# Calculating RMS current with a modified sine wave

Discussion in 'General Electronics Chat' started by bitrex, Oct 7, 2010.

1. ### bitrex Thread Starter Member

Dec 13, 2009
79
4
If you have a chance, please look at the following PDF file, page 11:

http://www.tube-tester.com/sites/nixie/dat_arch/ITT_app_notes.pdf

They're calculating the RMS value of current through a NIXIE tube driven from a full-wave rectified supply. Since the tube "strikes" at a certain voltage, and then drops out of conduction when the waveform drops back to that voltage, the calculation of the RMS voltage has to be done via an integration of the voltage between the starting angle $-\theta$ and the stopping angle $\theta$. What I don't understand is why they also evaluated R between the limits of $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Since the equation for RMS current is simply $I = \frac{Vrms}{R}$, shouldn't the R just move outside the integral as a constant? Thanks for any insight!

2. ### timrobbins Active Member

Aug 29, 2009
318
16
Are you talking about the equations on the documents page 8 ? If so, then R isn't inside the integral (!?!?) - only angular time is.

3. ### bitrex Thread Starter Member

Dec 13, 2009
79
4
I guess what I'm asking is, I don't understand why the integral of angular time in the denominator has to be done at all. Since the integral in the numerator gives the RMS voltage, why wouldn't the RMS current just be 1/R^2 times that?

4. ### timrobbins Active Member

Aug 29, 2009
318
16
Calculating rms means integrating over the complete cycle (or half cycle in the case), so you have to sweep time from -pi/2 to +pi/2, even though the numerator is non-zero during just a portion. If you didn't sweep time over the full cycle, but just the non-zero portion of the numerator, then the time weighted averaging effect would give a higher value.