[Amps] load capacitor ratings?

[Will Matney]

Rich,
Yep, your correct, I wasn't thinking about the differences in the 
readings. So in turn it would double the voltage values. Don't know why 
I was thinking that, must be getting senile?

Will

R. Measures wrote:

>
> On Aug 30, 2004, at 10:42 PM, Will Matney wrote:
>
>> Rich,
>> 1500 watts PEP on a 300 ohm load. Sqrt of 1500 X 300 = 670.8 Volts. 
>> 1500 v  x 300 ohms = 450,000 and the sqrt of 450,000 = 670.8. On a 50 
>> ohm load, 1500 watts PEP x 50 = 75,000 and the sqrt of 75,000 = 
>> 273.86 V. That would be in peak volts though as I would figure the 
>> power measured with a true peak reading wattmeter. The peak voltage 
>> would then have to yield a peak voltage in this circumstance wouldn't 
>> it? An RMS reading watt meter would be 1.414 times the voltage 
>> calculated yielding peak voltage?
>
>
> For a sinewave, RMS V is 0.707 of peak V.   When talking about AC, the 
> type of V must be specified as either peak, peak-to-peak, or RMS.  
> Also, PEP is not peak power, it is based on the RMS value of the 
> observed peak.  Thus, 1500W-PEP = 3000W-peak.
>
>>
>> Will
>>
>> R. Measures wrote:
>>
>>>
>>> On Aug 30, 2004, at 6:20 PM, Will Matney wrote:
>>>
>>>> I failed to mention that the values I were mentioning earlier, in 
>>>> the designs I was using, were for a Pi tank circuit and not a Pi-L. 
>>>> Thus the load C would have seen a different load resistance of 
>>>> about 300 ohms or so. For 1500 watts PEP on a 300 ohm load, a 
>>>> voltage of 671 volts would be incurred. At the 50 ohm load for a Pi 
>>>> tank, load C would see about 274 volts. So a capacitor with a 
>>>> rating of at least 2.5 : 1 difference would be needed. Where V = 
>>>> sqrt of W x R.
>>>>
>>> -  Peak-V or RMS-V?
>>>
>>>> Will Matney
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>>>>
>>>>
>>> Richard L. Measures, AG6K, 805.386.3734.  www.somis.org
>>>
>>>
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>>
>>
>>
> Richard L. Measures, AG6K, 805.386.3734.  www.somis.org
>
>
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