 RC low-pass output loading

A real electrical load can draw electric current from the output. For a load resistor $R_L>0$ across the capacitor, Kirchhoff's current law instead givesTherefore the loaded continuous-time frequency response isLoading reduces the eventual gain and shortens the time constant. With the constructed values and $R_L=10\mathrm{k}\Omega$, $K_L\approx 0.81967$, ${\tau }_L\approx 84.754\mu \mathrm{s}$, and the corner relative to the loaded low-frequency gain moves to $1.8778\mathrm{kHz}$. A $3\mathrm{V}$ step now approaches $2.4590\mathrm{V}$. A quicker rise alone would conceal the error in the final voltage.

With the ideal input voltage source set to zero, the output impedance seen by the electrical load isThis follows from Thévenin's theorem: the resistor and capacitor are then in parallel. A sufficient small-loading condition is $|Z_L(\omega )|\gg |Z_{\mathrm{out}}(\omega )|$ over the frequencies of interest. An oscilloscope is itself an electrical load: its input electrical resistance and cable or probe capacitance can both change the response. Pure extra parallel capacitance increases $C$ and lowers the corner, whereas a finite parallel electrical resistance gives the different gain-and-corner changes above.

The definition of input also matters. If voltage is prescribed inside a source with series electrical resistance $R_s$, the unloaded response to that internal source uses $R+R_s$. If the actual voltage at the filter input terminal is measured, the ratio of output to that measured input still uses $R$. Measuring both terminals avoids mistaking a source setting for the voltage delivered to the filter.