Extension-Cord Losses and Fuse Selection for a Heater
A household uses a 1500 W, 230 V electric heater through a 25 m extension cord made of two 1.0 mm² copper conductors (resistivity 1.7 × 10⁻⁸ Ω m). The heater is marked 1500 W at 230 V, so its resistance is taken as constant at the rated value. The cord is fitted with a 6 A fuse. On a cold morning the user notices that the heater seems slower to warm the room, and the cord feels slightly warm along its whole length. A technician measures the supply voltage as exactly 230 V at the socket and the current as 6.4 A. The user asks whether the 6 A fuse is appropriate, whether a 1.5 mm² cord would improve heater output, and why the cord warms even though copper is a good conductor. Assume the supply is ideal and the heater resistance does not change with temperature.
1. Calculate the heater resistance, the actual current, the power delivered to the heater, the power dissipated in the cord, and the voltage at the heater.
Read the worked answer
R_heater = V²/P = 230²/1500 = 35.27 Ω.
The cord has two conductors, so the total copper length is L = 2 × 25 = 50 m. With A = 1.0 × 10⁻⁶ m²: R_cord = ρL/A = (1.7 × 10⁻⁸)(50)/(1.0 × 10⁻⁶) = 0.85 Ω.
Total resistance = 35.27 + 0.85 = 36.12 Ω. I = 230/36.12 = 6.37 A, in agreement with the measured 6.4 A.
Heater power = I²R_heater = (6.37)²(35.27) = 1.43 × 10³ W. Cord loss = I²R_cord = (6.37)²(0.85) = 34.5 W. Heater voltage = IR_heater = 6.37 × 35.27 = 224.7 V.
2. Is the 6 A fuse appropriate? Explain what the fuse is protecting and what should be changed.
Read the worked answer
The steady current of 6.37 A exceeds the 6 A fuse rating. A fuse is a slow thermal device, so it may not open immediately, but on a long run it will eventually open or nuisance-blow. The fuse is there mainly to protect the cord, not the heater.
A replacement must not exceed the cord’s current rating. For a 1.0 mm² flexible cord, a 10 A fuse is a common safe choice if the cord is rated for it. The heater’s reduced output is explained by the 34.5 W lost in the cord and the reduced terminal voltage, not by a faulty heater.
3. Estimate the change if a 1.5 mm² copper cord were used instead. Does the cord size restore the full 1500 W?
Read the worked answer
Resistance scales inversely with cross-sectional area: R_cord(new) = 0.85 × (1.0/1.5) = 0.567 Ω.
Total resistance = 35.27 + 0.567 = 35.84 Ω. I = 230/35.84 = 6.42 A. Heater power = (6.42)²(35.27) = 1.453 × 10³ W. Cord loss = (6.42)²(0.567) = 23.4 W.
The gain is about 22 W, roughly 1.5%. It reduces cord heating but does not restore the full 1500 W. The cord warms because even low-resistivity copper has finite resistance, and the small cross-section makes the 34.5 W appear along the cord’s length.