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∙10^{4}∙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 we can prove
that:

**Z _{R+L} =
√(R^{2}+X_{L}^{2})**.

In our case Z_{R+L} =
√(1k^{2}+62.8^{2}) =
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∙10^{7}∙10^{-3} =
62.8kΩ. So Z_{R+L} =
√(1k^{2}+62.8k^{2}) =
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.