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Electricity and Magnetism - Practical electricity

Grade 12IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Electrical Power (PP): The rate at which electrical energy is transferred by a circuit. The SI unit is the Watt (WW), where 1W=1J/s1 W = 1 J/s.

Electrical Energy (EE): The total energy consumed over a period of time tt. In domestic settings, it is often measured in kilowatt-hours (kWhkWh) instead of Joules (JJ).

Mains Electricity: Most domestic supplies use Alternating Current (ACAC). In many IGCSE contexts, the standard supply is approximately 230V230 V at a frequency of 50Hz50 Hz.

The Three-Pin Plug: Consists of the Live wire (Brown, carries high voltage), Neutral wire (Blue, completes the circuit at 0V0 V), and Earth wire (Green/Yellow striped, safety wire).

Safety Components: Fuses and Circuit Breakers (MCBs) protect circuits from overheating by breaking the connection when the current exceeds a specific rating. The fuse must be placed in the Live wire.

Earthing and Double Insulation: The Earth wire provides a low-resistance path to the ground for fault currents. Appliances with plastic casings that do not require an earth wire are 'Double Insulated' and marked with the \square symbol.

Cost of Electricity: Calculated by multiplying the energy used in kWhkWh by the cost per unit. 1kWh1 kWh is the energy used by a 1kW1 kW appliance running for 11 hour (1kWh=3.6×106J1 kWh = 3.6 \times 10^6 J).

📐Formulae

P=I×VP = I \times V

E=P×t=I×V×tE = P \times t = I \times V \times t

P=I2×RP = I^2 \times R

P=V2RP = \frac{V^2}{R}

Energy(kWh)=Power(kW)×Time(h)Energy (kWh) = Power (kW) \times Time (h)

💡Examples

Problem 1:

An electric kettle is rated at 2.3kW2.3 kW and is connected to a 230V230 V mains supply. Calculate the current flowing through the kettle and suggest an appropriate fuse rating (3A,5A,13A3 A, 5 A, 13 A).

Solution:

I=PV=2300W230V=10AI = \frac{P}{V} = \frac{2300 W}{230 V} = 10 A

Explanation:

Since the operating current is 10A10 A, a fuse with a rating slightly higher than the operating current must be chosen to prevent it from blowing during normal use. Therefore, a 13A13 A fuse is the most appropriate choice.

Problem 2:

A 60W60 W light bulb is left on for 88 hours. If the cost of electricity is 0.150.15 per kWhkWh, calculate the total cost of using the bulb.

Solution:

Energy=0.06kW×8h=0.48kWhEnergy = 0.06 kW \times 8 h = 0.48 kWh Cost=0.48kWh×0.15=0.072Cost = 0.48 kWh \times 0.15 = 0.072

Explanation:

First, convert the power from Watts to kilowatts (60W=0.06kW60 W = 0.06 kW). Then, multiply the power by time in hours to find the energy in kWhkWh. Finally, multiply the energy by the price per unit.

Practical electricity - Revision Notes & Key Formulas | IGCSE Grade 12 Physics