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The World of Metals and Non-metals - Conduction of electricity

Grade 7CBSE

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

🔑Concepts

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Electrical conductivity is the property of a material to allow electric current to flow through it. Metals are generally excellent conductors of electricity.

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Metals like Silver (AgAg), Copper (CuCu), and Aluminum (AlAl) are widely used because they contain free electrons that act as charge carriers.

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Silver (AgAg) is the best conductor of electricity, but Copper (CuCu) is more commonly used in household wiring due to its cost-effectiveness.

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Most non-metals are insulators (poor conductors) because they lack free electrons. Examples include Sulfur (SS) and Phosphorus (PP).

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Graphite, an allotrope of the non-metal Carbon (CC), is a notable exception as it is a good conductor of electricity.

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Electrical wires are typically coated with insulators like Polyvinyl Chloride (PVC) or rubber to prevent electric shocks.

📐Formulae

I=QtI = \frac{Q}{t}

V=I×RV = I \times R

R=VIR = \frac{V}{I}

G=1RG = \frac{1}{R}

💡Examples

Problem 1:

Why is Copper (CuCu) preferred over Silver (AgAg) for making electrical wires even though Silver is a better conductor?

Solution:

Copper is preferred because it is significantly cheaper and more abundant than Silver.

Explanation:

While Silver (AgAg) has the highest electrical conductivity, its high cost makes it impractical for large-scale use. Copper (CuCu) provides a high level of conductivity (GG) at a much lower price point.

Problem 2:

A circuit is broken by a gap. Which of the following materials will complete the circuit and make the bulb glow: a piece of wood, a plastic scale, or a graphite pencil lead?

Solution:

The graphite pencil lead.

Explanation:

Wood and plastic are non-metals and insulators. Graphite is a form of Carbon (CC) that possesses free electrons, allowing it to conduct electricity and complete the circuit.

Problem 3:

If the resistance (RR) in a wire is doubled while keeping the Voltage (VV) constant, what happens to the current (II)?

Solution:

The current (II) becomes half of its original value.

Explanation:

According to the formula I=VRI = \frac{V}{R}, the current is inversely proportional to the resistance. If RR increases to 2R2R, then the new current Inew=V2R=12II_{new} = \frac{V}{2R} = \frac{1}{2}I.