Two Circuits, One Overloaded Fuse
A household is supplied at 220 V through two circuits: a 5 A lighting circuit and a 15 A power circuit. On the lighting circuit run four 60 W bulbs and two 70 W fans. On the power circuit a 2000 W geyser and a 1000 W room heater are connected, and the family also wants to use a 1000 W iron at the same time. The fuse on the power circuit is rated 15 A. After two years the geyser's heating element fails; the repairer replaces the original 40 cm nichrome wire with 30 cm of the same wire and assures the family that it will heat faster. The metal body of the geyser is connected to an earth wire, and the switchboard has a fuse only in the live wire.
1. Calculate the current in each circuit with the appliances listed, and decide whether the 15 A fuse will blow when the iron is switched on. Show your working.
Read the worked answer
Lighting circuit: P = (4 × 60) + (2 × 70) = 240 + 140 = 380 W. I = P/V = 380/220 = 1.73 A. This is well within the 5 A rating.
Power circuit without the iron: P = 2000 + 1000 = 3000 W. I = 3000/220 = 13.6 A, just below 15 A, so the geyser and heater can run together.
With the iron: P = 3000 + 1000 = 4000 W. I = 4000/220 = 18.2 A.
This exceeds the 15 A rating, so the fuse wire will heat and melt, breaking the circuit. The fuse is chosen to protect the cable, not the appliance: the cable of the power circuit is designed for about 15 A. Fitting a 20 A fuse instead would let the cable overheat and possibly start a fire before the fuse acted. The correct solution is to use the iron with the geyser switched off, or on a separate circuit.
2. Calculate the power and current of the geyser after the element is shortened from 40 cm to 30 cm. Will the 15 A fuse protect it? Explain what risk remains.
Read the worked answer
For a wire of the same material and thickness, resistance is proportional to length: R = ρL/A. Reducing the length from 40 cm to 30 cm makes the resistance 30/40 = 0.75 of its original value.
Original resistance: R = V²/P = 220²/2000 = 48400/2000 = 24.2 Ω. New resistance: 0.75 × 24.2 = 18.2 Ω.
New power: P = V²/R = 48400/18.2 ≈ 2660 W. New current: I = 2660/220 ≈ 12.1 A.
This is still below 15 A, so the fuse will not blow in normal use. The remaining risk is that the element now runs at about one-third more power and a higher temperature; it will glow hotter, oxidise and break sooner, and the thermostat and internal wiring may be overloaded even though the circuit fuse sees nothing wrong. A fuse protects against a gross overload, not against a component being run beyond its design, so the element should be replaced with one of the specified rating.
3. Explain why the bulbs and fans are connected in parallel rather than in series, and why the fuse is placed in the live wire and the metal body is earthed.
Read the worked answer
In parallel, each appliance is connected directly across the 220 V supply, so each receives its full rated voltage and works at its rated power. Each has its own switch and can be turned on or off independently, and if one burns out the others continue. In series, the appliances would share the supply voltage, so bulbs and fans would get much less than 220 V, run dim or slow, and one break would stop the whole circuit. The same current would also have to pass through appliances of very different power ratings, which is not how they are designed.
The fuse is in the live wire so that when it melts it disconnects the appliance from the high potential. If it were in the neutral, the appliance would remain connected to the live wire and anyone touching it could still receive a shock, even with the fuse blown. Earthing connects the metal body to the ground through a low-resistance wire: if a live wire touches the body, current flows to earth and blows the fuse, and during the fault the body is held at earth potential instead of at 220 V.