Question #200461

A 2-meter copper conductor which has a cross-sectional area of 8 x 10 – 7 m2 has a potential energy difference of 6 volts between its ends. The resistivity of copper is 1.7 x 10 – 8 Ω-m. Find:

a. The resistance of the conductor, and

b. The current passing through it



1
Expert's answer
2021-05-31T02:24:41-0400

R=ρLAR = \rho\frac{L}{A}


Where

R = is the resistance

ρ\rho = is the resistivity

L= is the length

A = is the cross sectional area


R=1.7×108Ωm[2m8×107m2]R= 1.7×10^{-8}\Omega m[\frac{2m}{8×10^{-7}m^2}]


R=3.4×108Ωm28×107m2R=\frac{3.4×10^{-8}\Omega m^2}{8×10^{-7}m^2}


R=0.0425ΩR= 0.0425\Omega





Current passing


V=IRV= IR

Where

V = is the potential

I = is the current

R = is the resistance


I=VRI = \frac{V}{R}


I=6V0.0425Ω=14.11I =\frac {6V}{0.0425\Omega}= 14.11 amp





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