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Electricity - Factors on which the resistance of a conductor depends

Grade 10CBSE

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

🔑Concepts

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Resistance (RR) of a conductor is directly proportional to its length (ll). Doubling the length of a wire doubles its resistance (R∝lR \propto l).

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Resistance is inversely proportional to the area of cross-section (AA). A thicker wire provides less resistance to the flow of electrons compared to a thinner wire (R∝1AR \propto \frac{1}{A}).

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Resistance depends on the nature of the material. This characteristic property is known as electrical resistivity (ρ\rho). Metals have low resistivity, while insulators have very high resistivity.

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The resistivity (ρ\rho) of a material is defined as the resistance of a conductor of unit length (1m1 m) and unit area of cross-section (1m21 m^2). Its SI unit is Ω⋅m\Omega \cdot m.

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Resistivity is an intrinsic property of the material and does not change with the dimensions (length or area) of the object, though it does change with temperature.

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For most metallic conductors, resistance increases with an increase in temperature.

📐Formulae

R∝lR \propto l

R∝1AR \propto \frac{1}{A}

R=ρlAR = \rho \frac{l}{A}

ρ=R⋅Al\rho = \frac{R \cdot A}{l}

💡Examples

Problem 1:

A wire of resistance 10Ω10 \Omega is stretched so that its length becomes three times its original length. If the volume remains constant, calculate the new resistance.

Solution:

Let initial length be ll and area be AA. Resistance R=ρlA=10ΩR = \rho \frac{l}{A} = 10 \Omega. When stretched to 3l3l, the area becomes A3\frac{A}{3} (since Volume V=l×AV = l \times A is constant). New resistance R′=ρ3lA/3=9(ρlA)=9×10=90ΩR' = \rho \frac{3l}{A/3} = 9 \left( \rho \frac{l}{A} \right) = 9 \times 10 = 90 \Omega.

Explanation:

Stretching a wire increases its length and simultaneously decreases its cross-sectional area. Because RR is proportional to ll and inversely proportional to AA, the resistance increases by the square of the change in length.

Problem 2:

Compare the resistance of two wires of the same material: Wire A has length LL and radius rr, Wire B has length 2L2L and radius 2r2r.

Solution:

Resistance of Wire A: RA=ρLπr2R_A = \rho \frac{L}{\pi r^2}. Resistance of Wire B: RB=ρ2Lπ(2r)2=ρ2L4πr2=12(ρLπr2)=12RAR_B = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{4 \pi r^2} = \frac{1}{2} \left( \rho \frac{L}{\pi r^2} \right) = \frac{1}{2} R_A.

Explanation:

Even though Wire B is twice as long (which increases resistance), its radius is doubled, making its area four times larger (which decreases resistance). The net effect is that Wire B has half the resistance of Wire A.