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Bruce Trump

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

          Index to all The Signal blogs.

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?

Anonymous

  • Sorry, my last comment should have said "1/sqrt(2)" for the RMS cap voltages.  A senior moment.

  • My terrible memory was driving me crazy so I figured out the "general" solution for the series RC circuit IF the voltage is applied at t=0 (but not before).  ASSUMING a 1 Vrms, 1 Hz sinusoid input voltage starting at t=0, and C=1/(2*pi), the full solution for t>0 is Vcap(t) = [exp(-2*pi*t)]/sqrt(2) + sin(2*pi*t - pi/4).  For the 1st cycle of applied input the AC reactance calculation yields an RMS cap voltage of sqrt(2) whereas the exact solution gives an RMS value over this cycle of 0.735.  By the 9th cycle they both yield the same value: sqrt(2).  There's surprisingly small differences even by the 2nd cycle!  But the differing calculated cap voltage waveforms do look pretty different during that 1st cycle (ONLY!).  If an interviewee gave me this answer, I think I'd hire him on the spot with no more questions asked.

  • This concept may be a good way to create a very efficient led lighting system; however the power companies may not like it. JF

  • Perhaps a better question to ask, rather than the frequency, is "How long has the AC source been applied to the circuit?".  If it's been applied "forever", then your answer is correct.  However, it was only raised from zero VAC to 1 VAC very recently, the steady state solution given by an AC reactance calculation is not valid.  There is both a general and a particular solution to the differential equation for the capacitor voltage.  I don't recall the "particular" solution for this simple system (it's been a LONG time since I've been in a classroom), but I suspect that the initial "AC voltage", defined perhaps as the Fourier component at the applied frequency, will actually be different from your answer.  Of course, your answer will approach perfection as time increases far past the RC time constant value (but is it a nanosecond or a million years?)!