A 1Ω resistor and a 1Ω capacitor are connected in series. What percentage of the AC signal is on the capacitor?

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A 1Ω resistor and a 1Ω capacitor are connected in series. What percentage of the AC signal is on the capacitor? [Copy link]

A 1V AC signal is connected to a 1Ω resistor and a 1Ω capacitor. What is the AC voltage across the capacitor?

　　I have shared this question with many engineers over the years. The most common response is, "What is the frequency". But why do we need to know the frequency? We already know the reactance of the capacitor, so the frequency is redundant. Others ask if the signal source could be DC, but this is not a brain teaser. I have marked AC (alternating current) in my diagram, and the reactance of the capacitor is finite, so it cannot be DC.

　　Some people fall into the trap of 0.5V, and a purely resistive voltage divider, 1Ω-1Ω, will give you 0.5V output. That is not the case here. I did some simple vector calculations and answered the question correctly. But now to add a few more points:

　　The R/C circuit creates a pole. The frequency where the resistance and impedance are equal is the "knee point" or cutoff frequency. The response at this point is -3dB (0.707V) attenuation, and 45° phase lag. Simple as that, no math required, the Bode plot is as follows:

　　In addition, there is a voltage of the same magnitude across the resistor as across the capacitor, both of which are 0.707V. Of course, the phase is different.