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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

  • No. Sorry, Gene, but VAC should always be assumed to be RMS. For example, 120VAC is the nominal line voltage in USA. That's 120 volts RMS. AC voltmeters read in RMS. Even though they may respond to average-rectified, averaged-peak, or averaged-peak-to-peak, they read in RMS.

  • What?!!!

    Vdc = Vrms _they_ have the same meaning.

    Vac = A sin wt

    Vrms = A / sqrt(2)

    If Vac = 1 then Vrms = 0.707.

    And the steady state solution is 0.707 Vac, which is 0.5 Vrms.

    Also let me define period:

    period = 1 / frequency

    I meant any n period (where n is an integer < 10).

  • I’ll jump in here with some clarification. There was nothing in the original question indicating a transient condition. This is a steady-state AC situation. I fear that the suggestion of otherwise may shake the confidence of some who might read this whole thread. You are all welcome to continue with the discussion of a transient case (waveform starting at t=0) if you wish but it’s probably best to avoid terminology involving “RMS” during the transient. The RMS value is continuously changing during the transient (it’s not yet a sinusoid) and you must define the period over which you are averaging. Going back to the original question--the steady-state solution is 0.707VAC = 0.707 VRMS (they have the same meaning). With that clarification—have fun, all!

  • Oops, got ahead of myself, that should have read:

    However, with the source at 1V over the first cycles the rms voltage across the _source_ will be less than [sqrt(2)/2]Vrms, further the rms voltage across the capacitor will be more than 0.5 Vrms and resistor will be less than 0.5 Vrms.

    Note: The rms voltage across the resistor will be slightly _less_, over any n periods, than the steady state until it reaches steady state.

    The rms voltage across the capacitor will be slightly _more_ than the steady state until it reaches steady state.

  • What Robert Scott explains is incorrect.

    The waveform of the source is 1VAC not 1Vrms.

    At the specified 1V magnitude for the source, the voltage across the capacitor (and resistor) is 0.707V or 0.5 Vrms (steady state).

    At the RS specified 1Vrms, the voltage across the the source is 0.707, and the voltage across the capacitor and resistor is 0.5V or 0.354Vrms (steady state).

    It is correct the the first ten cycles or so will not exactly line up with the steady state.

    However, with the source at 1V over the first cycles the rms voltage across the _source_ will be less than [sqrt(2)/2]Vrms, further the rms voltage across the capacitor and resistor will be less than 0.5 Vrms.

    Note: The rms voltage across the capacitor will be _less_, over any n periods, than the steady state until it reaches steady state.

    Robert, I am certain that with a 1 Vrms source no voltage in this circuit could exceed the peak of 0.707 [sqrt(2)/2] V generated.

    Would you hire me on the spot if I told you the above?