Two heating elements of resistances `R_1` and `R_2` when operated at a constant supply of voltage, V, consume powers `P_1` and `P_2` respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in (i) series and (ii) parallel across the same voltage supply.

Explanation

When heating elements of resistances `R_1` and `R_2` are operated at a constant supply of voltage V, power consumed by these are respectively
`P_1 = (V^2)/(R_1) " and " P_2 = (V^2)/(R_2)`
(i) When the two heating elements are connected in series across the same supply voltage V, then resultant resistance `R_s = R_1 + R_2`
` therefore ` Resultant power of series combination `P_s = (V^2)/(R_s) = (V^2)/((R_1 + R_2)`
`(1)/(P_s) = (R_1 + R_2)/(V^2) = (R_1)/(V^2) + (R_2)/(V^2) = (1)/(P_1)+ (1)/(P_2)`
(ii) When the two heating elements are connected in parallel across the same supply voltage V, then resultant resistance `R_P` is given by the relation
`(1)/(R_P) = (1)/(R_1) + (1)/(R_2) rArr (V^2)/(R_P) = (V^2)/(R_1) + (V^2)/(R_2) ` or `P_P = P_1 + P_2`