An electric current of 5 Amp is divided into three branches, the length of the resistance wires in three branches being proportional to 1 2 and 3. Find the current in each branch

R is directly proportional to the length of the conductor. So R¹:R²:R³ = 1:2:3Let R¹ be = xSo, R² = 2xAnd, R³ = 3xNow, R(eq.) = (R¹R²R³)/(R¹R² + R²R³ + R³R¹) = 6x/11Now, I = V/R=> 5 = V/(6x/11) = 11x/6 * V=> V = 30x/11 Now I² = V/R = 30x/11 * 1/2x = 15/11So, the current flowing in the middle branch will be 15/11 A.