Question #232454

300-volt power station is connected to a factory through two power lines. If the factory is 5 Km away from the station and electric potential difference at factory 220 V, current intensity 100 A, calculate: a) The resistance of unit length. b) Radius of CSA of the wire. given that its resistivity 1.5x10-8 2.m.

Expert's answer

The potential drop on the power line is equal

ΔV=300 V−220 V=80 V\Delta V=300\:\rm V-220\: V=80\: V

Hence, the resistance of the power line

R=ΔVI=80 V100 A=0.8 ΩR=\frac{\Delta V}{I}=\frac{80\:\rm V}{100\:\rm A}=0.8\:\Omega

a) The resistance of unit length

R/l=0.8 Ω/(2∗5000 m)=8∗10−5 Ω/mR/l=0.8\:\Omega/(2*5000\:\rm m)=8*10^{-5}\:\Omega/m

b) Radius of CSA of the wire

r=Aπ=ρl/Rπ=1.5∗10−8 Ω⋅m3.14∗8∗10−5 Ω/m=7.7∗10−3 m=7.7 mmr=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{\rho l/R}{\pi}}\\ =\sqrt{\frac{1.5*10^{-8}\:\Omega\cdot\rm m}{3.14*8*10^{-5}\:\rm\Omega/m}}=7.7*10^{-3}\:\rm m=7.7\: mm


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