Two cylindrical wires, A and B, are made of the same material and have identical lengths. The radius of wire A is twice the radius of wire B. If these two wires are connected in series across a constant voltage source, what will be the ratio of the heat dissipated in wire A to that in wire B during the same time interval?

Step-by-step solution

Resistance of a cylindrical wire is given by R = L r^2 . Since both wires have the same material ( ) and length ( L ), the resistance is inversely proportional to the square of the radius: R 1 r^2 . Given r_A = 2r_B , the ratio of their resistances is: R_A R_B = ( r_B r_A )^2 = ( r_B 2r_B )^2 = 1 4 When connected in series, the same current I flows through both wires. The heat dissipated in time t is H = I^2 R t . Since I and t are constant, H R . Therefore, the ratio of heat dissipated is: H_A H_B = R_A R_B = 1 4