A plug is used to connect an electrical appliance in the home to the 230 volt mains supply - Leaving Cert Physics - Question 8 - 2012

A fuse is present in a plug to provide protection against current overloads. It acts as a safety mechanism that prevents potential fire hazards by breaking the circuit if the current exceeds safe levels.

A fuse operates by melting its internal wire when the current flowing through it exceeds a certain threshold. This disruption in the circuit prevents excess current from passing through, thereby protecting the appliance and reducing fire risks.

To determine the suitable fuse rating, we can calculate the current using the formula:
$I = \frac{P}{V}$
where P is the power (900 W) and V is the voltage (230 V). Thus,
$I = \frac{900}{230} \approx 3.91 A$
The most suitable fuse to use would be a 5 A fuse, as it is the closest higher rating that can accommodate the vacuum cleaner's current.

Using a fuse of a lower rating would be unsuitable because it may blow frequently under normal operating conditions, rendering the appliance inoperable. A fuse must be rated appropriately to handle the maximum current safely.

An example of a device that serves the same function as a fuse in modern circuits is a Miniature Circuit Breaker (MCB). It automatically interrupts the circuit when an overload is detected, providing similar protection.

To calculate the units of electricity used, we convert 90 minutes into hours: 90 minutes = 1.5 hours. The formula for energy used in kilowatt-hours is:
$E = P \times t$
where E is energy in kWh, P is power in kW, and t is time in hours. Therefore,
$E = 0.9 \times 1.5 = 1.35 \text{ kWh}$
This corresponds to 1.35 units of electricity.

To find the cost of the energy consumed, we calculate:
$\text{Cost} = \text{units used} \times \text{price per unit}$
Inputting the values:
$\text{Cost} = 1.35 \text{ units} \times 22 \text{ cent/unit} = 29.7 \text{ cents}$