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Resistance in a Wire Simulator

Ready. Change length, diameter or material to observe resistance.
Resistance of a Uniform Wire Length L Diameter d Area A = πr² Formula R = ρL / A Longer wire → more R | Thicker wire → less R

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6.0 V
0.00 VVoltmeter
0.00 AAmmeter
5 ΩRheostat
60 cm
0.8 mm
Nichrome wire
Heating: Low
Resistance of a wire R = ρL / A Length ↑ → R ↑ | Area ↑ → R ↓
Wire resistance0 Ω
Current0 A
Wire voltage0 V
Area0 mm²

Observation Table

Trial Material Length Diameter Area Wire Resistance Battery Voltage Rheostat Current Observation

Solved Problem 4

Question

The resistance of a wire of length 10 m is 2 Ω. If the area of cross-section of the wire is 2 × 10⁻⁷ m², determine its (i) resistivity, (ii) conductance and (iii) conductivity.

Length L10 m
Resistance R2 Ω
Area A2 × 10⁻⁷ m²
Solved Problem Wire Model
L = 10 m
A = 2 × 10⁻⁷ m²
R = 2 Ω
Resistivityρ = RA / L
ConductanceG = 1 / R
Conductivityσ = 1 / ρ
Resistivity ρ4 × 10⁻⁸ Ω m
Conductance G0.5 mho
Conductivity σ0.25 × 10⁸ mho m⁻¹