Re: Power Measuring!! Need Assistance

On Tue, 13 Dec 2005 13:00:58 +0000 (GMT), Tony Williams
<tonyw@xxxxxxxxxxxxxxxxxxx> wrote:

>In article <codtp11u3ip3c717i5e7e7cnok269kc0t9@xxxxxxx>,
> theJackal <f_a_s_t_g_s_f_r_i_d_e_r@xxxxxxxxxxx> wrote:
>
>> On Tue, 13 Dec 2005 15:51:14 +1100, "Phil Allison"
>> <philallison@xxxxxxxxxx> wrote:
>> >True Power = average value of V x I
>> >
>> >where V = instantaneous voltage and I = instantaneous current.
>
>
>> Totally wrong! The average value of a sinusoidal waveform is
>> Zero ... so according to your answer the True Power = 0!
>
> Umm......
>
> The product, Vinst*Iinst, is a sine wave which has
> an average dc-value proportional to the true power.
>
> The product only has an average value of Zero when
> Iinst is exactly 90 degrees away from Vinst.
> ie, When the load is a pure reactance, = zero power.

I know its quite easy to see but this Phil chap has been bugging me
ever since i started writing here ... so I was just teasing him.

Let me give you my version of the facts .
To actually talk about the average value of a sine wave over ONE
period is senseless as its zero. Thats why when we mention 115V or
250V we are actually talking about a different kind of average the rms
which is the square root of the sum of the squares of the averages
(over 1 period) of the instantaneous voltages . Square the Sine
function and its positive all over 1 period ... and so it makes sense
finding its average which is called the rms. Its also interesting to
note that this rms value would actually give the same power
dissipation as a direct current or voltage of the same value through
a resistive load.

"Go easy on the whisky"

theJackal
.

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