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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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>
>