The Workshop Fuse That Keeps Blowing
A small workshop runs on a 220 V supply through a single main fuse. The connected appliances are: an electric iron of 1000 W, a room heater of 1500 W, three LED lamps of 10 W each, and a 750 W motor. The main fuse is marked 10 A. On a cold morning, when everything is switched on, the fuse blows. The owner replaces it with a 20 A fuse and says the problem is solved. The heater is plugged into a long, thin extension cord; during use the cord becomes warm to the touch, and the lamps dim for a moment whenever the motor starts. The workshop wiring is rated for a maximum safe current of 15 A. Electricity costs ₹5.00 per unit, and the owner plans to add a 2000 W room heater next month.
1. Calculate the total current when the four existing appliances run together on 220 V. Is the 10 A fuse adequate? Why is fitting a 20 A fuse dangerous, and what should be done instead? Include the effect of the planned extra heater.
Read the worked answer
Total power = 1000 + 1500 + (3 × 10) + 750 = 3280 W.
Current I = P/V = 3280/220 = 14.9 A.
The 10 A fuse is therefore inadequate: the normal running current is about 15 A, so it blows even when nothing is faulty. That is exactly what happened.
A 20 A fuse is dangerous because a fuse protects the wiring, not the appliance. The workshop wiring is rated for 15 A. A 20 A fuse will carry 16-19 A without melting, so the wires overheat, insulation chars, and a short circuit or fire can start. The fuse must be the weakest link, not the strongest.
The correct action is not to uprate the fuse. Split the load into separate circuits - for example, the heater on its own 16 A circuit with thicker wire - use an MCB of the correct rating, and avoid running the iron, heater and motor together on one line. With the extra 2000 W heater, total power would be 5280 W, i.e. 24 A, which certainly requires a separate circuit.
2. Explain why the thin extension cord becomes warm and why the lamps dim when the motor starts. Use R = ρL/A and the idea of voltage drop. Why is a long thin cord particularly risky for a 1500 W heater?
Read the worked answer
Resistance is R = ρL/A. A long, thin cord has a large length L and a small cross-section A, so its resistance is much higher than that of the house wiring.
The heater draws about 1500/220 ≈ 6.8 A. If the cord has a resistance of 0.5 Ω, the heat produced in it is I²R = (6.8)² × 0.5 ≈ 23 W, enough to make the cord noticeably warm. The voltage drop is IR ≈ 6.8 × 0.5 ≈ 3.4 V, so the heater receives less than 220 V.
When the motor starts, its starting current is several times its running current. This larger current increases the voltage drop across the cord and the supply wiring, so the voltage available to the lamps falls for a moment and they dim.
For a 1500 W heater the risk is continuous heating: the cord carries about 6.8 A for a long time, its insulation can soften and melt, exposed conductors can touch, and a short circuit or fire can follow. The cord must be rated for at least the heater current, kept as short as possible, and fully uncoiled.
3. The workshop runs the heater for 2 hours and the iron for 1 hour every day for 30 days. Calculate the monthly energy consumption and the cost at ₹5.00 per unit. Which appliance costs more to run?
Read the worked answer
Heater: 1.5 kW × 2 h = 3.0 kWh per day. Iron: 1.0 kW × 1 h = 1.0 kWh per day. Total = 4.0 kWh per day.
For 30 days: 4.0 × 30 = 120 kWh (units).
Cost = 120 × ₹5.00 = ₹600 per month.
The heater uses 3/4 of the energy, so it costs about ₹450, while the iron costs about ₹150. The heater is more expensive to run even though the iron is used for a full hour, because power matters as much as time. The three LED lamps together are only 0.03 kW and add a few rupees at most, so they can be ignored in this comparison.