A 2.0-m-long, 1.0-mm-diameter wire has a variable resistivity given by
ρ(x)=(2.5×10−6)[1+(x1.0m)2]Ωm
where x is measured from one end of the wire.
What is the current if this wire is connected to the terminals of a 18.0 V battery?
billrussell42
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ρ(x)=(2.5×10−6)[1+(x1.0m)2]Ωm ??
I have to guess this is
ρ(x) = (2.5e−6)[1 + x²] Ωm
where x is in meters
1 mm diameter is 0.0005 m radius
A = πr² = π(0.0005)² = 7.85e-7 m²
R = ρL/A
but with ρ variable, this becomes
R = (1/A) ∫ ρx dx from 0 to 2
R = (2.5e-6/A) ∫ (1 + x²)x dx [0-2]
R = (2.5e-6/A) ∫ (x + x³) dx [0-2]
R = (2.5e-6/7.85e-7) ∫ x dx + ∫ x³ dx [0-2]
R = (3.183) ( (x²/2) + (x⁴/4) [0-2]
R = (3.183) ( 4+ 16)
R = 63.7 Ω
Resistance of a wire in Ω
R = ρL/A
ρ is resistivity of the material in Ω-m
L is length in meters
A is cross-sectional area in m²
A = πr², r is radius of wire in m