Question

Show that the effective value of a sinusoidal alternating current is
independent of frequency and
phase angle.​

Answer 1

Let the current be :

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

Now, effective current means RMS CURRENT:

$\rm i_{rms} = \sqrt{ \frac{ \displaystyle \int_{0}^{T} {i}^{2} \: dt }{ \displaystyle\int_{0}^{T} \: dt} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle \int_{0}^{T} { \sin}^{2}( \omega t) \: dt }{T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle \int_{0}^{T} \{1 - \cos( 2\omega t) \} \: dt }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle \int_{0}^{T} dt - \int_{0}^{T}\cos(2 \omega t)dt }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle T - \frac{ \sin(2 \omega T) }{2 \omega} }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle T - \frac{ \sin(2 \times \frac{2\pi}{T}\times T) }{2 \omega} }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle T - \frac{ \sin(4\pi) }{2 \omega} }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ \displaystyle T - \frac{ 0 }{2 \omega} }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ T }{2T} }$

$\rm \implies i_{rms} = i_{0} \sqrt{ \dfrac{ 1 }{2} }$

$\rm \implies i_{rms} = \dfrac{i_{0}}{ \sqrt{2} }$

So, the effective current is independent of frequency and phase angle.

[Hence proved].