Question

Household wiring often uses 2.0 mm diameter copper wires. The wires can get rather long as they snake through the walls from the fuse box to the farthest corners of your house.

What is the potential difference across a 16 m long, 2.0 mm diameter copper wire carrying a 7.3 A current?

Answer 1

Concepts and reason

The concepts required to solve this problem are resistivity and Ohm’s law.

First use the resistivity equation to solve for the resistance of the wire.

Later, calculate the potential difference by using the Ohm’s law.

Fundamentals

The resistivity of a wire is given as,

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

Here, $L$ is the length of the wire, $A$ is the area of cross section of the wire, and $R$ is the resistance of wire.

The Ohm’s law states that the potential difference in a resistor is directly proportional to the current flowing through the circuit keeping all other conditions constant. The constant of proportionality is called resistance. The equation of Ohm’s law is,

$V = IR$

Here, $V$ is the applied potential difference, $I$ is the current, and $R$ is the resistance.

(A)

Use the resistivity equation to solve for the resistance.

Substitute $\pi \frac{{{d^2}}}{4}$ for $A$ in the equation $\rho = \frac{A}{L}R$ and solve for $R$.

$\begin{array}{c}\\\rho = \frac{{\pi \left( {\frac{{{d^2}}}{4}} \right)}}{L}R\\\\R = \frac{{4L\rho }}{{\pi {d^2}}}\\\end{array}$

Use the Ohms law to solve for potential difference.

Substitute $\frac{{4L\rho }}{{\pi {d^2}}}$ for $R$ in the equation$V = IR$.

$V = \frac{{4L\rho I}}{{\pi {d^2}}}$

Substitute $1.68 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for$\rho$, ${\rm{7}}{\rm{.3 A}}$ for$I$, $16{\rm{ m}}$ for$L$, and ${\rm{2}}{\rm{.0 mm}}$ for $d$ in the equation $V = \frac{{4L\rho I}}{{\pi {d^2}}}$ and calculate the potential difference.

$\begin{array}{c}\\V = \frac{{4\left( {1.68 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}} \right)\left( {{\rm{7}}{\rm{.3 A}}} \right)\left( {16{\rm{ m}}} \right)}}{{\pi {{\left( {{\rm{2}}{\rm{.0 mm}}} \right)}^2}}}\\\\ = \frac{{4\left( {1.68 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}} \right)\left( {{\rm{7}}{\rm{.3 A}}} \right)\left( {16{\rm{ m}}} \right)}}{{\pi {{\left( {{\rm{2}}{\rm{.0 mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)} \right)}^2}}}\\\\ = 0.625{\rm{ V}}\\\end{array}$

Ans:

The potential difference across the wire is${\rm{0}}{\rm{.625 V}}$.