Two electric bulbs rated `P_(1)` and `P_(2)` watt at `V` volt are connected in series across `V` volt mains then their total power consumption `P` is

Answer

C

Explanation

Resistance of `I^(st)`=bulb|  `R_(1)`=`(V_(2))/(P_(1))`     Resistance `II^(nd)`bulb `R_(2)` =`(V_(2))/(P_(2))`
When both bulbs are connected in series       `R_(eq)`=`V^(2)[(1)/(P_(1))+(1)/(P_(2))]`=`(V^(2)(_(1)+P_(2)))/(P_(1)P_(2))`
Hence power consumed  ``P=(V^(2))/(R)`=`(V^(2))/(V^(2)((P_(1)+P_(2))/(P_(1)P_(2))))`