For an ideal inductor under the passive sign convention,

$$v(t)=L\frac{di}{dt}.$$

The voltage therefore depends on how quickly the current changes rather than directly on the current value.

Voltage leads current by 90°

Suppose

$$i(t)=I_m\cos\omega t.$$

Then

$$v(t)=L\frac{di}{dt} =-\omega LI_m\sin\omega t.$$

Using

$$-\sin\omega t=\cos\left(\omega t+\frac{\pi}{2}\right),$$

we obtain

$$v(t)=\omega LI_m\cos\left(\omega t+\frac{\pi}{2}\right).$$

Thus inductor voltage leads current by $90^\circ$, or equivalently current lags voltage by $90^\circ$.

The peak voltage is

$$V_m=\omega L I_m,$$

so

$$\boxed{V_m=I_mX_L},$$

where

$$\boxed{X_L=\omega L}$$

is the inductive reactance.

Reactance has units of ohms and measures the amplitude ratio between sinusoidal voltage and current for an ideal reactive element.

Frequency dependence

Because

$$X_L=\omega L,$$

inductive reactance increases with frequency. For the same sinusoidal voltage amplitude, an ideal inductor therefore supports a smaller current amplitude at higher frequency.

In the ideal steady-DC limit, $\omega\to0$ and the inductor behaves as a short circuit after transients have disappeared.

Example

For

$$L=20,\mathrm{mH}$$

at

$$f=1.00,\mathrm{kHz},$$

$$\omega=2\pi f\approx6283,\mathrm{rad/s}.$$

Then

$$X_L=(6283)(20\times10^{-3}) \approx126,\Omega.$$

If the current amplitude is $0.050,\mathrm A$, the voltage amplitude is

$$V_m=I_mX_L\approx6.28,\mathrm V.$$

The voltage waveform reaches corresponding points in its cycle one quarter-period before the current waveform.

An ideal inductor stores energy in its magnetic field and returns it to the circuit rather than dissipating net energy over a complete sinusoidal cycle.