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Electricity - FA 11.5 FA11.5 FA 11.5 FA11.5 FA CTORS ON WHICH THE RESISTCTORS ON WHICH THE RESISTCTORS ON WHICH THE RESISTCTORS ON WHICH THE RESISTCTORS ON WHICH THE RESIST ANCE OF AANCE OF AANCE OF AANCE OF AANCE OF A

Grade 10CBSE

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

🔑Concepts

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Resistance of a uniform metallic conductor is directly proportional to its length (ll): R∝lR \propto l.

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Resistance is inversely proportional to the area of cross-section (AA): R∝1AR \propto \frac{1}{A}.

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Resistance depends on the nature of the material of the conductor.

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Resistance of a conductor also changes with change in temperature; for pure metals, it increases with temperature.

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The constant of proportionality ρ\rho (rho) is called electrical resistivity of the material of the conductor.

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Resistivity is a characteristic property of the material and does not depend on the dimensions (length or area) of the object.

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The SI unit of resistivity is Ohm-metre (Ω⋅m\Omega \cdot m).

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Metals and alloys have very low resistivity in the range of 10−8Ω⋅m10^{-8} \Omega \cdot m to 10−6Ω⋅m10^{-6} \Omega \cdot m, while insulators like rubber have resistivity of 1012Ω⋅m10^{12} \Omega \cdot m to 1017Ω⋅m10^{17} \Omega \cdot m.

📐Formulae

R∝lAR \propto \frac{l}{A}

R=ρlAR = \rho \frac{l}{A}

ρ=R⋅Al\rho = \frac{R \cdot A}{l}

A=πr2=π(d2)2A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2

💡Examples

Problem 1:

Resistance of a metal wire of length 1 m1\text{ m} is 26Ω26 \Omega at 20∘C20^{\circ}C. If the diameter of the wire is 0.3 mm0.3\text{ mm}, what will be the resistivity of the metal at that temperature?

Solution:

Given: l=1 ml = 1\text{ m}, R=26ΩR = 26 \Omega, d=0.3 mm=3×10−4 md = 0.3\text{ mm} = 3 \times 10^{-4}\text{ m}. Area A=πd24A = \frac{\pi d^2}{4}. Resistivity ρ=R⋅Al=Rπd24l\rho = \frac{R \cdot A}{l} = \frac{R \pi d^2}{4l}. Substituting values: ρ=26×3.14×(3×10−4)24×1\rho = \frac{26 \times 3.14 \times (3 \times 10^{-4})^2}{4 \times 1} ρ=26×3.14×9×10−84\rho = \frac{26 \times 3.14 \times 9 \times 10^{-8}}{4} ρ=1.84×10−6Ω⋅m\rho = 1.84 \times 10^{-6} \Omega \cdot m.

Explanation:

We first convert all units to SI (diameter to meters). We use the resistivity formula derived from the factors affecting resistance. The diameter allows us to calculate the cross-sectional area of the wire.

Problem 2:

A wire of given material having length ll and area of cross-section AA has a resistance of 4Ω4 \Omega. What would be the resistance of another wire of the same material having length l2\frac{l}{2} and area of cross-section 2A2A?

Solution:

For the first wire: R1=ρlA=4ΩR_1 = \rho \frac{l}{A} = 4 \Omega. For the second wire: l2=l2l_2 = \frac{l}{2}, A2=2AA_2 = 2A. New resistance R2=ρl2A2=ρl/22A=14(ρlA)R_2 = \rho \frac{l_2}{A_2} = \rho \frac{l/2}{2A} = \frac{1}{4} \left( \rho \frac{l}{A} \right). R2=14R1=14×4=1ΩR_2 = \frac{1}{4} R_1 = \frac{1}{4} \times 4 = 1 \Omega.

Explanation:

Since the material is the same, ρ\rho remains constant. By substituting the new dimensions into the resistance formula, we find that the resistance becomes one-fourth of the original value.