A Blown Fuse in a Kitchen Circuit: Protection or Overload?
A household kitchen circuit is wired with 1.5 mm² copper cable, which the electrician's chart rates at 14 A continuous. The circuit is protected by a 15 A cartridge fuse and feeds three socket outlets. Plugged into it are a 2500 W immersion water heater, a 1000 W clothes iron and 300 W of LED lighting. On a cold morning all three are switched on at once from a 230 V supply. About four seconds later the fuse blows. The owner replaces it with a 20 A fuse; this one holds, but the cable at the wall box now becomes noticeably warm to the touch. He argues that the 20 A fuse is safer 'because it does not blow'. The heater element is nichrome wire, whose resistance when cold is roughly one-eighth of its resistance when red-hot, and the iron uses the same kind of element. The house has a spare 13 A fuse and a second, unused 14 A circuit to the kitchen.
1. Find the steady current drawn when all three appliances run together, and decide what rating of protective device this circuit should actually have.
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Total power = 2500 + 1000 + 300 = 3800 W, so at 230 V the steady current is I = P/V = 3800/230 ≈ 16.5 A.
So the 15 A fuse blowing is not a fault - the load genuinely exceeds it. But the cable is rated only 14 A, which means even the original 15 A fuse was already too large: a fuse protects the cable, and its rating must not exceed the cable's current-carrying capacity.
Correct choice: a 13 A fuse (or 13 A MCB) on this circuit, with the heater moved to the second, unused 14 A circuit. Then neither cable carries more than its rating.
The 20 A fuse is the dangerous option: it lets 16.5 A flow indefinitely in a 14 A cable. Heating in the conductor goes as I²R, so the cable is running about (16.5/14)² ≈ 1.4 times its rated heating - enough to soften PVC insulation over months and start a fire inside the wall, where nobody sees it.
2. The steady current is only 16.5 A, barely above the 15 A rating, yet the fuse ruptured within about four seconds. Explain the timing.
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A fuse is a thermal device: it melts when it has absorbed enough energy, so the time depends on I²t. At only 1.1 times its rating a 15 A fuse would take many minutes to operate - four seconds is far too fast for a 1.1× overload. Something much larger flowed briefly.
That is the cold-resistance effect. The heater's hot resistance is R = V²/P = 230²/2500 ≈ 21 Ω, but cold nichrome has about one-eighth of that, ≈ 2.6 Ω, giving an initial current near 230/2.6 ≈ 88 A. The iron behaves the same way, adding tens of amperes for a fraction of a second.
So the sequence is: an inrush surge of the order of a hundred amperes melts the element in a few seconds, the elements heat up, their resistance rises, and the current settles to 16.5 A - which is still above the cable rating, so the circuit was overloaded even after the surge ended.
3. The owner now proposes rewiring the three socket outlets with 2.5 mm² cable (rated 20 A) and keeping the 20 A fuse. Evaluate this proposal as an electrician would.
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It fixes one problem and creates another. With 2.5 mm² cable, a 20 A fuse does correctly protect the cable, so the wall-warming problem disappears.
But the fuse must protect the weakest thing in the chain, and that is no longer the cable. Socket outlets, the 13 A plugs on appliances and the flexible cords of the iron and heater are rated around 10–13 A. Under a 20 A fuse, a damaged flex carrying 15–18 A would glow and burn without the fuse ever operating.
A defensible design keeps 13 A protection on each socket outlet (a 13 A fused plug or fused spur does exactly this), or splits the load over two 14 A circuits. Relying on the user never to run all three appliances at once is not protection - it is a hope. The rule is: rate the fuse for the cable, and rate each outlet for its own flex.