The Dimming Hostel Room: Breaker, Cable or Supply?
A student volunteer is asked to audit the electrical wiring of a hostel common room before a college fest. The room is fed from a distribution board by a single-phase 230 V circuit wired with 1.0 mm² copper cable (resistivity 1.7 × 10⁻⁸ Ω·m), which the manufacturer rates for 11 A continuous. The cable run from the board to the room is 28 m one way. The circuit is protected by a 6 A miniature circuit breaker. Connected loads are six LED tube lights rated 20 W each, one 1500 W immersion heater and one 750 W microwave oven. With only the lights on, a multimeter at the socket reads 231 V. When the heater is switched on, the tubes visibly dim and the same socket reads 214 V; the breaker does not trip even after ten minutes. The warden wants to simply replace the 6 A breaker with a 16 A unit so that everything can run together. The student must decide whether that is safe, what the 17 V drop actually indicates, and what the correct remedy is.
1. Compute the current drawn when all loads run together, and judge whether the warden's 16 A breaker swap is acceptable.
Read the worked answer
Total connected load = 6 × 20 + 1500 + 750 = 2370 W. At 230 V the nominal current is 2370/230 ≈ 10.3 A, so the 6 A breaker is genuinely undersized for the room's equipment as a group.
That does not make the swap safe. The breaker's job is to protect the cable, not the appliances. A 1.0 mm² conductor rated 11 A protected by a 16 A breaker can sit at 1.5 times its rating indefinitely without the breaker ever acting; the insulation then degrades and the risk moves from a tripped breaker to a fire inside a wall.
The refusal to trip is itself a clue. At 214 V the heater's resistance (≈35.3 Ω) draws only about 6.1 A, and the six tubes add ≈0.6 A, so the real current is just over 6 A - a marginal overload that a thermal-magnetic MCB can sustain for many minutes. The load is therefore smaller than nameplate suggests, which means the fix is not 'more current', it is 'a properly sized dedicated circuit'.
2. Estimate the voltage drop caused by the 28 m cable run and decide whether the cable explains the measured 17 V drop.
Read the worked answer
Round-trip length = 2 × 28 = 56 m.
R = ρL/A = (1.7 × 10⁻⁸ × 56)/(1.0 × 10⁻⁶) ≈ 0.95 Ω.
At the actual current of roughly 6.7 A, the cable accounts for only 0.95 × 6.7 ≈ 6.4 V - well short of the 17 V observed. So about 10–11 V originates upstream: incoming service, meter tails, board busbar, or other rooms sharing the feeder.
A single socket measurement cannot separate the two contributions. The decisive test is to log the socket voltage and the board's outgoing terminal voltage simultaneously while the heater cycles:
- Both fall by ~17 V → the cable is innocent; the fault is upstream, and a loose service joint is a genuine fire hazard.
- Board holds ~231 V while the socket falls → the run is at fault, but then a healthy 0.95 Ω of copper cannot produce 17 V at 6.7 A, so suspect a bad joint, corrosion or broken strands rather than the conductor length.
3. State the remedy you would recommend and justify it in terms of what each protective device is protecting.
Read the worked answer
Recommend a dedicated radial circuit for the heater: 2.5 mm² copper (≈18–20 A rating) on its own 16 A MCB, taken directly from the board. Keep the lighting socket circuit on the existing 1.0 mm² cable with a 6 A device, and move the microwave to the dedicated circuit as well or give it a third 6 A radial.
Reasoning: selectivity. Each breaker must be chosen so that its rated current lies below the cable's continuous rating but above the intended load current. For the 1.0 mm² lighting run, load ≈ 0.6 A, so 6 A is generous and correct. For the heater, load ≈ 6.1–6.5 A, so a 6 A device is marginal; 16 A with 2.5 mm² satisfies both conditions. Uprating the breaker alone satisfies the load but violates the cable condition - exactly the failure mode the warden's shortcut invites. Also flag the 17 V drop for investigation, because a 7% drop at the socket already exceeds the 3–6% band normally recommended and suggests an upstream defect independent of the heater circuit.