Physics - EMI/AC Question with Solution | TestHub

Given, $I=6+10\sin \omega t$ $$\begin{array}{rl}{I}_{\mathrm{eft}}& =[\frac{{{\int }_0^T{I}^{2}dt]}^{1/2}}{{\int }_0^{T}dt}={\left[\frac{1}{T}{\int }_0^T(6+10\sin \omega t{)}^{2}dt\right]}^{1/2}\\ & ={\left[\frac{1}{T}{\int }_0^T\left(36+120\sin \omega t+100{\sin }^2\omega t\right)dt\right]}^{1/2}\end{array}$$ But as, $\frac{1}{T}{\int }_0^T\sin \omega tdt=0$ and $\frac{1}{T}{\int }_0^T{\sin }^2\omega t=\frac{1}{2}$ $$\Rightarrow I_{\mathrm{eff}}={\left[36+\frac{1}{2}\times 100\right]}^{1/2}=9.27$$ Thus, $I_{\text{eff }}=9.27\mathrm{A}$.