Physics - EMI/AC Question with Solution | TestHub

If the maximum value of induced EMF is$\mathrm{V}$then$50\mathrm{\%}$of induced EMF is$\frac{\mathrm{V}}{2}$. Also, at this moment voltage across resistor will be$\frac{\mathrm{V}}{2}$.

$\mathrm{E}=\frac{1}{2}\mathrm{L}{\mathrm{i}}^2$

$\mathrm{E}=\frac{1}{2}\mathrm{L}{\left(\frac{\frac{\mathrm{V}}{2}}{\mathrm{R}}\right)}^2$

$\mathrm{E}=\frac{1}{2}\left(5\times 10^{-3}\right){\left(\frac{2}{2\left(1\right)}\right)}^2$

$\mathrm{E}=2.5\mathrm{ }\mathrm{m}\mathrm{J}$