A bulb with a rating of 250 V , 100 W is connected to a 220 V power supply situated 10 m away, utilizing a copper wire with a cross-sectional area of 5 mm ^2 . Determine the power consumed by the connecting wires. The resistivity of copper is 1.7 10⁻⁸ m .

Step-by-step solution

The resistance of the bulb is given by: R = V^2 P = 250^2 100 = 625 Since the power supply is situated 10 m away, a complete circuit requires both a forward and a return path. Thus, the total length of the connecting wire is: l = 2 10 = 20 m The resistance of the connecting wire is: R_w = l A = 1.7 10⁻⁸ 20 5 10⁻⁶ = 0.068 The total resistance of the circuit is: R_ eq = R + R_w = 625 + 0.068 625 The current flowing through the circuit is: I = V' R_ eq = 220 625 = 0.352 A The power consumed by the connecting wires is: