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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

  • Isuru-- TI's blog site is not public and all its blogs are written by TI employees (or perhaps occasionally by invited guests). You might consider whether your comments are better suited to our E2E forums, directed to a specific product area.

  • can someone tell me as to how I can start writing a blog on TI.com?

  • My point there is _no_ phase angle to an rms solution. And most people would have to think 1 V amplitude from your academic question. If you are talking about lagging 45 deg you are looking at the sinusoid and not your rms meter. I would bet most would think V = 1 sin wt at that point.

    The math will work the same either way... So it is irrelevant to your question.

    Robert's point was the transient is different, which is true and worth noting.

    I got some things muddled up in my initial post as well. I think the main point is old men shouldn't blog:) (To be clear, I'm including myself)

  • Gene--  I believe you may still be thinking of a time-domain solution with a starting transient. Mine is a steady-state solution. The input signal is 1VAC = 1V RMS. The voltage on the capacitor is -3dB relative to the input signal which is 0.707VAC = 0.707 V RMS. The phase of the voltage on the capacitor 45 degrees lagging relative to the input signal. I'm not sure what you mean by "rms phase."

  • Didn't you write: The response at this point is -3dB (0.707V) at 45° lagging.  The phase is different, of course.

    What is the rms phase of your solution?