Chapter 2, Problem 63E

To determine

Find the resistive power loss in aluminum $-$ clad steel $\text{B415}$ wire, and the resistive power reduced by using $\text{B75}$.

Answer to Problem 63E

The power wasted in aluminum $-$ clad steel $\text{B415}$ wire is $\text{248}\text{.35 nW}$ and the resistive power reduced by using $\text{B75}$.is $79.67\text{ \%}$.

Explanation of Solution

Given Data:

Refer to the $\text{Table 2}\text{.4}$ in the textbook.

The current flowing in the wire is $1\text{ mA}$,

The diameter of wire is $1\text{ mm}$,

The length of wire is $2.3\text{ m}$,

The resistivity $\text{B75}$ wire is $1.7241\text{ μ}\Omega .\text{cm}$ and

The resistivity of $\text{B415}$ wire is $8.4805\text{ μ}\Omega .\text{cm}$.

Formula used:

The expression for cross sectional area of wire is as follows.

$A=\pi {\left(\frac{d}{2}\right)}^2$ (1)

Here,

$A$ is the cross sectional area of wire and

$d$ is the diameter of wire.

The expression for resistance of $\text{B415}$ wire is as follows.

$R_{\text{B415}}=\left({\rho }_{\text{B415}}\frac{l}{A}\right)$ (2)

 Here,

$R_{\text{B415}}$ is the resistance of $\text{B415}$ wire,

$l$ is the length of wire,

${\rho }_{\text{B415}}$ is the resistivity of $\text{B415}$ wire.

The expression for resistance of $\text{B75}$ wire is as follows.

$R_{\text{B75}}=\left({\rho }_{\text{B75}}\frac{l}{A}\right)$ (3)

 Here,

$R_{\text{B75}}$ is the resistance of $\text{B75}$ wire,

${\rho }_{\text{B75}}$ is the resistivity of $\text{B75}$ wire.

The expression for the power wasted in $\text{B415}$ wire is as follows.

$p_{\text{B415}}=i^2{R}_{\text{B415}}$ (4)

Here,

$p_{\text{B415}}$ is the resistive power loss in $\text{B415}$ wire,

$i$ is the current flowing in $\text{B415}$ wire.

The expression for the power wasted in $\text{B75}$ wire is as follows.

$p_{\text{B75}}=i^2{R}_{\text{B75}}$ (5)

Here,

$p_{\text{B75}}$ is the resistive power loss in $\text{B75}$ wire.

The expression for the percentage resistive power loss reduced by using $\text{B75}$ is as follows.

$p_t=\left(\frac{p_{\text{B415}}-p_{\text{B75}}}{p_{\text{B415}}}\times 100\right)\text{ \%}$ (6)

Here,

$p_t$ is the resistive power loss reduced by using $\text{B75}$.

Calculation:

Substitute $1\text{ mm}$ for $d$ in the equation (1).

$\begin{array}{c}A=\pi {\left(\frac{1\text{ mm}}{2}\right)}^2\\ =0.7854{\text{ mm}}^2\\ =0.7854\times {10}^{-6}{\text{ m}}^2\left\{\because 1{\text{ mm}}^2={10}^{-6}{\text{ m}}^2\right\}\end{array}$

Substitute $8.4805\text{ μ}\Omega .\text{cm}$ for ${\rho }_{\text{B415}}$, $2.3\text{ m}$ for $l$ and $0.7854\times {10}^{-6}{\text{ m}}^2$ for $A$ in equation (2).

$\begin{array}{c}{R}_{\text{B415}}=\left(8.4805\text{ μ}\Omega .\text{cm}\times \frac{2.3\text{ m}}{0.7854\times {10}^{-6}{\text{ m}}^2}\right)\\ =\left(8.4805\times {10}^{-2}\text{ μ}\Omega .\text{m}\times \frac{2.3\text{ m}}{0.7854\times {10}^{-6}{\text{ m}}^2}\right)\left\{\because 1\text{ cm}={10}^{-2}\text{m}\right\}\\ =\left(8.4805\times {10}^{-8}\Omega .\text{m}\times \frac{2.3\text{ m}}{0.7854\times {10}^{-6}{\text{ m}}^2}\right)\left\{\because 1\text{ μ}\Omega ={10}^{-6}\Omega \right\}\\ =248.83\times {10}^{-3}\Omega \end{array}$

Substitute $1.7241\text{ μ}\Omega .\text{cm}$ for ${\rho }_{\text{B75}}$, $2.3\text{ m}$ for $l$ and $0.7854\times {10}^{-6}{\text{ m}}^2$ for $A$ in equation (3).

$\begin{array}{c}{R}_{\text{B75}}=\left(1.7241\text{ μ}\Omega .\text{cm}\times \frac{2.3\text{ m}}{0.7854\times {10}^{-6}{\text{ m}}^2}\right)\\ =\left(1.7241\times {10}^{-2}\text{ μ}\Omega .\text{m}\times \frac{2.3\text{ m}}{0.7854\times {10}^{-6}{\text{ m}}^2}\right)\left\{\because 1\text{ cm}={10}^{-2}\text{m}\right\}\\ =\left(1.7241\times {10}^{-8}\Omega .\text{m}\times \frac{2.3\text{ m}}{0.7854\times {10}^{-6}{\text{ m}}^2}\right)\left\{\because 1\text{ μ}\Omega ={10}^{-6}\Omega \right\}\\ =50.48\times {10}^{-3}\Omega \end{array}$

Substitute $1\text{ mA}$ for $i$ and $248.83\times {10}^{-3}\Omega$ for $R_{\text{B415}}$ in equation (4).

$\begin{array}{c}{p}_{\text{B415}}=\left({\left(1\text{ mA}\right)}^2\times 248.83\times {10}^{-3}\Omega \right)\\ =\left({\left(1\times {10}^{-3}\text{ A}\right)}^2\times 248.83\times {10}^{-3}\Omega \right)\left\{\because 1{\text{ mA=10}}^{-3}\text{ A}\right\}\\ =0.24835\times {10}^{-6}\text{ W}\\ \text{=248}\text{.35 nW }\left\{\because 1{\text{ W=10}}^9\text{ nW}\right\}\end{array}$

Substitute $1\text{ mA}$ for $i$ and $50.48\times {10}^{-3}\Omega$ for $R_{\text{B75}}$ in equation (5).

$\begin{array}{c}{p}_{\text{B75}}=\left({\left(1\text{ mA}\right)}^2\times 50.48\times {10}^{-3}\Omega \right)\\ =\left({\left(1\times {10}^{-3}\text{ A}\right)}^2\times 50.48\times {10}^{-3}\Omega \right)\left\{\because 1{\text{ mA=10}}^{-3}\text{ A}\right\}\\ =5.048\times {10}^{-8}\text{ W}\\ \text{=50}\text{.48 nW }\left\{\because 1{\text{ W=10}}^9\text{ nW}\right\}\end{array}$

Substitute $\text{248}\text{.35 nW}$ for $p_{\text{B415}}$ and $\text{50}\text{.48 nW }$ for $p_{\text{B75}}$ in equation (6).

$\begin{array}{c}{p}_t=\left(\frac{\text{248}\text{.35 nW}-\text{50}\text{.48 nW}}{\text{248}\text{.35 nW}}\times 100\right)\text{ \%}\\ =\left(\frac{197.87\text{ nW}}{\text{248}\text{.35 nW}}\times 100\right)\text{ \%}\\ =79.67\text{ \%}\end{array}$

Conclusion:

Thus, the power wasted in aluminum $-$ clad steel $\text{B415}$ wire is $\text{248}\text{.35 nW}$ and the resistive power reduced by using $\text{B75}$.is $79.67\text{ \%}$.