Saved Bookmarks
| 1. |
The following table gives the lengths of four aluminium rods of uniform cross-section their diameters and potential difference maintained across the opposite ends of the rods. All the four rods are maintained at the same temperature. Now answer the following questions :(a) Which rod has maximum resistance ? (b) Which rod carries the maximum current ? (c) In which rod is the drift speed of electrons maximum ? (d) In which rod does the dissipation of electric energy take place at a maximum rate ? Give reason for each of your answer. |
|
Answer» Solution :(a) ` becauseR = (rho L)/(A) = (4 rho L)/(pi D^2) rArr R PROP (L)/(D^2)` ` therefore R_1 : R_2 : R_3 : R_4 = (L)/(9D^2) : (2L)/(D^2) : (3L)/(4D^2) : (3L)/(D^2) = 1/9 : 2 : 3/4 : 3` It shows that `R_4` is maximum (B) Current `I= V/R` ` therefore I_1 : I_2 : I_3 : I_4 = (V)/(R_1) : (3V)/(R_2) : (2V)/(R_3) : (V)/(R_4) = (V.9 D^2)/(L) : (3V.D^2)/(2L) : (2V.4D^2)/(3L) : (V.D^2)/(3L)` ` = (9VD^2)/(L) : (3VD^2)/(2L) : (8VD^2)/(3L) : (VD^2)/(3L) = 9 : 3/2 : 8/3 : 1/3` It shows that `I_1` is maximum ( c) Drift speed `v_d prop 1/A "or" v_d prop V/L` ` therefore (v_d)_1 : (v_d)_2 : (v_d)_3 : (v_d)_4 = V/L : (3V)/(2L) : (2V)/(3L) : (V)/(3L) = 1 : 3/2 : 2/3 : 1/3` It shows that `(v_d)_2` is maximum (d) Rate of electrical energy dissipated `P = (V^2)/(R ) = (pi V^2 D^2)/(4 rho L) = P prop (V^2 D^2)/(L)` ` therefore P_1 : P_2 :P_3 :P_4 = (V^2 (3D)^2)/(L) : ((3V)^2 .D^2)/(2L) : ((2V)^2.(2D)^2)/(3L) : (V^2 D^2)/(3L) = 9 : 9/2 : 16/3 : 1/3 ` It shows that `P_1` is maximum. |
|