Author Topic: Derivation of RMS (Read 1832 times) Tweet Share

Mao · « **on:** May 03, 2009, 10:27:51 pm »

WARNING - requires VCE Specialist Mathematics knowledge

(posting by request)

A sinusoidal AC supply has the voltage $V(t) = V_m \sin (\omega t + \phi)$ , and current $I(t) = I_m \sin (\omega t + \phi)$ (voltage and current have the same phase and frequency) [ $\omega = \frac{2\pi}{T}$ , where T is the period, $\phi$ is the initial phase and $V_m,\; I_m$ are the peak voltage and current]

Its power is hence $P(t) = V_m I_m \sin^2 (\omega t + \phi)$ .

The average power output can be given by $P_{ave} = \frac{\int_0^T P(t)\; dt}{T}$ , where T is the period.

$\implies P_{ave} = \frac{V_m I_m}{2T} \cdot \left[ t - \frac{1}{4\omega} \sin (2(\omega t + \phi)) \right]_0^T = \frac{V_m I_m}{2}$

That is, a DC signal with $V_{RMS}$ and $I_{RMS}$ could produce the same power output. $\implies V_{RMS}I_{RMS} = \frac{V_mI_m}{2}$
However, in an AC signal, the voltage varies with the current ( $V = kI$ ), hence, this becomes

$V_{RMS}(k\cdot V_{RMS}) = \frac{V_m (k\cdot V_m)}{2}$

$\implies V_{RMS} = \frac{V_m}{\sqrt{2}}$ , where V_m is the peak voltage

The word 'root mean squared' comes from this process, the voltage is squared, the mean of the squared voltage is taken, the square rooted to give the 'average' voltage. (which essentially is what's happening here, as the current is directly proportional to voltage)

Therefore, $V_{RMS} = \frac{\sqrt{2}}{4} V_{p-p}$ , $V_{p-p} = 2\sqrt{2}V_{RMS}$ (and the same for current)

TrueTears · « **Reply #1 on:** May 03, 2009, 10:32:31 pm »

Thank you very very much Mao. 8-) 8-) :coolsmiley: :coolsmiley:

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Author Topic: Derivation of RMS (Read 1832 times) Tweet Share

Mao

Derivation of RMS

TrueTears

Re: Derivation of RMS

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