 at 4:24
• @SredniVashtar “If you accept the fact that sinusoidal signals are the building blocks of all other signals” ... not necessarily. We just like to say so because it makes the math easier. – Blair Fonville May 1 '18 at 4:51

Because inductors are made to react against the change in current, it causes it to lag behind the voltage. When you apply a voltage to an inductor, you make a magnetic field. Henceforth, if the magnetic field varies with respect to time, there is an electric field that opposes the magnetic field inside the inductor. In other words, the electric field generated by the voltage behaves like a wall to the magnetic field generated by the current.

This behavior is described by Len'z Law.

$\displaystyle EMF=-\frac{\partial \Phi_B}{\partial t}$, which happens to be the negative of Faraday's Law of Induction.

I didn't draw this, I just found it on Wikipedia. But as you can see, the magnetic field $B$ is being pushed back.

The 90 degrees that you're thinking of comes from the sine function... The sine function represents the reactive current. When you plot the cosine and sine functions, the graphs are 90 degrees of a phase difference.

If you're asking why it's 90 degrees rather than something else like 45 degrees, then again, it's because of the sine and cosine functions.

So why is there a sine function? Because when you talk about the voltage and current, there's a derivative relationship.

$\displaystyle V(t)_L = L\frac{di(t)_L}{dt} \rightarrow \cos(t)=\frac{d \sin(t)}{dt}=\sin(t-90°)$

Typical AC voltage is described by a cosine sinusoidal function.

The voltage across an inductor is the derivative of the current. So when the current is 0, its derivative is at a maximum. When the current is at the peak of a sinusoid, the derivative is 0.

• Well he said he understand the mathematics – Claudio Avi Chami May 1 '18 at 3:50

If you have a volt meter, a coil of wire and a magnet and statically place the magnet close to the coil you do not get any DC voltage shown on the voltmeter. However, if you move the magnet around you get a voltage induced in the coil whose magnitude is proportional to the change in magnetic flux experienced by the coil. In other words: -

$$V = N\dfrac{d\phi}{dt}$$

But you can replace the magnet with a current source. If you apply a ramping waveform of current to the coil it produces a ramping magnetic flux and across the terminals of the inductor the voltage will be constant. That constant voltage is proportional to the rate of change of flux.

You might be able to see that this is mathematical differentiation so, applying a sinusoidal current through the coil (from the current source) produces a voltage that is the differential of the sine wave across the terminals. That means the voltage is a cosine and thus, in AC analysis the voltage and current are displaced by 90 degrees.