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

Options

A. `(P_(1) + P_(2))`
B. `sqrt(P_(1).P_(2))`
C. `P_(1)P_(2)//(P_(1) +P_(2))`
D. `(P_(1)+P_(2))//P_(1)P_(2)`

Answer

D

Explanation

Resistance of` I^(st)` bulb = `R_(1)=(V^(2)/(P_(1)`
Resistance of `11^(nt)` 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)(P_(1)+P^(2)))/(P_(1)P_(2))`
Hence, powerconsumed`P=(V^(2))/(R)=(V^(2))/(V^(2))(P_(1)+P_(2))/(P_(1)P_(2))=(P_(1)P_(2))/(p_(1)+p_(2))`
Hence, (C) is correct