Question 1.18: A wire having a resistivity of 1.724 × 10^−8 Ωm, length 20 m...
A wire having a resistivity of 1.724 × 10^{−8} Ωm, length 20 m and cross sectional area 1 mm² carries a current of 5 A. Determine the voltage drop between the ends of the wire.
First we must find the resistance of the wire (as in Example 1.17):
R=\frac{\rho l}{A}=\frac{1.6\times 10^{-8}\times 20}{1\times 10^{-6}} =32\times 10^{-2}=0.32 \ \OmegaThe voltage drop can now be calculated using Ohm’s Law:
V = I × R = 5A × 0.32 Ω = 1.6 V
This calculation shows that a potential of 1.6 V will be dropped between the ends of the wire.
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