An electric bulb rated as 200W at 100V is used in a circuit having 200V supply. The resistance “R” that must be put in series with the bulb so that the bulb delivers the same power is ___  $\Omega$
 

$\begin{array}{l}\mathrm{power}\mathrm{p}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\\ \mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{p}}=\frac{{\left(100\right)}^2}{200}=\frac{100\times 100}{200}\\ \mathrm{R}=50\mathrm{\Omega }\end{array}$

For the bulb to deliver same power, it should have its rated voltage across it. 

$\begin{array}{l}\therefore \mathrm{current}\mathrm{through}\mathrm{bulb}\mathrm{rated}200\mathrm{W}\mathrm{and}100\mathrm{V}\\ \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{100}{50}=2\mathrm{A}\\ \mathrm{load}\mathrm{resistance}{\mathrm{R}}_{\mathrm{L}}=\frac{{\mathrm{V}}_{\mathrm{L}}}{\mathrm{i}}=\frac{100}{2}=50\mathrm{\Omega }\end{array}$