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Resistivity Calculator

Calculate conductor resistance using R = ρL/A

Parameters

Ω·mⓘ
mⓘ
m²ⓘ
Show Trail

Controls

xⓘ

Calculated Values

Resistance:
0.03;Ω0.03;Ω
Conductivity σ:
59523809.52;S/m59523809.52;S/m

Examples

Copper wire 2 m, 1 mm²

ρ = 1.68×10⁻⁸ Ω·m.

  • Resistance: 0.030.03

Nichrome 0.5 m, 0.2 mm²

Higher ρ heater wire.

  • Resistance: 2.502.50

Visualization

Resistivity — Macroscopic Resistance of Conductors

The resistance of a uniform conductor depends on material, length, and cross-section: R = ρL/A, where ρ is resistivity (Ω·m), L is length (m), and A is cross-sectional area (m²).

Resistivity is a material property. Copper ≈ 1.68×10⁻⁸ Ω·m at 20°C; nichrome ≈ 1×10⁻⁶ Ω·m (used in heaters); silicon doped semiconductors have much higher ρ.

Conductivity σ = 1/ρ (S/m, siemens per meter). Good conductors have large σ; insulators have very small σ.

Microscopic picture (Drude model): electrons drift under electric field, collide with ions; more collisions → higher ρ. In metals, ρ usually increases with temperature.

Geometry: doubling length doubles R; doubling diameter (A ∝ d²) quarters R. AWG wire gauge tables give A for standard wire sizes.

Temperature dependence: ρ(T) ≈ ρ₀[1 + α(T−T₀)] for metals, with temperature coefficient α. Heaters use high-α materials.

Key Concepts

  • R = ρL/A
  • σ = 1/ρ — conductivity
  • L ↑ → R ↑ ; A ↑ → R ↓
  • ρ depends on material and temperature
  • AWG → cross-section for wire sizing
  • α — temperature coefficient of resistivity

Real-World Applications

  • Wire gauge selection for circuits
  • Nichrome and tungsten heating elements
  • Four-point probe resistivity of semiconductors
  • Earth resistance and grounding design
  • Class 12 current electricity numericals

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Physics Equations

Resistance:
R=ρLAR = \rho \frac{L}{A}
Conductivity:
σ=1ρ\sigma = \frac{1}{\rho}

Step-by-Step Solution

See how the main results are calculated.

1

Step 1: Identify Material and Geometry

List resistivity ρ, conductor length L, and cross-sectional area A.

Result:

ρ=1.6800e−8 Ω\cdotpm,L=2 m,A=1.0000e−6 m2\rho = 1.6800e-8 \text{ Ω·m}, \quad L = 2 \text{ m}, \quad A = 1.0000e-6 \text{ m}^2

Explanation:

Formula assumes uniform cross-section and homogeneous material at uniform temperature.

2

Step 2: Write Resistance Formula

Equation:

R=ρLAR = \rho \frac{L}{A}

Explanation:

Macroscopic form of resistance: longer path (L↑) increases R; larger area (A↑) decreases R.

3

Step 3: Compute Geometry Factor L/A

Calculation:

LA=21.0000e−6=2.0000e+6 m−1\frac{L}{A} = \frac{2}{1.0000e-6} = 2.0000e+6 \text{ m}^{-1}

Result:

L/A=2.0000e+6m−1L/A = 2.0000e+6 m⁻¹

Explanation:

This factor has units of inverse area — converts resistivity to resistance.

4

Step 4: Calculate Resistance

Calculation:

R=1.6800e−8×2.0000e+6=0.033600 ΩR = 1.6800e-8 \times 2.0000e+6 = 0.033600 \text{ Ω}

Result:

R=0.033600ΩR = 0.033600 Ω

Explanation:

Compare with expected scale: copper wire of few meters and mm² area is often milliohms to ohms.

5

Step 5: Conductivity (Related Quantity)

Equation:

σ=1ρ\sigma = \frac{1}{\rho}

Calculation:

σ=11.6800e−8=5.9524e+7 S/m\sigma = \frac{1}{1.6800e-8} = 5.9524e+7 \text{ S/m}

Result:

σ=5.9524e+7S/mσ = 5.9524e+7 S/m

Explanation:

Good conductors have large σ (small ρ). Semiconductors and insulators have much smaller σ.

6

Step 6: Physical Interpretation

Summarize how geometry changes would affect R.

Explanation:

If length doubled → R would be 0.0672 Ω. If area doubled → R would be 0.0168 Ω.

Frequently Asked Questions (FAQ)

Why does longer wire have more resistance?

Electrons travel farther between collisions; more scattering path length increases opposition to drift.

Does shape matter besides L and A?

For uniform cross-section, R = ρL/A. Non-uniform shapes need integration dR = ρ dl/A(l).

What is superconductivity?

Below critical T_c, ρ → 0 — zero resistance. Not described by R = ρL/A in normal state.

How to measure ρ of a sample?

Measure R, L, A carefully; ρ = RA/L. Four-probe method avoids contact resistance.

Effect of temperature on copper wire?

ρ increases with T; a hot wire has higher R than when cold.

Practice MCQs

  1. Doubling the length of a wire (same material, same area) changes R by:
  2. Doubling the diameter (same length, material) changes R by:
  3. Unit of resistivity ρ is:
  4. A material with high resistivity is generally:
  5. Copper vs nichrome for a heater — better choice is:
  6. Conductivity σ and resistivity ρ relate as:

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