Question

Find the Voltage drop (in mV) across an 46.6 m long copper wire with diameter of 1.11 mm and with 47.6 miliAmps of current running through it. (The resistivity of copper at room temperature is 1.68×10-8 Ohm×meter). Express the answer (only numerical value) to the nearest whole number.

Answer 1

First we need to calculate the resistance of this piece of wire. For a wire with resistivity $\rho$, length L and cross section A, the resistance is
$R= \frac{\rho L}{A}$
The diameter d of the wire is $d=1.11 mm=1.11 \cdot 10^{-3} m$, so the cross sectional area is
$A=\pi ( \frac{d}{2} )^2=9.7 \cdot 10^{-7} m^2$
Now, using $L=46.6 m$ and $\rho=1.68 \cdot 10^{-8} \Omega m$, we can calculate the resistance:
$R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(46.6 m)}{9.7 \cdot 10^{-7}m^2} =0.807 \Omega$

And now we can calculate the voltage drop across the resistor, by using Ohm's law, since we know the current flowing through it: $I=47.6 mA=47.6 \cdot 10^{-3} A$
$V=IR=(47.6 \cdot 10^{-3} A)(0.807 \Omega)=0.038 V=38 mV$