Physics - Current Electricity Question with Solution | TestHub

Resistance of bulb${\mathrm{B}}_1,\mathrm{ }{\mathrm{R}}_1=\frac{{\left(250\right)}^2}{100}=\mathrm{ }625\mathrm{ }\mathrm{\Omega }$

Resistance of bulb${\mathrm{B}}_2,\mathrm{ }{\mathrm{R}}_2=\frac{{\left(200\right)}^2}{100}=\mathrm{ }400\mathrm{ }\mathrm{\Omega }$



${\mathrm{V}}_{{\mathrm{R}}_2}=\mathrm{ }\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}\times \mathrm{ }250=\frac{400}{625+400}\times \mathrm{ }250=\mathrm{98 }\mathrm{V}$