Two bulbs are rated `(P_1, V)" and "(P_2, V)` . If they are connected (i) in series and (ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of `P_1` and `P_2`.

Explanation

Resistance of given bulbs are `R_1 = (V^2)/(P_1) " and " R_2 = (V^2)/(P_2)`
(i) In series grouping net resistance `R_s = R_1 + R_2`. If `P_s` be the power dissipated in series arrangement then `R_s = (V^2)/(P_s)`
`rArr (V^2)/(P_s) = (V^2)/(P_1) + (V^2)/(P_2) rArr (1)/(P_s) = (1)/(P_1) + (1)/(P_2) rArr P_s = (P_1 P_2)/(P_1 + P_2)`
(ii) In parallel grouping equivalent resistance `R_P = (V^2)/(P_P)` is given as per relation :
`(1)/(R_P) = (1)/(R_1) + (1)/(R_2) rArr (P_P)/(V^2) = (P_1)/(V^2) + (P_2)/(V^2) rArr P_P = P_1 + P_2`