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∙10⁴∙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 = √(R²+X_C²).

In our case Z_R+C = √(1k²+15.9²) = 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/√(1k²+15.9k²)) = 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∙10³∙10^-6) = 159Hz.