Physics - Current Electricity Question with Solution | TestHub

Resistance of lamp ${\mathrm{R}}_0=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{(30{)}^2}{90}=10\Omega$ Current in the lamp $\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_0}=\frac{30}{10}=3\mathrm{A}$ As the lamp is operated on $120\mathrm{V}\mathrm{DC}$, then resistance becomes ${\mathrm{R}}^{\prime }=\frac{{\mathrm{V}}^{\prime }}{\mathrm{i}}=\frac{120}{3}=40\Omega$ For proper glow, a resistance $\mathrm{R}$ is joined in series with the bulb $\begin{array}{rl}& {\mathrm{R}}^{\prime }=\mathrm{R}+{\mathrm{R}}_0\\ \Rightarrow & {\mathrm{R}}^{\alpha }={\mathrm{R}}^{\prime }-{\mathrm{R}}_0=40-10=30\Omega \end{array}$