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Re: [Amps] Measuring RF Power

To: "R.Measures" <r@somis.org>
Subject: Re: [Amps] Measuring RF Power
From: Gary Schafer <garyschafer@comcast.net>
Reply-to: garyschafer@comcast.net
Date: Tue, 22 Mar 2005 14:18:23 -0500
List-post: <mailto:amps@contesting.com>

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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