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Re: Average, RMS and Power Factor made easy!



Original poster: "Terry Fritz" <twftesla-at-uswest-dot-net>

Hi,

Here is a simple example were RMS voltage and RMS current readings of the
signals going into a load will not tell you the power the load is using.

Imagine a 1 volt battery that puts out 1 amp of current.  Put a RMS
voltmeter across it and an RMS current meter in series with one terminal of
it.  Now feed the output to a switch that is switching from on to off at
say 100Hz with a 50% duty cycle.  We'll assume the switch is a really good
fast one.

When the switch is open the voltage is 1 volt but the current is zero.
When the switch is closed the current is one amp but the voltage is now
zero (it shorts the battery's output).  So there is no time when both the
voltage and current are both non-zero.  Thus, the power going into the
switch is zero since P = V x I will always be zero at every instant in time.

However.....  What will the meters say??  The volt meter will read 0.5 VRMS
and the current meter will read 0.5 ARMS!!  If you multiply them the power
appears to be 0.25 watts.  But, of course, that is not true.

RMS meters are wonderful things but you just have to be a bit careful that
you don't try and use them in the wrong situations.  It is critical that
one knows the timing relationship between the voltage and current to
determine power.  Thus is where the COS(theta) function comes in for steady
state AC sine wave situations and where the P = integral (0,T) (v(t) x i(t)
dt) formula is needed when things really get nasty.  All the last formula
does is add up every tiny V x I chunk for every tiny instant of time.  Thus
by doing the calculation in tiny pieces and adding them all up, you can't
miss anything.  The last formula is needed even in this simple case but it
also has the power to handle any case if you can figure out the equations
and math (computers can if all else fails ;-))

Cheers,

	Terry


At 04:15 PM 1/10/2001 +0000, you wrote:
>Read the spec sheet on your Multimeter, and see if it doesn't just give
>you a readout in RMS.
>
>Or, use an oscilliscope which reads peak to peak.  Then find the peak
>to peak voltage, divide by two, and multiply by seven tenths, and seven
>hundredths.
>
>10 volts peak to peak translates to
>5 volts peak, and 5 volts peak translates to
>3.535 Volts RMS.
>
>The RMS voltage gives you the DC equivalent of AC voltage.  Since, a
>sine wave is not continually on, or off the length of time, and the
>amount of power over time is reduced to a value equal an amount of
>power from a continuous source.
>
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