Chapter 18, Problem 37P

(a)

To determine

the ratio of resistance of silver wire to the copper wire.

Answer to Problem 37P

The ratio of resistance of silver wire to the copper wire is $1.1$.

Explanation of Solution

Write the equation for resistance.

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

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

Write the equation for mass density.

$D=\frac{m}{V}=\frac{m}{AL}$ (II)

Here, $D$  is mass density, $V$  is volume, $M$  is mass, $A$  is area and $L$  is length.

Substitute value of $A$ from equation II to I and write for silver and copper wires and If the wires have the same length and same mass, then resistance can be written as,

$\frac{R_{Ag}}{R_{Cu}}=\frac{{\rho }_{Ag}{D}_{Ag}}{{\rho }_{Cu}{D}_{Cu}}$ (III)

Here, $R_{Ag}$ is the resistance of silver, $R_{Cu}$ is the resistance of copper, ${\rho }_{Ag}$ is the resistivity of silver, ${\rho }_{Cu}$ is the resistivity of copper, $D_{Ag}$ is the mass density of silver and $D_{Cu}$ is the mass density of copper.

Conclusion:

Substitute $1.59\times {10}^{-8}\Omega \text{m}$ for ${\rho }_{Ag}$, $1.67\times {10}^{-8}\Omega \text{m}$ for ${\rho }_{Cu}$, $10.1\times {10}^3{\text{ kg/m}}^3$ for $D_{Ag}$ and $8.9\times {10}^3{\text{ kg/m}}^3$ for $D_{Cu}$ in equation III.

$\begin{array}{c}\frac{R_{Ag}}{R_{Cu}}=\frac{\left(1.59\times {10}^{-8}\Omega \text{m}\right)\left(10.1\times {10}^3{\text{ kg/m}}^3\right)}{\left(1.67\times {10}^{-8}\Omega \text{m}\right)\left(8.9\times {10}^3{\text{ kg/m}}^3\right)}\\ =1.1\end{array}$

Therefore, the ratio of resistance of silver wire to the copper wire is $1.1$.

(b)

To determine

the ratio of resistance of aluminum wire to the copper wire.

Answer to Problem 37P

The ratio of resistance of aluminum wire to the copper wire is $0.48$.

Explanation of Solution

If the wires have the same length and same mass, then resistance can be written as,

$\frac{R_{Al}}{R_{Cu}}=\frac{{\rho }_{Al}{D}_{Al}}{{\rho }_{Cu}{D}_{Cu}}$ (IV)

Here, $R_{Al}$ is the resistance of aluminum, $R_{Cu}$ is the resistance of copper, ${\rho }_{Al}$ is the resistivity of aluminum, ${\rho }_{Cu}$  is the resistivity of copper, $D_{Al}$ is the mass density of silver and $D_{Cu}$ is the mass density of copper.

Conclusion:

Substitute $2.65\times {10}^{-8}\Omega \text{m}$ for ${\rho }_{Al}$, $1.67\times {10}^{-8}\Omega \text{m}$ for ${\rho }_{Cu}$, $2.7\times {10}^3{\text{ kg/m}}^3$ for $D_{Al}$ and $8.9\times {10}^3{\text{ kg/m}}^3$ for $D_{Cu}$ in equation IV.

$\begin{array}{c}\frac{R_{Ag}}{R_{Cu}}=\frac{\left(2.65\times {10}^{-8}\Omega \text{m}\right)\left(2.7\times {10}^3{\text{ kg/m}}^3\right)}{\left(1.67\times {10}^{-8}\Omega \text{m}\right)\left(8.9\times {10}^3{\text{ kg/m}}^3\right)}\\ =0.48\end{array}$

Therefore, the ratio of resistance of aluminum wire to the copper wire is $0.48$.

(c)

To determine

the best conducting material among silver and aluminum.

Answer to Problem 37P

The best conducting material among silver and aluminum is aluminum.

Explanation of Solution

For the wires with equal length and equal mass, the ratio of resistance shows that aluminum is the best conductor.

Conclusion:

Among the materials aluminum has the lowest resistance value.

Therefore, the best conducting material among silver and aluminum is aluminum.