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Electricity - Ohm’s law; Resistance, Resistivity

Grade 10CBSE

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

🔑Concepts

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Ohm's Law: It states that the potential difference (VV) across the ends of a given metallic wire in an electric circuit is directly proportional to the current (II) flowing through it, provided its temperature remains the same (V∝IV \propto I).

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Resistance (RR): It is the property of a conductor to resist the flow of charges through it. The SI unit of resistance is Ohm (ΩΩ).

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One Ohm (1Ω1 \Omega): If the potential difference across the two ends of a conductor is 1V1 V and the current through it is 1A1 A, then the resistance RR of the conductor is 1Ω1 \Omega.

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Factors affecting Resistance: Resistance of a uniform metallic conductor is directly proportional to its length (ll), inversely proportional to the area of cross-section (AA), and depends on the nature of the material.

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Resistivity (\rho): It is a characteristic property of the material. Metals and alloys have very low resistivity in the range of 10−8Ωm10^{-8} \Omega m to 10−6Ωm10^{-6} \Omega m, while insulators have resistivity of the order of 1012Ωm10^{12} \Omega m to 1017Ωm10^{17} \Omega m.

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Temperature Dependence: The resistance and resistivity of a material increase with an increase in temperature for pure metals.

📐Formulae

V=IRV = IR

R=ρlAR = \rho \frac{l}{A}

ρ=RAl\rho = \frac{R A}{l}

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

A=πr2A = \pi r^2

💡Examples

Problem 1:

The potential difference between the terminals of an electric heater is 60V60 V when it draws a current of 4A4 A from the source. What current will the heater draw if the potential difference is increased to 120V120 V?

Solution:

Given V=60VV = 60 V and I=4AI = 4 A. From Ohm's law, R=VI=60V4A=15ΩR = \frac{V}{I} = \frac{60 V}{4 A} = 15 \Omega. When potential difference is increased to 120V120 V, the new current I′I' is I′=V′R=120V15Ω=8AI' = \frac{V'}{R} = \frac{120 V}{15 \Omega} = 8 A.

Explanation:

Resistance remains constant for the same heater. By doubling the voltage, the current also doubles according to V∝IV \propto I.

Problem 2:

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

Solution:

Given l=1ml = 1 m, R=26ΩR = 26 \Omega, and diameter d=0.3mm=3×10−4md = 0.3 mm = 3 \times 10^{-4} m. Radius r=1.5×10−4mr = 1.5 \times 10^{-4} m. Area A=πr2=3.14×(1.5×10−4)2m2A = \pi r^2 = 3.14 \times (1.5 \times 10^{-4})^2 m^2. Resistivity ρ=RAl=26×3.14×(1.5×10−4)21≈1.84×10−6Ωm\rho = \frac{RA}{l} = \frac{26 \times 3.14 \times (1.5 \times 10^{-4})^2}{1} \approx 1.84 \times 10^{-6} \Omega m.

Explanation:

First, convert diameter to meters and find the cross-sectional area. Then use the resistivity formula ρ=RAl\rho = \frac{RA}{l} to solve for the unknown.