Mean Value and Root Mean Square Value of Alternating Current

Derivation of Mean Or Average Value of Alternating Current:

Let us consider alternating current $i$ propagating in a circuit then the average value of current.

$ i_{mean}=\frac{1}{\left ( \frac{T}{2} \right )}\int_{0}^{\frac{T}{2}}i \:dt \qquad (1)$

$ where \quad i = i_{0}sin \omega t\quad(2)$

Now substitute the value of current $i$ in above equation $(1)$ 

$ i_{mean}= \frac{2}{T}\int_{0}^{\frac{T}{2}}i_0. sin \omega t.dt$

$ i_{mean}= \frac{2.i_{0}}{T}\int_{0}^{\frac{T}{2}}\sin \omega t.dt$

$ i_{mean}= \frac{2 i_{0}}{T}[\frac{-cos\:\omega t}{\omega} ]_{0}^{\frac{T}{2}}$

The value of $\omega$ is $\frac{2 \pi}{T}$ i.e $\omega=\frac{2\pi}{T}$

$ i_{mean}= \frac{2 i_{0}}{T \left (\frac{2\pi}{T} \right )}\left [ -cos \left (\frac{2 \pi}{T} \right ) \left ( \frac{T}{2} \right ) \\ \qquad \qquad \qquad +cos0^\circ \right ] $

$ i_{mean}= \frac{i_{0}}{\pi}\left [ - cos\pi+cos0^{\circ} \right ]$

$ i_{mean}=\frac{i_{0}}{\pi}\left [1+1 \right ]$

$ i_{mean} =\frac{2 i_{0}}{\pi}$

$ i_{mean} = 0.637\: i_{0}$

Thus, The mean (or average) value of alternating current for the cycle is $0.637$ times or $63.7 \%$ of the peak value.

Root mean square value of Alternating Current:

Let us consider current $i$ propagating in a circuit then the mean square value of alternating current.

$ \left (i_{mean} \right )^{2} = \frac{1}{T}\int_{0}^{T}i^{2}.dt \qquad(1)$

$ where\quad i= i_{0} \dot sin\omega t\qquad (2)$

Now substitute the value of current $i$ in above equation $(1)$

$ \left ( i_{mean} \right )^{2} = \frac{1}{T}\int_{0}^{T} \left ( i_{0} \:sin\omega t\right )^{2} dt $

$ \left (i_{mean} \right )^{2} = \frac{i_{0}^2}{T}\quad\int_{0}^{T} sin^{2}\omega t.dt$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{T}\int_{0}^{T} \frac{1-cos2\omega t}{2}\ dt$

$ \left (i_{mean} \right )^{2} = \frac{i_o^{2}}{2T}\int_{0}^{T}\left ( 1-cos2\omega t \right )dt$

$ \left (i_{mean} \right )^{2} = \frac{i_{0}^{2}}{2T}\left [ \left ( t \right )_{0}^{T} - \left ( \frac{sin2 \omega t}{2 \omega} \right )_{0}^{T} \right ] $

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ \left ( T-0 \right ) \\ \qquad\qquad\qquad -\frac{1}{2\omega}\left ( sin2\omega T-sin 0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ \left ( T-0 \right ) \\ \qquad\qquad\qquad -\frac{1}{2\omega}\left ( sin2\omega T-sin0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ T-\frac{1}{2\omega}\left ( sin4\pi \\ \qquad \qquad\qquad  -sin0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ T-\frac{1}{2\omega}\left ( 0-0 \right ) \right ]$

$ \left ( i_{mean} \right )^{2}=\frac{i_0^{2}}{2}$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2}$

So root mean square value of above equation:

$ i_{rms} = \sqrt{i_{mean}^{2}}$

$i_{rms} = \frac{i_{0}}{\sqrt{2}}$

$i_{rms} = 0.707\:i_{0}$

Thus, the root mean square value of an alternating current is $0.707$ times or $70.7 \%$ of the peak value.

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