12. Find the potential drop along a copper wire 500 m long and 2 mm in diameter if the current in it is equal to 2 A
Solution
Length of wire l=500m
Radius R=1mm
So area A=πr2=3.14×1mm×1mmA=\pi r^2=3.14\times1mm\times1mmA=πr2=3.14×1mm×1mm
A=3.14×10−6m2A=3.14\times10^{-6}m^2A=3.14×10−6m2
Resistivity of copper wire
ρ=1.68∗10−8Ω−m\rho=1.68 * 10^{-8} Ω -mρ=1.68∗10−8Ω−m
So potential difference of wire
V=IR=IρlAV=IR\\=\frac{I\rho l}{A}V=IR=AIρl
Putting all values
V=2×1.68∗10−8×5003.14×10−6=5.35VoltsV=\frac{2\times1.68 * 10^{-8}\times500 }{3.14\times10^{-6}}\\=5.35VoltsV=3.14×10−62×1.68∗10−8×500=5.35Volts
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