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Electricity and Magnetism - Practical electricity

Grade 12A LevelPhysics

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

🔑Concepts

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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.

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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).

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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.

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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).

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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.

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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.

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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.