Question

a.) A 33.9-m length of copper wire at 20.0°C has a radius of 0.31 mm. If a potential difference of 7.00 V is applied across the length of the wire, determine the current in the wire.

b.) If the wire is heated to 40.0°C while the 7.00-V potential difference is maintained, what is the resulting current in the wire?

Answer 1

a) the resistance of a conductor can be defined like

$R=\frac{\rho\,L}{A}$

where $\rho$ ​ is its resistivity, the lenght and the cross-section area

The copper resistivity at $\mathrm{20\,^{\circ}C\,\left( 293\,^{\circ}K \right)}$ is

$\rho_{\mathrm{cu}}=\mathrm{1.72\times10^{-8}\,\Omega.m}$

then, the resistance of the wire is

$R=\frac{\rho_{\mathrm{cu}}\,L}{\pi\,r^{2}}$

$R=\mathrm{\frac{\left( 1.72\times10^{-8}\,\Omega.m \right)\left( 33.9\,m \right )}{\pi\left( 0.31\times10^{-3}\,m \right )^{2}}}$

$R=\mathrm{1.93\,\Omega}$

Hence, from the equation

$V=i\,R$

the current through the wire is

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

$i=\mathrm{\frac{7.00\,V}{1.93\,\Omega}}$

$i=\mathrm{3.63\,A}$

b) The resistivity of a conductor changes with temperature according to the equation.

$\rho=\rho_{0}\left[ 1+\alpha\left( T-T_{0} \right) \right]$

where $\alpha$ is the coefficient of temperature of resistivity. For a copper conductor

$\alpha=\mathrm{3.90\times10^{-3}\,^{\circ}K^{-1}}$

then, the resistivity of the copper wire at $\mathrm{40\,^{\circ}C\,\left( 313\,^{\circ}K \right)}$   is

$\rho=\mathrm{\left( 1.72\times10^{-8}\,\Omega.m \right)\left[ 1+\left( 3.90\times10^{-3}\,^{\circ}K^{-1} \right)\left( 313\,^{\circ}K-293\,^{\circ}K \right) \right]}$

$\rho_{\mathrm{cu}}=\mathrm{1.85\times10^{-8}\,\Omega.m}$

and the resistance

$R=\frac{\rho_{\mathrm{cu}}\,L}{\pi\,r^{2}}$

$R=\mathrm{\frac{\left( 1.85\times10^{-8}\,\Omega.m \right)\left( 33.9\,m \right )}{\pi\left( 0.31\times10^{-3}\,m \right )^{2}}}$

$R=\mathrm{2.08\,\Omega}$

Hence, the current through the wire is

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

$i=\mathrm{\frac{7.00\,V}{2.08\,\Omega}}$

$i=\mathrm{3.37\,A}$