Question

Show that Irms= 0.707 I of an AC where symbols have there usual meaning​

Answer 1

Proof:

$\therefore \: I_{rms} = \sqrt{ \dfrac{ \displaystyle\int \: I_{ac} \: dt}{ \displaystyle\int \: dt} }$

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

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

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

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

$\implies \: I_{rms} = I \times \sqrt{ \dfrac{ \displaystyle T - \bigg \{\sin(2 \omega t) \bigg \}_{0}^{T} \: dt}{2T}}$

$\implies \: I_{rms} = I \times \sqrt{ \dfrac{ \displaystyle T - \bigg \{\sin(2 \times \frac{2\pi}{T} \times t) \bigg \}_{0}^{T}}{2T}}$

$\implies \: I_{rms} = I \times \sqrt{ \dfrac{ \displaystyle T - \bigg \{\sin(\frac{4\pi}{T} \times t) \bigg \}_{0}^{T} }{2T}}$

$\implies \: I_{rms} = I \times \sqrt{ \dfrac{ \displaystyle T - \bigg \{\sin(4\pi) - 0 \bigg \}}{2T}}$

$\implies \: I_{rms} = I \times \sqrt{ \dfrac{T}{2T}}$

$\implies \: I_{rms} = I \times \sqrt{ \dfrac{1}{2}}$

$\implies \: I_{rms} =0.707( I )$