When three identical bulbs of $$60\,W,200\,V$$   rating are connected in series to a $$200\,V$$  supply, the power drawn by them will be

Solution :
Let $${R_1},{R_2}$$  and $${R_3}$$ are the resistances of three bulbs respectively.
In series order, $$R = {R_1} + {R_2} + {R_3}$$
but, $$R = \frac{{{V^2}}}{P}$$  and supply voltage in series order is the same as the rated voltage.
$$\eqalign{ & \therefore \frac{{{V^2}}}{P} = \frac{{{V^2}}}{{{P_1}}} + \frac{{{V^2}}}{{{P_2}}} + \frac{{{V^2}}}{{{P_3}}} \cr & {\text{or}}\,\,\frac{1}{P} = \frac{1}{{60}} + \frac{1}{{60}} + \frac{1}{{60}} \cr & {\text{or}}\,\,P = \frac{{60}}{3} = 20\,W \cr} $$
Alternative
As three bulbs have same power and voltage so, they have equal resistance. So, power equivalent when connected in series is given by,
$$\eqalign{ & \frac{1}{{{P_{EQ}}}} = \frac{1}{{{P_1}}} + \frac{1}{{{P_2}}} + \frac{1}{{{P_3}}} \cr & \frac{1}{{{P_{EQ}}}} = \frac{1}{P} + \frac{1}{P} + \frac{1}{P}\,\,\left( {{\text{as}}\,\,{P_1} = {P_2} = {P_3} = P} \right) \cr & {\text{So,}}\,\,{P_{EQ}} = \frac{P}{3} \Rightarrow \frac{{60}}{3} = 20\,W \cr} $$