Physics - EMI/AC Question with Solution | TestHub

Impedance of the circuit is given by

$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(\omega L{)}^2}$,

$Z=\sqrt{(50{)}^2+(0.2\times 250{)}^2}$

$Z=\sqrt{2500+(50{)}^2}=50\sqrt{2}$

Average power is given by

$\therefore$$P=V_{rms}{I}_{rms}\mathit{cos}\varphi$



$P=$$\frac{V_{rms}^2}{Z^2}R$

$P=\frac{{\left(\frac{200}{\sqrt{2}}\right)}^2}{{\left(50\sqrt{2}\right)}^2}\times 50=200 \mathrm{w}\mathrm{a}\mathrm{t}\mathrm{t}$