The inductor is an ideal one, the phase difference is with respect to current, and the voltage varies by the law $V=V_Lsin(\omega t).$ One can prove that the current function will come out to be $I=I_Lsin(\omega t-\frac{\pi}{2}).$ It can also be proved that the phase difference b/w the functions wrt to the later function will be $\frac{T}{4}$ where $T$ is the fundamental period of the function $cos(\omega t).$ All books mention this (tough without giving proof). All books also mention that the phase difference is also $\frac{\pi}{2}.$ Some hint that they think this because there is $-\frac{\pi}{2}$ in the current function. But that's stupid.

If one accepts that both are phase difference then $\frac{T}{4}=\frac{\pi}{2}$ then $\omega=1.$ So $\frac{\pi}{2}$ will be phase difference only when $\omega=1.$ So it is clear that , in general, $\frac{\pi}{2}$ must not be phase difference; it is not true for $\omega=2.$

You can judge this graphically as well. As you move the slider for $w$ in this graph you will note that the phase difference changes and does not remain constant to $\frac{\pi}{2}.$

But again all the books have written that hence there should be mistake in my reasoning but where?

  • $\begingroup$ I think you have an error in your initial example. $\endgroup$ May 15, 2022 at 3:18
  • $\begingroup$ @BioPhysicist which exmaple? $\endgroup$
    – Osmium
    May 15, 2022 at 5:48
  • 1
    $\begingroup$ $I=I_L\sin(\omega t-\frac{\pi}{2})$ is the same as $I=-I_L\cos(\omega t)$ which is a $\pi$ phase difference. $\endgroup$
    – Farcher
    May 15, 2022 at 5:51
  • $\begingroup$ @Farcher I changed the voltage function. Thanks. $\endgroup$
    – Osmium
    May 15, 2022 at 5:57

3 Answers 3


It's because the voltage across an inductor is related to the current going through it by a time derivative

$$V_L=L\frac{\text dI}{\text dt}$$

and the derivative of sines and cosines give you functions that are $\pi/2$ out of phase. This is independent of $\omega$ for $\sin(\omega t)$ or $\cos(\omega t)$.

As an aside, note that this is also why the lag for an ideal capacitor goes the other way

$$V_C=\frac1C\int I\,\text dt$$

  • $\begingroup$ I don't see why the phase difference should be independent of the frequency; according to my post, it does depend on it. Is it wrong that $T/4$ is not a phase difference as well? Is it not true that $T$ depends upon $\omega?$ $\endgroup$
    – Osmium
    May 15, 2022 at 5:55
  • $\begingroup$ The phase difference changes as you move the slider for $w$ which is contrary to what you say: desmos.com/calculator/eklxduxw3k $\endgroup$
    – Osmium
    May 15, 2022 at 6:06
  • 3
    $\begingroup$ @Osmium No, it doesn't. The phase difference between the functions $\cos(\omega t + \phi_1)$ and $\cos(\omega t + \phi_2)$ is $\phi_2 - \phi_1$. The change you are speaking of is the time offset $(\phi_2 - \phi_1)/\omega$ between the two functions, which is different from phase difference and obviously depends on the frequency if the phase difference is constant. $\endgroup$
    – Puk
    May 15, 2022 at 8:17
  • $\begingroup$ @Puk The latter one is what I call the phase difference. The first quantity doesn't seem useful; the latter one tells us by how much moving one graph can we make it to coincide with the other. Using this definition the phase difference should be $T/4.$ But using the other definition it should be $\pi/2$ which remains independent of $\omega.$ But why on earth do books mention both as phase differences. Which definition are they using? Both ($T/4$, $\pi/2$) cant be phase difference at the same time, in general, using either of the deifnation. $\endgroup$
    – Osmium
    May 15, 2022 at 11:29
  • $\begingroup$ @Osmium Sorry, I didn't know you were using unconventional terminology $\endgroup$ May 15, 2022 at 19:19

The sinusoidal voltage and current for an inductor vary as shown below.

enter image description here

Both variations are of the same frequency with the voltage leading the current by a phase of $\pi/2$ which is consistent with the relationship between voltage and current for an inductor, $v\propto \frac{di}{dt}$ both mathematically (as shown by others) and by inspection of the graphs.

  • $\begingroup$ How can you draw a graph if $\omega$ is variable? Can you draw a graph for $y=ax+b$ without knowing what $a$ and $b$ are? If you draw the graph for different values of $\omega$ (desmos.com/calculator/eklxduxw3k), the phase changes with $\omega$-angular frequency. $\endgroup$
    – Osmium
    May 15, 2022 at 7:34
  • $\begingroup$ @Osmium that's why the axes are not labeled. You can get whatever frequency you want by writing your own numbers on the axes. But they all have this shape. The phase shift does not change with $\omega$ $\endgroup$
    – user253751
    May 17, 2022 at 11:13
  • $\begingroup$ @user253751 I didn't see that. Thank you. Have a good day. $\endgroup$
    – Osmium
    May 17, 2022 at 13:52

To find the phase difference between two quantities, their equations must have same trigonometric ratio and same sign.

Say you want to find the phase difference between two following quantities $x_{1}=a \sin⁡(\omega t)$ and $x_{2}=b \cos⁡(\omega t)$

Firstly, both equations have same signs. Now, use the quadrant formulas to have sine ratio in the second equation. Now, we can write, $x_{1}=a \sin⁡(ωt)$ and $x_{2}=b \sin⁡(\omega t+\frac{\pi}{2})$

Now we can say that the phase of $x_{1}$ is $\omega t$ and that of $x_{2}$ is $\omega t+\frac{\pi}{2}$ .

So, the phase difference between $x_{1}$ and $x_{2}$ is $π/2$ and $x_2$ is leading ahead of $x_{1}$.

Let’s see another example.

$y_{1}=p \sin⁡(\omega t)$ and $y_2=-q \sin⁡(\omega t)$

Here both the equations must have same trigonometric ratios and same sign. So, we’ll rewrite the equation of $y_{2}$ as, $y_{2}=q \sin⁡(\omega t + \pi)$

Now the phase of $y_{1}$ is $\omega t$ and that of $y_{2}$ is $\omega t + \pi$.

So, the phase difference between $y_1$ and $y_2$ is $\pi$ and $y_{2}$ is leading ahead of $y_{1}$.


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