the resistance of an electric wire of an alloy is 10 ohm .if the thickness of wire is 0.001 m and length is 1m .find its resistivity
Answers
Resistance(R)=10Ω,
length(L)=1m,
Diameter=0.001m,
Radius (Diameter/2)=0.005m.
The area of the cross section of the wire is
π r^2 Or, A = 3.14 x (0.005)^2 m^2 Or,
A = 0.0000785 m^2.
We know that, ρ= (R*A)/L.
Therefore,
ρ= (10Ω*0.0000785)sqm / 1m.
ρ=10Ω*0.0000785. ρ=0.000785Ωm.
Therefore, the resistance is 0.000785Ωm.
Resistivity of an Electric Wire
Explanation:
GIVEN:
Resistance (R) = 10 Ω
Length (L) = 1 m
Diameter (D) = 0.001 m (mentioned as thickness)
The resistivity of an Electric Wire is given by the equation:
R = ρ * (L / A)
Resistivity is measured in Ω.m
Where, ρ is the resistivity of the wire
R is the resistance of the wire
L is the length of the wire
A is the area of cross section of the wire.
The equation is rearranged as
ρ = R * (A / L) ; where A = π r^2 … (π = 3.14)
Substituting the values,
A = 3.14 (0.0005)^2 …………D = r / 2 ; 0.001 / 2 = 0.0005
= 0.000000785 = 0.785 * 10^-6
ρ = 10 / (0.785 * 10^-6)
ρ = 0.785 * 10^-5 Ω.m
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