Two cylindrical wires, A and B, made of the same material and having the same length, are connected in series across a voltage source. The ratio of the thermal energy developed in wire A to that in wire B in a given time is 9:1 . The ratio of the radius of wire A to the radius of wire B is

Step-by-step solution

In a series combination, the current I flowing through both wires is the same. The thermal energy developed in a given time t is given by H = I^2 R t . Therefore, the ratio of heat produced is H_A H_B = R_A R_B = 9 1 . The resistance of a wire is given by R = L A = L r^2 . Since both wires have the same material (same ) and same length L , their resistance is inversely proportional to the square of their radius: R 1 r^2 . Thus, R_A R_B = ( r_B r_A )^2 = 9 1 . Taking the square root gives r_B r_A = 3 1 , which means