RE: Average, RMS and Power Factor made easy!

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-uswest-dot-net>" <couturejh-at-worldnet.att-dot-net>


Terry -

I agree with your calculations now. The voltage is not zero and Mr. Ohm
rests in peace.

It should be noted that the electromagnetic ammeter normally used to measure
the TC input current actually measures average current but the meter is
calibrated in RMS amps.

John Couture

-------------------------------


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Thursday, January 11, 2001 7:51 PM
To: tesla-at-pupman-dot-com
Subject: RE: Average, RMS and Power Factor made easy!


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

Hi John,

	Ok John, well say the switch's resistance is 1E-1000 Ohms, the voltage is
1E-1000 Volts, the current is 1 Amp and the power is 1E-1000000 Watts :-)))
 Such an infinitesimally small amount of power will not wake Georg.

Cheers,

	Terry


At 04:45 PM 1/11/2001 -0800, you wrote:
>
>Terry, -
>
>If I understand you correctly this is better than over unity energy. How
can
>the current be 1 amp with no voltage? What happened to Ohm's law? I
>understand Mr.Ohm is shaking in his grave.
>
>John Couture
>
>--------------------------------------
>
>
>-----Original Message-----
>From: Tesla list [mailto:tesla-at-pupman-dot-com]
>Sent: Wednesday, January 10, 2001 11:40 AM
>To: tesla-at-pupman-dot-com
>Subject: 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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>
>