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Specific Resistance or Electrical Resistivity

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Estimated time: 7 minutes
Maharashtra State Board: Class 11
CISCE: Class 12

Definition: Electrical Resistivity

Electrical Resistivity (ρ) is defined as the resistance offered by a conductor of unit length and unit cross-sectional area at a given temperature. It is a characteristic property of the material, independent of its dimensions.

Maharashtra State Board: Class 11
CISCE: Class 12

Formula & Derivation

The resistance R of a uniform conductor is:

  • Directly proportional to its length l: R ∝ l
  • Inversely proportional to its area of cross-section A: R ∝ 1/A

Combining:

  • R = ρ\[\frac {l}{A}\]

Rearranging for resistivity:

  • ρ = \[\frac {RA}{l}\]

SI Unit: ohm-metre (Ω ⋅ m)

Dimensional Formula: [ML3T−3A−2]

Maharashtra State Board: Class 11
CISCE: Class 12

Conductivity — The Inverse Relationship

Electrical Conductivity (σ) is the reciprocal of resistivity:

  • σ = \[\frac {1}{ρ}\]

SI Unit: siemens per metre (S m−1)

Property Symbol Formula SI Unit
Resistivity ρ RA/l Ω·m
Conductivity σ 1/ρ S·m⁻¹
Maharashtra State Board: Class 11
CISCE: Class 12

Vector (Microscopic) Form of Ohm's Law

  • \[\vec E\] = ρ\[\vec J\] or equivalently \[\vec J\] = σ\[\vec E\]

Where \[\vec{E}\] is the electric field and \[\vec{J}\] is the current density.

  • This microscopic form connects macroscopic resistivity to electron-level behaviour and is often tested in assertion-reason questions.

Free-Electron Model Expression:

  • ρ = \[\frac {m}{ne^2τ}\]

Where m = electron mass, n = free electron density, e = electron charge, τ = relaxation time.

Maharashtra State Board: Class 11

Resistivity Comparison Table

Material Category Approx. Resistivity (Ω·m)
Silver Conductor 1.6 × 10-8
Copper Conductor 1.7 × 10-8
Aluminium Conductor 2.7 × 10-8
Nichrome Alloy (used in heaters) 1.1 × 10-6
Constantan Alloy (low temp. coefficient) 5.0 × 10-7
Carbon Semiconductor 3.5 × 10-5
Silicon Semiconductor ~2300
Glass Insulator 1010 – 1014
Rubber Insulator ~1013
Wood (dry) Insulator 108 – 1011
Maharashtra State Board: Class 11

Example 1

Q. A constantan wire of diameter 1.25 mm has resistivity 5.0 × 10−7 Ω ⋅ m. Find its resistance per metre.

Solution:

A = πr2 = π(0.625 × 10−3)2 ≈ 1.227 × 10−6 m2
\[\frac {R}{l}\] = \[\frac {ρ}{A}\] = \[\frac {5.0×10^{−7}}{1.227×10^{−6}}\] ≈ 0.41 Ω m−1

Answer: ≈ 0.41 Ω per metre

CISCE: Class 12

Example 2

Q. A wire is stretched to increase its length by 10%, keeping volume constant. Find the percentage change in resistance.

Solution:
Since volume V = Al is constant, if l′ = 1.1l, then A′ = A/1.1.

R′ = ρ\[\frac {l′}{A′}\] = ρ\[\frac {1.1l}{A/1.1}\] = 1.21 × ρ\[\frac {l}{A}\] = 1.21R

Answer: Resistance increases by 21%

Shaalaa.com | Current Electricity part 14 (Resistivity)

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