Three `60 W`, `120 V` light bulbs are connected across a `120 V` power source. If resistance of each bulb does not change with current then find out total power deliver to the three bulbs.

Answer

`40 watt`

Explanation

Let `R` be the resistance of each bulb ,Here bulb `B and C` are in parallel .Their effective resistance
`R_(p) = (R xx R)/(R + R) = (R )/(2)`
These bulb togather are in series with bulb A since in series combination of resistance potential divides in proportion to resistance so
`V_(A) = (R xx V)/(R + R//2) = (2)/(3) V = (2)/(3) xx 120 = 80 V`
`V_(B) = V_(C) = (R//2 xxV)/(R+ R//2) = (1)/(3) V = (1)/(3) xx 120 = 4 0 V`
Total power of bulb B and C
`P' = 60 + 60 = 120 W`
Total power of all the there bulb will be
`(1)/(P) = (1)/(P_(A)) + (1)/(P') = (1)/(60) + (1)/(120) = (3)/(120) = (1)/(40)`
or `P= 40` watt