Two bulbs A and B of ratings forty W – two hundred V and one hundred W and two hundred V respectively are connected in series to a four hundred V d.c. supply. Then

the resistance of bulbs A and B are

${\mathrm{R}}_{\mathrm{A}}=\frac{{\mathrm{V}}^2}{{\mathrm{P}}_{\mathrm{A}}}=\frac{{\left(200\right)}^2}{40}=1000\mathrm{\Omega }$

and $R_B=\frac{V^2}{P_B}=\frac{{\left(200\right)}^2}{100}=400\Omega$

The  current stings (i.e maximum current they can withstand ) are

$$\begin{gathered} {\mathrm{I}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}}{\mathrm{V}}=\frac{40}{200}=0.2\mathrm{A} \\ {\mathrm{I}}_{\mathrm{B}}=\frac{{\mathrm{P}}_{\mathrm{B}}}{\mathrm{V}}=\frac{100}{200}=0.5\mathrm{A} \end{gathered}$$

when the bulbs are connected in series , their combined resistance is $R=R_A+R_B=1000+400=1400\Omega .$ . The current through each bulb is

$\mathrm{I}=\frac{{\mathrm{V}}^{\text{'}}}{\mathrm{R}}=\frac{400}{1400}=0.286\mathrm{A}$

since I is greater than ${\mathrm{I}}_{\mathrm{A}}$ but less than ${\mathrm{I}}_{\mathrm{B}}$; bulb A will fuse but bulb B will not.