After some Facebook chatter on last week’s blog, I think it deserves some follow-up. Here is the interview question that has bugged me for 41+ years:

A 1V AC source is connected to a 1Ω resistor in series with a 1Ω reactance capacitor. What is the AC voltage across the capacitor?

I’ve shared this question with various engineers through the years. The most common response is, “What’s the frequency?” But why would you need to know? We’re already given the reactance of the capacitor. The frequency would be superfluous. A few others have questioned whether the source was perhaps DC, but, it’s not a trick question. In my diagram, I made it clear that the source was AC, and furthermore, the capacitor has a finite reactance. It can’t be DC.

A few folks fall for the 0.5V trap. That would be the output voltage if it were a purely resistive divider, 1Ω-1Ω. That’s not the case here.

I correctly answered the question back then by doing some simple phasor math. No big deal. But, as I said last week, I thought I could have shown more insight. So here is the way I wish I had answered:

The R/C circuit creates a simple real pole. The case of equal resistance and capacitive reactance is the “corner” or cutoff frequency. The response at this point is -3dB (0.707V) at 45° lagging. Simple as that. No need for math. Here’s the Bode plot:

Another observation: The voltage across the resistor is the same magnitude as across the capacitor—they both have 0.707V. The phase is different, of course.

Now I bet that my original answer didn’t cost me that job. After all, it was correct. So why does this bother me? I appreciate intuitive understanding. I think it’s a crucial ingredient in creativity.

Comments welcome.

Bruce thesignal@list.ti.com

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p.s. I created the Bode plot (above) in TINA-TI with a 1Hz corner frequency. Using a 1Ω resistor, it required a large capacitor value. It should be a non-polar capacitor. I have a good stock of 1uF polyester capacitors. How many would it require?