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Electricity and Magnetism - Electrical safety and mains circuits

Grade 11IGCSEPhysics

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

🔑Concepts

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Mains electricity is an Alternating Current (A.C.A.C.), meaning the current periodically reverses direction. In most IGCSE contexts, the mains supply is approximately 230 V230 \text{ V} with a frequency of 50 Hz50 \text{ Hz}.

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Direct Current (D.C.D.C.) flows in one direction only and is typically supplied by cells and batteries.

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The three-pin plug consists of: the Live wire (Brown) which carries the high voltage, the Neutral wire (Blue) which completes the circuit, and the Earth wire (Yellow/Green) which is a safety wire.

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A Fuse is a safety device containing a thin wire that melts and breaks the circuit if the current exceeds a specific rating. It must always be connected to the Live wire.

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Circuit Breakers (such as MCBsMCBs) are electromagnetic switches that open ('trip') when the current is too high. They act faster than fuses and can be reset.

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Earthing protects users from electric shocks. If a fault occurs and the live wire touches a metal casing, a large current flows through the Earth wire to the ground, blowing the fuse.

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Double Insulation is used for appliances with plastic casings (indicated by the □\square symbol). These do not require an Earth wire because the outer casing cannot become live.

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Electrical hazards include damaged insulation (exposing live wires), overheating of cables (due to overloads), and damp conditions (which lower the resistance of the human body).

📐Formulae

P=I×VP = I \times V

P=I2×RP = I^2 \times R

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

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

V=I×RV = I \times R

💡Examples

Problem 1:

An electric heater is rated at 2.3 kW2.3 \text{ kW} and is connected to a 230 V230 \text{ V} mains supply. Calculate the current flowing through the heater and suggest an appropriate fuse rating from the following: 3 A3 \text{ A}, 5 A5 \text{ A}, 13 A13 \text{ A}.

Solution:

I=PV=2300 W230 V=10 AI = \frac{P}{V} = \frac{2300 \text{ W}}{230 \text{ V}} = 10 \text{ A}

Explanation:

Since the operating current is 10 A10 \text{ A}, the fuse must have a rating slightly higher than this to allow normal operation but blow during a fault. Therefore, a 13 A13 \text{ A} fuse is the correct choice.

Problem 2:

Calculate the electrical energy transferred in Joules by a 100 W100 \text{ W} light bulb left on for 22 hours.

Solution:

E=P×t=100 W×(2×3600 s)=720,000 J=7.2×105 JE = P \times t = 100 \text{ W} \times (2 \times 3600 \text{ s}) = 720,000 \text{ J} = 7.2 \times 10^5 \text{ J}

Explanation:

Energy is the product of power and time. Time must be converted from hours to seconds (1 hour=3600 s1 \text{ hour} = 3600 \text{ s}) to get the result in Joules (JJ).

Electrical safety and mains circuits Revision - Grade 11 Physics IGCSE