Plugs are used to connect electrical appliances in the home to the 230 volt ESB supply - Leaving Cert Physics - Question 8 - 2009

The function of a fuse is to act as a safety device that prevents excessive current flow. It breaks the circuit if the current exceeds a safe level, protecting the electrical appliance and reducing the risk of fire.

A fuse works by having a thin metal wire that melts when the current flowing through it becomes too high. This melting breaks the circuit, stopping the flow of electricity and preventing damage to the appliance or wiring.

Another device with the same function as a fuse is an MCB (Miniature Circuit Breaker).

To find the most suitable fuse, we first need to calculate the current using the formula:
$I = \frac{P}{V}$
Where:

• $P$ is the power rating (800 W)
• $V$ is the voltage (230 V)

$I = \frac{800 W}{230 V} \approx 3.48 A$
Thus, the most suitable fuse rating would be a 5 A fuse.

Using a fuse with too high a rating can be dangerous because it may not blow when the current exceeds the safe limits for the appliance. This can lead to overheating of wires or devices, increasing the risk of fire or equipment damage.

To calculate the units of electricity used, we convert the power rating into kilowatts and multiply by the time in hours:

1. Convert power:
   $P = \frac{800 W}{1000} = 0.8 kW$
2. Convert time:
   $150 \text{ minutes} = \frac{150}{60} = 2.5 \text{ hours}$
3. Calculate units:
   $\text{Units} = P \times t = 0.8 kW \times 2.5 h = 2.0 \text{ units}$

To calculate the cost:

1. Cost per unit = 15 cents
2. Total cost = Number of units × Cost per unit:
   $\text{Total Cost} = 2.0 \times 15 = 30 \text{ cents}$