Two bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a source of constant voltage. What will be the ratio of power dissipation in these?

Explanation

Given `R_(1) : R_(2) = 1 : 2` In parallel series
Dissipated power `P = (V^(2))/(R )`
`implies P prop (1)/(R )` [`:'` V = constant]
The ratio of dissipated powers in two bulbs is given by,
`(P_(1))/(P_(2)) = (R_(2))/(R_(1)) = (2)/(1)`
`:. P_(1) : P_(2) = 2 : 1`