Take a look on the diagram above. 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 capacitor C will be X_{C} =
1/(2∙π∙10^{4}∙10^{-6}) =
15,9Ω. The output voltage (voltage across capacitor C) will be
1V∙(X_{C}/Z_{R+C}), where
Z_{R+C} is the total impedance of R and C. Because a
capacitor causes a phase shift in the current flow, we cannot just state
that Z_{R+C} = R + X_{C}. Using
some complex math we can prove
that:

**Z _{R+C} =
√(R^{2}+X_{C}^{2})**.

In our case Z_{R+C} =
√(1k^{2}+15.9^{2}) =
1000.13Ω. So the output voltage becomes 1V∙(15.9/1000.13) =
0.0159V.

Now assume that the voltage source supplies a 1V/10Hz signal. The
impedance of capacitor C will then be X_{C} =
1/(2∙π∙10∙10^{-6}) = 15,9kΩ. The output voltage
will be 1V∙(X_{C}/(Z_{R+C})) =
1V∙(15.9k/√(1k^{2}+15.9k^{2}))
= 0.998V. So we created a very simple frequency filter with just a
resistor and a capacitor.

In this case we created a so called low pass filter (LPF) since it passes low frequency signals and suppresses high frequency signals. If you swap R and C, you create a high pass filter (HPF).

Let's calculate the cut-off frequency of our filter. The cut-off
frequency is the frequency at which R=X_{C} => R =
1/(2∙π∙f∙C) =>

In our case f =
1/(2∙π∙10^{3}∙10^{-6}) =
159Hz.