When three identical bulbs of 60W, 200 V rating are connected in series to a 200 V supply,

            the power drawn by them will be

: Let, R₁,R₂,R₃ are the resistances of three bulbs respectively.

In series order,

$$\begin{gathered} \text{R}={\text{R}}_{\text{1}}+{\text{R}}_{\text{2}}+{\text{\hspace{0.17em}R}}_{\text{3}} \\ But \text{R}={\text{V}}_{\text{2}}\text{/P and supply voltage in series order is the same as the rated voltage} \\ \therefore \frac{{\text{V}}_{\text{2}}}{\text{P}}=\frac{{\text{V}}_{\text{2}}}{{\text{P}}_1}+\frac{{\text{V}}_{\text{2}}}{{\text{P}}_2}+\frac{{\text{V}}_{\text{2}}}{{\text{P}}_3} \\ OR \\ \frac{1}{\text{P}}=\frac{1}{60}+\frac{1}{60}+\frac{1}{60} \\ \text{P}=\frac{60}{3}=20\text{\hspace{0.17em}W} \end{gathered}$$