Low pass and high pass \$RC\$ filters have a cutoff frequency that is equal to $$f_c=\frac{1}{2\pi RC}$$ This should be the frequency at which the ratio \$|V_{out}/V_{in}|\$ is equal to \$1/\sqrt{2}\$.

But how can I have at the same frequency the same ratio \$|V_{out}/V_{in}|\$ measuring \$V_{out}\$ firstly across capacitor and then across resistance?

KVL states that \$V_{in}=V_{resistance}+V_{capacitor}\$ at any time, but if at \$f_{c}\$ we have \$V_{resistance}=V_{capacitor}=(1/\sqrt{2}) \, V_{in}\$, then KVL would not be respected!

I'm surely missing something but I really can't see how the definition of cutoff frequency is consistent with KVL. Any help is highly appreciated

  • 1
    \$\begingroup\$ Suppose \$R=1\:\textrm{k}\Omega\$ and \$C=1\:\mu\textrm{F}\$. Then \$f_c=159.155\:\textrm{Hz}\$. At that frequency the impedance magnitudes of both are the same. In series, currents are the same, but the voltage across the capacitor is 90 degrees out of phase with the voltage across the resistor. In parallel, the voltages are the same, but the currents through the capacitor are 90 degrees out of phase with currents through the resistor. Do you see how this helps resolve your ratio question? \$\endgroup\$
    – jonk
    Mar 24, 2017 at 2:32
  • 1
    \$\begingroup\$ KVL applies to instantaneous voltages. At any instant, it is true. You are trying to add RMS voltage magnitudes of sine waves. There is a way to do AC steady state analysis so that KVL applies to the steady state result, but you will have to take voltage and current phase angles into account using the "phasor" concept. Google phasor for more information. If you are using a textbook, phasors are probably in there. \$\endgroup\$
    – user57037
    Mar 24, 2017 at 3:29
  • \$\begingroup\$ You know what might help the OP? A simulator. Put the circuit in a simulator and run it. Or build it in a lab and look at it with an oscilloscope. Then you can see how the phase shift allows both things to be true. \$\endgroup\$
    – user57037
    Mar 28, 2017 at 5:18

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Applying KVL,

$$ V_{in} = V_{R} + V_{C} = I\cdot Z_C + I\cdot Z_R \\ \implies I = \frac{V_{in}}{Z_R+Z_C} $$

Setting \$f=\frac{1}{2\pi RC}\$, we have

$$ Z_C=\frac{1}{j2\pi fC}=\frac{1}{j2\pi \frac{1}{2\pi RC} C} = \frac{R}{j} = -jR $$

In the low-pass filter, the output voltage is taken across the capacitor:

$$ V_{out,lpf}=I\cdot Z_C = V_{in} \cdot \frac{Z_C}{Z_R+Z_C} = V_{in} \cdot \frac{-jR}{R-jR} = V_{in} \cdot \frac{-j}{1-j} \\ \implies \left|\frac{V_{out,lpf}}{V_{in}}\right| = \left|\frac{-j}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} $$

In the high-pass filter, the output voltage is taken across the resistor:

$$ V_{out,hpf}=I\cdot Z_R = V_{in} \cdot \frac{Z_R}{Z_R+Z_C} = V_{in} \cdot \frac{R}{R-jR} = V_{in} \cdot \frac{1}{1-j} \\ \implies \left|\frac{V_{out,hpf}}{V_{in}}\right| = \left|\frac{1}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} $$


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