R.Measures wrote:
> On Mar 22, 2005, at 8:16 AM, Gary Schafer wrote:
>
>
>>
>>R.Measures wrote:
>>
>>>On Mar 21, 2005, at 10:06 AM, Gary Schafer wrote:
>>>
>>>>R. Measures wrote:
>>>>
>>>>
>>>>>On Mar 21, 2005, at 9:08 AM, Gary Schafer wrote:
>>>>>
>>>>>
>>>>>>Tony King - W4ZT wrote:
>>>>>>
>>>>>>
>>>>>>
>>>>>>>Harold B. Mandel wrote:
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>><snip> PEP is different
>>>>>>>>than RMS, and RMS is different than Average. <snip>
>>>>>>>>
>>>>>>>>
>>>>>>>It may be worth mentioning that RMS is the usual method of
>>>>>>>defining a DC
>>>>>>>equivalent which, given a pure waveform, will be about the same
>>>>>>>as the
>>>>>>>average. With distorted waveforms RMS will not define the average
>>>>>>>properly.
>>>>>>
>>>>>>
>>>>>>It may also be worth mentioning that there really is no such thing
>>>>>>as
>>>>>>"RMS power". The proper term is average power. RMS is only valid in
>>>>>>terms of voltage or current.
>>>>>>
>>>>>>You can find an RMS value of power but it is not useful for
>>>>>>anything.
>>>>>
>>>>>It's fairly useful for heating water.
>>>>
>>>>Only average power is useful.
>>>
>>>Average power is not a measure of heating ability. RMS power is a
>>>measure of heating ability.
>>>cr
>>
>>That's a common misconception.
>>Only RMS voltage or RMS current through a resistor produces the
>>equivalent heating that the same value DC voltage or current provide
>>into that same resistor. When calculated for power they are all
>>average power.
>>
>>2 amps DC into 1 ohm = 4 watts average power.
>>2 amps RMS into 1 ohm = 4 watts average power.
>>
>>You can not find RMS power by multiplying RMS voltage by RMS current.
>
>
> I disagree. .
>
>
>>That gives average power.
>>
>>If you square RMS voltage and divide by resistance that gives average
>>power.
>>
>>If you square RMS current and multiply by resistance that gives
>>average power.
>>
>>73
>>Gary K4FMX
>>
>>
>>
>>
>
>
> Richard L. Measures, AG6K, 805.386.3734. www.somis.org
1. How much peak power do you get when you have 10 volts peak across a
50 ohm resistor?
@ 2 watts.
2. What is the average power across that resistor?
@ 1 watt. (average power = 1/2 peak power)
3. If you find the RMS voltage of that 10 volts peak that = 7.07 volts.
What do you get when you find power from the RMS voltage?
@ 1 watt.
Why would the one watt be average power in one case and RMS power in the
other ?
73
Gary K4FMX
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