How is definition of cutoff frequency consistent with KVL?

Question

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

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:

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