Saved Bookmarks
| 1. |
Two heating elements of resistances R_1 and R_2when 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. |
|
Answer» SOLUTION :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` |
|