Question

When the starter motor on a car is engaged, there is a 290 A current in the wires between the battery and the motor. Suppose the wires are made of copper and have a total length of 1.5 m.What minimum diameter can the wires have if the voltage drop along the wires is to be less than 0.55 V?

Answer 1

Answer:

$33.6\times10^{-4}}\text{ m}$ = 3.36 mm

Explanation:

From Ohm's law,

$V = IR$ (Voltage = Current * Resistance)

$R = \dfrac{V}{I}$

The geometric definition of resistance is

$R = \rho\dfrac{l}{A}$

where $\rho$ is the resistivity of the material, $l$  and $A$ are the length and cross-sectional area, respectively.

$A = \rho\dfrac{l}{R}$

$A = \rho\dfrac{l\timesI}{V}$

Since the wire is assumed to have a circular cross-section, its area is given by

$A = \pi\dfrac{d^2}{4}$ where $d$ is the diameter.

$\pi\dfrac{d^2}{4} = \rho\dfrac{l\timesI}{V}$

$d = \sqrt{\dfrac{4\rho l I}{\pi\times V}}$

Resistivity of copper = $1.68\times10^{-8}$. With these and other given values,

$d = \sqrt{\dfrac{4\times1.68\times10^{-8}\times1.5\times290}{3.14\times 0.55}}$

$d = \sqrt{1128.43\times10^{-8}}$

$d = 33.6\times10^{-4} \text{ m}$