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.
Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig
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