We Know that,
I = I0 sinωt
\(I^2=\frac{1}{T}\int\limits_0^TI^2\,dt\)
\(=\frac{1}{T}\int\limits_0^T I^2_0\,sin^2ωt\,dt\)
\(=\frac{I_0^2}{T}\int\limits_0^T(\frac{1-cos2ωt}{2})dt\)
\(=\frac{I_0^2}{2T}[t-\frac{sin\,2ωt}{2ω}]^T_0\)
\(=\frac{I_0^2}{2T}[T-0]\)
\(=\frac{I_0^2}{2}\)
Roots mean square value
\(I_{rms}=\sqrt{I^2}\)
\(⇒ \sqrt{\frac{I_0^2}{2}}⇒ \frac{I_0}{\sqrt{2}}\)
