Two elelctric bulbs P and Q have their resistance in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.

Explanation

Let P=x and Q= 2x
Given `P : Q = x : 2x - 1:2`
For bulb P power loss `P_(1) = (V_(2))/(R_(1)) = (V_(2))/(P)`
`P_(1) = (V^(2))/(x)`
Similarly `P_(2) = (V^(2))/(Q) = (V^(2))/(2x)`
Ratio of power loss `(P_(1))/(P_(2)) = (cancel(V)^(cancel(2))/(cancel(H)))/((cancel(V)^(2))/(2cancel(H))`
`P_(1) : P_(2) = 2:1`