Uprating a Workshop Breaker Without Rewiring
A small workshop is supplied by a single 230 V branch circuit protected by a 15 A miniature circuit breaker. The branch uses 2.5 mm² copper conductors with a total out-and-back length of 18 m; the resistivity of copper is 1.7 × 10^-8 Ω m. Three loads can be switched on together: a 3.0 kW heater, a 1.5 kW kettle, and 0.4 kW of lighting. The cable is run inside thermal insulation, and the manufacturer's current-carrying capacity for this installation is 18 A. The breaker trips during normal use. The owner proposes replacing the 15 A breaker with a 25 A breaker, which is cheaper than rewiring, so that the circuit stops tripping. A 25 A breaker of the same type is available. The supply voltage remains close to 230 V. The workshop owner asks whether this substitution is safe and, if not, what should be done instead.
1. Calculate the total load current and the voltage drop in the 2.5 mm² cable at that current. Does the 25 A breaker solve the problem safely?
Read the worked answer
Total power = 3000 + 1500 + 400 = 4900 W. Assuming unity power factor, I = P/V = 4900/230 ≈ 21.3 A. This already exceeds the 15 A breaker rating and the cable's 18 A installed capacity.
The cable resistance is R = ρL/A = (1.7 × 10^-8)(18)/(2.5 × 10^-6) ≈ 0.122 Ω. The voltage drop is IR ≈ 21.3 × 0.122 ≈ 2.6 V, about 1.1% of 230 V. That is small, so voltage drop is not the main issue. The cable would dissipate I²R ≈ (21.3)²(0.122) ≈ 55 W along its length, and more importantly it would carry 21.3 A continuously through insulation rated for 18 A.
A 25 A breaker would not trip at 21.3 A, so it would allow the cable to overheat. A breaker protects the cable, not just the load. The substitution is unsafe.
2. Propose a safe way to run the same three loads. Check the branch currents if the heater is placed on one 15 A circuit and the kettle plus lighting on another.
Read the worked answer
Split the loads onto separate circuits, each wired with cable of adequate installed capacity. On circuit A, the heater alone draws 3000/230 ≈ 13.0 A, below 15 A. On circuit B, the kettle plus lighting draw (1500 + 400)/230 = 1900/230 ≈ 8.3 A, also below 15 A. If each circuit has its own 2.5 mm² cable with an 18 A installed rating, both are acceptable.
Alternatively, keep one circuit but replace the cable with a size whose tabulated installed capacity is at least 25 A, and verify voltage drop and earthing. Simply changing the breaker is not enough. The key check is that the protective device rating does not exceed the cable's installed current-carrying capacity.
3. The owner argues that because 21.3 A is below 25 A, the 25 A breaker will never be overloaded. Explain why this argument is irrelevant to safety.
Read the worked answer
The argument confuses protecting the load with protecting the cable. The 25 A breaker is indeed rated above the normal 21.3 A load current, so it may not trip. That is exactly the danger: the cable is rated for only 18 A in its insulated installation, so it can carry 21.3 A for a long time while its conductor and insulation exceed their rated temperature. Insulation degrades, connections loosen, and the risk of fire increases.
A correct protective device must satisfy I_breaker ≤ I_cable,rated. Here 25 A > 18 A, so the combination is non-compliant. The breaker's job is to interrupt current before the cable is damaged, not to match the appliance current.