the resistivity of copper is 1.7×10^-8ohm metre.(a) what length of copper wire of diameter 0.1 mm will have a resistance of 34 Ohm?
(b) another copper wire of the same length but of half the diameter as the first is taken. what is the ratio of its resistance to that of the first wire?
Answers
Resistance of a wire is given as,
R = ρL/A
where,
ρ = resistivity = 1.7 x 10⁻⁸
L = length
A = area
= πr²
= π(0.05 x 10⁻³)²
= 25π x 10⁻¹⁰ m²
i) Given R = 34
=> ρL/A = 34
=> (1.7 x 10⁻⁸ x L)/25π x 10⁻¹⁰ = 34
=> L = 34 x 25π x 10⁻²/1.7
=> L = 5π = 15.7 meters
Hence the length of the wire should be 15.7 meters.
ii) since R = ρL/A
and A = πd²/4
=> R ∝ 1/d₂
=> R₂/R₁ = (d₁/d₂)²
=> R₂/R₁ = 2² = 4
Hence R₂ : R₁ = 4 : 1
the resistivity of copper is 1.7×10^-8ohm metre.(a) what length of copper wire of diameter 0.1 mm will have a resistance of 34 Ohm?
(b) another copper wire of the same length but of half the diameter as the first is taken. what is the ratio of its resistance to that of the first wire?
R = ρ L/A
ρ = 1.7×10^-8 Ωm
L = ?
Diameter = 0.1 mm
Radius = Diameter /2 = 0.05 mm
A = π (0.05/1000)² ( 1mm = 1/1000 m)
π = 3.14
A = 3.14 × 25 * 10⁻¹⁰
R = 34 Ω
34 = 1.7×10^-8 * L / (25 × 3.14 × 10⁻¹⁰)
=> L = 5 * 3.14
=> L = 15.7 m
New Diameter = half
A ∝ Radius² or Diameter²
So new Area will be 1/4 of older area
R ∝ 1/A
so Resistance of new wire will be 4 times
R = 4 * 15.7
R = 62.8 Ω
New R : old R :: 4:1