Somewhat OT Knob & Tube wiring (was Re: Power and the RA82)

[der Mouse]

> Or -- some meters when set to DC and fed AC will measure the RMS
> [kinda DC-equivalent value] voltage of the AC...

> 100+V AC * .707 = 70+V DC.

Not quite.  When one sees (say) "120V AC", that 120V *is* the RMS
value; "120V" is actually 120/sqrt(.5) = about 170V peak.  (Assuming a
pure sine wave, that is, which is close enough for practical purposes
when dealing with mains power.)

As for "kinda DC-equivalent", it *is* equivalent as far as mean power
delivered into a purely resistive load goes, which is why that measure
is used: 120 RMS volts AC into (say) 120 ohms will dissipate as much
power, long-term average, as 120 volts DC into the same load.  (RMS
here stands for Root-Mean-Square, which is a thumbnail description of
the mathematics involved: it's the square root of the mean value of the
square of the instantaneous voltage.  This is derived thus: let the
voltage, as a function of time, be v(t).  The current into a load of
resistance r will then, by Ohm's law, be i(t)=v(t)/r and the power will
be i(t)v(t), or v(t)^2/r.  If we now work let c be the constant DC
voltage necessary to dissipate the same amount of power, average, we
find that c^2/r=mean(v(t)^2)/r, leading directly to the RMS formula.
sqrt(2) comes into it because that's what the RMS computation produces
when v(t) is a pure sine wave.)

> The equation for figuring RMS voltage on sine-wave AC is: 1/sqrt(2)

Only if you start with the peak AC voltage, which is not what
voltmeters normally indicate when measuring AC voltages.

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