Take a look at the diagram on the right. Assume that the voltage source supplies a 1V/10kHz signal (this means: the amplitude is 1V and the frequency is 10kHz = 10000Hz).

The impedance of inductor L will be X_L = 2∙π∙10⁴∙10^-3 = 62.8Ω. The output voltage (voltage across inductor L) will be 1V∙(X_L/(Z_R+L)), where Z_R+L is the total impedance of R and L. Because an inductor, just like a capacitor, causes a phase shift in in the current flow, we cannot just state that Z_R+L = R + X_L. Using some complex math^[4] we can prove that:

Z_R+L = √(R²+X_L²).

In our case Z_R+L = √(1k²+62.8²) = 1002Ω. Thus, the output voltage is 1V∙(62.8/1002) = 0.0627V.

(By the way, an inductor causes a +90 degrees phase shift, while a capacitor causes a -90 degrees phase shift.)

Now assume that the voltage source supplies a 1V/10MHz signal. The impedance of inductor L will then be X_L = 2∙π∙10⁷∙10^-3 = 62.8kΩ. So Z_R+L = √(1k²+62.8k²) = 62.8k. This is the same as X_L, so the output voltage will be 1V. So we created a very simple frequency filter with just a resistor and a inductor.

In this case we created a so called high pass filter (HPF) since high frequency signals pass this filter easily while frequency signals are suppressed. If you swap R and L, you create a low pass filter (LPF).

Let's calculate the cut-off frequency of our filter. The cut-off frequency is the frequency at which R = X_L => R = 2∙π∙f∙L =>

In our case f = 1k/(2∙π∙10^-3) = 159kHz.^[4]