Chapter 28, Problem 72PQ

To determine

The criterion for choosing the dimension of the lead wire so that it provides the same resistance of the gold wire.

Answer to Problem 72PQ

In order to result the same resistance as that of the gold wire, the lead wire should have dimensions such that the cross sectional area and the lengths of the wires are related as $\underset{\_}{\frac{l_{\text{Pb}}}{A_{\text{Pb}}}=\left(0.108\right)\frac{l_{\text{Au}}}{A_{\text{Au}}}}$. The lead wire can be made as either with $\underset{\_}{A_{\text{Pb}}=A_{\text{Au}}}$ and $\underset{\_}{l_{\text{Pb}}=\left(0.108\right)l_{\text{Au}}}$, or with $\underset{\_}{l_{\text{Pb}}=l_{\text{Au}}}$ and $\underset{\_}{A_{\text{Pb}}=\frac{A_{\text{Au}}}{\left(0.108\right)}}$.

Explanation of Solution

The resistivity of gold is $2.24\times {10}^{-8}\text{Ω}\cdot \text{m}$, and that of lead is $2.065\times {10}^{-7}\Omega \cdot \text{m}$.

Write the expression for the resistance.

  $R=\rho \frac{l}{A}$                                                                                                                (I)

Here, $R$ is the resistance, $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross sectional area of the wire.

Analogous to equation (I), write the expression for the resistance of gold (Au) and lead (Pb) wires.

  $R_{\text{Au}}={\rho }_{\text{Au}}\frac{l_{\text{Au}}}{A_{\text{Au}}}$                                                                                                      (II)

  $R_{\text{Pb}}={\rho }_{\text{Pb}}\frac{l_{\text{Pb}}}{A_{\text{Pb}}}$                                                                                                       (III)

Since both wires has to have same resistance, equate the right-hand sides of equations (II) and (III) and reduce.

  $\begin{array}{c}{\rho }_{\text{Au}}\frac{l_{\text{Au}}}{A_{\text{Au}}}={\rho }_{\text{Pb}}\frac{l_{\text{Pb}}}{A_{\text{Pb}}}\\ \frac{l_{\text{Pb}}}{A_{\text{Pb}}}=\left(\frac{{\rho }_{\text{Au}}}{{\rho }_{\text{Pb}}}\right)\frac{l_{\text{Au}}}{A_{\text{Au}}}\end{array}$                                                                                         (IV)

Conclusion:

Substitute $2.24\times {10}^{-8}\text{Ω}\cdot \text{m}$ for ${\rho }_{\text{Au}}$, and $2.065\times {10}^{-7}\Omega \cdot \text{m}$ for ${\rho }_{\text{Pb}}$ in equation (IV) to find the condition on the dimensions of the wires.

  $\begin{array}{c}\frac{l_{\text{Pb}}}{A_{\text{Pb}}}=\left(\frac{2.24\times {10}^{-8}\text{Ω}\cdot \text{m}}{2.065\times {10}^{-7}\Omega \cdot \text{m}}\right)\frac{l_{\text{Au}}}{A_{\text{Au}}}\\ =\left(0.108\right)\frac{l_{\text{Au}}}{A_{\text{Au}}}\end{array}$

This expression is the most general statement about the dimensions with which the lead wire has to be made such that it offers same resistance as that of the gold wire. The ratio of the length to cross sectional area of the lead wire must be $0.108$ times that of the gold wire. Thus, the new wire can be made either by keeping cross sectional area to be same and lengths vary or vice versa.

Therefore, in order to result the same resistance as that of the gold wire, the lead wire should have dimensions such that the cross sectional area and the lengths of the wires are related as $\underset{\_}{\frac{l_{\text{Pb}}}{A_{\text{Pb}}}=\left(0.108\right)\frac{l_{\text{Au}}}{A_{\text{Au}}}}$. The lead wire can be made as either with $\underset{\_}{A_{\text{Pb}}=A_{\text{Au}}}$ and $\underset{\_}{l_{\text{Pb}}=\left(0.108\right)l_{\text{Au}}}$, or with $\underset{\_}{l_{\text{Pb}}=l_{\text{Au}}}$ and $\underset{\_}{A_{\text{Pb}}=\frac{A_{\text{Au}}}{\left(0.108\right)}}$.