Question

a bulb is rated 220 V 100 W
a.what is its resistance
b.what safe current can be passe through it

Answer 1

Dear Student ,
$$\begin{gathered} R=V/I. \\ P\left(Power\right)=100W. \\ V=220V \\ P=VI \\ \Rightarrow 100=220I \\ \Rightarrow I=5/11 \\ R=\frac{220}{5/11}=\frac{220\times 11}{5}=484\Omega \end{gathered}$$
Regards