Physics - Modern Physics Question with Solution | TestHub

Power, $P=\frac{V^2}{R},\Rightarrow R=\frac{V^2}{P}$ For the first bulb, $$R_1=\left(\frac{V^2}{P_1}\right)=\left(\frac{(220{)}^2}{25}\right)=1936\Omega$$ For the second bulb, $$R_2=\left(\frac{V^2}{P_2}\right)=\left(\frac{(220{)}^2}{100}\right)=484\Omega$$ Current in series combination is the same in the two bulbs, $i=\frac{V}{R_1+R_2}=\frac{220}{1936+484}=\frac{220}{2420}=\frac{1}{11}A$ If the actual powers in the two bulbs be $P_1$ and $P_2$ then $P_1^{\prime }=i^2{R}_1={\left(\frac{1}{11}\right)}^2\times 1936=16\mathrm{W}$ and $P_2^{\prime }=i^2{R}_2={\left(\frac{1}{11}\right)}^2\times 484=4\mathrm{W}$ Since $P_1^{\prime }>P_2^{\prime }$, so, $25\mathrm{W}$ bulb will glow more brightly.