Question

A rear window defroster consists of a long, flat wire bonded to the inside surface of the window. When current passes through the wire, it heats up and melts ice and snow on the window. For one window the wire has a total length of 12.0 m, a width of 1.8 mm, and a thickness of 0.12 mm. The wire is connected to the car's 12.0 V battery and draws 12.0 A. What is the resistivity of the wire material?

Answer 1

Answer:

$\rho=1.8\times 10^{-8}\Omega.m$

Explanation:

Given:

width of the wire, $w=1.8\ mm=1.8\times 10^{-3}\ m$

thickness of the flat wire, $d=0.12\ mm=1.2\times 10^{-4}$

length of the wire, $l=12\ m$

voltage across the wire, $V=12\ V$

current through the wire, $I=12\ A$

Now the net resistance of the wire:

using ohm's law

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

$R=\frac{12}{12}$

$R=1\ \Omega$

We have the relation between the resistivity and the resistance as:

$R=\rho.\frac{l}{a}$

where:

a = cross sectional area of the wire

$\rho =$ resistivity of the wire material

$1=\rho\times \frac{12}{1.8\times 10^{-3}\times 1.2\times 10^{-4}}$

$\rho=1.8\times 10^{-8}\Omega.m$