Question

An alternating voltage is given by $e = {e_1}\sin \omega t + {e_2}\cos \omega t$. Then the root means the square value of voltage.
(A) $\sqrt {e_1^2 + e_2^2} $
(B) $\sqrt {{e_1}{e_2}} $
(C) $\sqrt {\dfrac{{{e_1}{e_2}}}{2}} $
(D) $\sqrt {\dfrac{{e_1^2 + e_2^2}}{2}} $

Answer 1

Hint: The root mean square value of any physical quantity is given as
Let A is the any physical quantity
So, ${A_{rms}} = \sqrt {{{\bar A}^2}} $
or $A_{rms}^2 = \dfrac{{\int\limits_0^T {{A^2}dt} }}{{\int\limits_0^T {dt} }}$ (for complete circle)

Complete step by step solution:
Given that alternating voltage
$e = {e_1}\sin \omega t + {e_2}\cos \omega t$
So, for complete circle cone cycle of AC
${e_{rms}} = \sqrt {{{\bar e}^2}} $
$e_{rms}^2 = {\bar e^2}\dfrac{{\int\limits_0^T {{e^2}dt} }}{{\int\limits_0^T {dt} }}$
$e_{rms}^2 = \dfrac{{\int\limits_0^T {{{({e_1}\sin \omega t + {e_2}\cos \omega t)}^2}dt} }}{{\int\limits_0^T {dt} }}$
$e_{rms}^2 = \dfrac{1}{T}\left[ {e_1^2\int\limits_0^T {({{\sin }^2}\omega t)dt + {e_1}{e_2}\int\limits_0^T {(\sin 2\omega t)dt} } + e_2^2\int\limits_0^T {({{\cos }^2}\omega t)dt} } \right]$
$e_{rms}^2 = \dfrac{1}{T}\left[ {e_1^2\int\limits_0^T {\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right)dt + {e_1}{e_2}\left[ {\dfrac{{ - \cos 2\omega t}}{{2\omega }}} \right]_0^T} + e_2^2\int\limits_0^T {\left( {\dfrac{{1 + \cos 2\omega t}}{2}} \right)} dt} \right]$
$e_{rms}^2 = \dfrac{1}{T}\left[ {\dfrac{{e_1^2}}{2}\left( {t - \dfrac{{\sin 2\omega t}}{{2\omega }}} \right)_0^T - \dfrac{{{e_1}{e_2}}}{{2\omega }}(\cos 2\omega T - \cos 0) + \dfrac{{e_2^2}}{2}\left( {t + \dfrac{{\sin 2\omega t}}{{2\omega }}} \right)_0^T} \right]$
$e_{rms}^2 = \dfrac{1}{T}\left[ {\dfrac{{e_1^2}}{2}\left[ {T - \dfrac{1}{{2\omega }}\left( {\sin 2\omega T - \sin 0} \right)} \right] - \dfrac{{{e_1}{e_2}}}{{2\omega }}\left( {\cos 2\omega T - 1} \right) + \dfrac{{e_2^2}}{2}\left[ {T + \dfrac{1}{{2\omega }}\left( {\sin 2\omega T - \sin 0} \right)} \right]} \right]$
$\because T = \dfrac{{2\pi }}{\omega }$
$e_{rms}^2 = \dfrac{1}{T}\left[ {\dfrac{{e_1^2}}{2}\left[ {\dfrac{{2\pi }}{\omega } - \dfrac{1}{{2\omega }}\left( {\sin 2\omega \dfrac{{2\pi }}{\omega } - 0} \right) - \dfrac{{{e_1}{e_2}}}{{2\omega }}\left( {\cos 2\omega \dfrac{{2\pi }}{\omega } - 1} \right) + \dfrac{{e_2^2}}{2}\left[ {\dfrac{{2\pi }}{\omega } + \dfrac{1}{{2\omega }}\left( {\sin 2\omega \dfrac{{2\pi }}{\omega }} \right)} \right]} \right]} \right]$
$e_{rms}^2 = \dfrac{1}{T}\left[ {\dfrac{{e_1^2}}{2}\left[ {\dfrac{{2\pi }}{\omega } - 0} \right] - \dfrac{{{e_1}{e_2}}}{{2\omega }}\left( {1 - 1} \right) + \dfrac{{e_2^2}}{2}\left( {\dfrac{{2\pi }}{\omega } - 0} \right)} \right]$
$\left( {\sin n\pi = 0} \right)$ where $n = 0,1,2,....$
$(\cos 4\pi = 1)$
$e_{rms}^2 = \dfrac{\omega }{{2\pi }}\left[ {\dfrac{{e_1^2}}{2} \times \dfrac{{2\pi }}{\omega } + \dfrac{{e_2^2}}{2} \times \dfrac{{2\pi }}{\omega }} \right]$
$e_{rms}^2 = \dfrac{{e_1^2 + e_2^2}}{2}$
$\boxed{{e_{rms}} = \sqrt {\dfrac{{e_1^2 + e_2^2}}{2}} }$

Hence option D is correct.

Note: We know that the mean or expectation value of sin 0 & cos 0 is 0 and ${\sin ^2}0$ & ${\cos ^2}0$ is $\dfrac{1}{2}$.

So, we can directly calculate the root mean square value of alternating voltage.
$e_{rms}^2 = < {({e_1}\sin \omega t + {e_2}\cos \omega t)^2} > $
$ = e_1^2 < {\sin ^2}\omega t > + 2{e_1}{e_2} < \sin \omega t > < \cos \omega t > + e_2^2 < {\cos ^2}\omega t > $
$e_{rms}^2 = \dfrac{{e_1^2}}{2} + 0 + \dfrac{{e_2^2}}{2}$
$e_{rms}^2 = \dfrac{{e_1^2 + e_2^2}}{2}$
$\boxed{{e_{rms}} = \sqrt {\dfrac{{e_1^2 + e_2^2}}{2}} }$