An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

Ans: Option D - 25 W

Energy consumed by an appliance is given by the expression,

                                                          $\mathrm{P}=\mathrm{VI}=\raisebox{1ex}{${\mathrm{V}}^2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{R}$}\right.$

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. 

So, resistance,

                                                         $\mathrm{R}=\raisebox{1ex}{${\mathrm{V}}^2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{P}$}\right.$

Given $\mathrm{P} = 100 \mathrm{W}$ and $\mathrm{V} = 220 \mathrm{V}$

                                              $\Rightarrow \mathrm{R}=\frac{ {\left(220\right)}^2}{100}= 484 \mathrm{\Omega }$

Now if the bulb is operated on ${\mathrm{V}}^{\text{'}}=110 \mathrm{V}$ , then the energy consumed by it is given by the expression for power as

                                                   ${\mathrm{P}}^{\text{'}}=\frac{ {( {\mathrm{V}}^{\text{'}})}^2}{\mathrm{R}}=\frac{ {( 110)}^2}{484}$

                                                         $\Rightarrow {\mathrm{P}}^{\text{'}}=25 \mathrm{W}$