Chapter 2, Problem 57E

Using the data in Table 2.4, calculate the resistance and conductance of 50 ft of wire with the following sizes: AWG 2, AWG 14, and AWG 28.

To determine

Find the resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 2}$, $\text{AWG 14}$ and $\text{AWG 28}$

Answer to Problem 57E

The resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 2}$ is $7.815\text{ m}\Omega$ and $127.96\text{ S}$ respectively.

The resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 14}$ is $\text{126 m}\Omega$ and $7.936\text{ S}$ respectively.

The resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 28}$ is $3.265\Omega$ and $306\text{ mS}$ respectively.

Explanation of Solution

Formula used:

The expression for resistance per feet for $\text{AWG 2}$ wire is as follows.

$R_{L\left(\text{AWG 2}\right)}=\frac{R_{S\left(\text{AWG 2}\right)}\Omega }{1000\text{ ft}}$ (1)

Here,

$R_{L\left(\text{AWG 2}\right)}$ is the resistance per feet for $\text{AWG 2}$ wire and $R_{S\left(\text{AWG 2}\right)}$ is the resistance per $1000\text{ ft}$ for $\text{AWG 2}$ wire.

Hence, the expression for the total resistance of given wire is as follows.

$R_{\left(\text{AWG 2}\right)}=L\times R_{L\left(\text{AWG 2}\right)}$ (2)

Here,

$R_{\left(\text{AWG 2}\right)}$ is the total resistance of given wire,

$L$ is the given length of the wire and

$R_{L\left(\text{AWG 2}\right)}$ is the, resistance per feet for $\text{AWG 2}$ wire.

Expression for the conductance of the wire is as follows.

$G_{\left(\text{AWG 2}\right)}=\frac{1}{R_{\left(\text{AWG 2}\right)}}$ (3)

Here,

$G_{\left(\text{AWG 2}\right)}$ is the conductance of the wire and

$R_{\left(\text{AWG 2}\right)}$ is the resistance of given wire.

The expression for resistance per feet for $\text{AWG 14}$ wire is as follows.

$R_{L\left(\text{AWG 14}\right)}=\frac{R_{S\left(\text{AWG 14}\right)}\Omega }{1000\text{ ft}}$ (4)

Here,

$R_{L\left(\text{AWG 14}\right)}$ is the resistance per feet for $\text{AWG 14}$ wire and

$R_{S\left(\text{AWG 14}\right)}$ is the resistance per $1000\text{ ft}$ for $\text{AWG 14}$ wire.

Hence, the expression for the total resistance of given wire is as follows.

$R_{\left(\text{AWG 14}\right)}=L\times R_{L\left(\text{AWG 14}\right)}$ (5)

Here,

$R_{\left(\text{AWG 14}\right)}$ is the total resistance of given wire,

$L$ is the given length of the wire and

$R_{L\left(\text{AWG 14}\right)}$ is the, resistance per feet for $\text{AWG 14}$ wire.

Expression for the conductance of the wire is as follows.

$G_{\left(\text{AWG 14}\right)}=\frac{1}{R_{\left(\text{AWG 14}\right)}}$ (6)

Here,

$G_{\left(\text{AWG 14}\right)}$ is the conductance of the wire and

$R_{\left(\text{AWG 14}\right)}$ is the resistance of given wire.

The expression for resistance per feet for $\text{AWG 28}$ wire is as follows.

$R_{L\left(\text{AWG 28}\right)}=\frac{R_{S\left(\text{AWG 28}\right)}\Omega }{1000\text{ ft}}$ (7)

Here,

$R_{L\left(\text{AWG 28}\right)}$ is the resistance per feet for $\text{AWG 28}$ wire and

$R_{S\left(\text{AWG 28}\right)}$ is the resistance per $1000\text{ ft}$ for $\text{AWG 28}$ wire.

Hence, the expression for the total resistance of given wire is as follows.

$R_{\left(\text{AWG 28}\right)}=L\times R_{L\left(\text{AWG 28}\right)}$ (8)

Here,

$R_{\left(\text{AWG 28}\right)}$ is the total resistance of given wire,

$L$ is the given length of the wire and

$R_{L\left(\text{AWG 28}\right)}$ is the, resistance per feet for $\text{AWG 28}$ wire.

Expression for the conductance of the wire is as follows.

$G_{\left(\text{AWG 28}\right)}=\frac{1}{R_{\left(\text{AWG 28}\right)}}$ (9)

Here,

$G_{\left(\text{AWG 28}\right)}$ is the conductance of the wire and

$R_{\left(\text{AWG 28}\right)}$ is the resistance of given wire.

Calculation:

Refer Table 2.4 in the textbook.

Substitute $0.1563\Omega$ for $R_{S\left(\text{AWG 2}\right)}$ in equation (1).

$\begin{array}{c}{R}_{L\left(\text{AWG 2}\right)}=\frac{\text{0}\text{.1563 }\Omega }{1000\text{ ft}}\\ =156.3\times {10}^{-6}\frac{\Omega }{\text{ft}}\\ =156.3\frac{\mu \Omega }{\text{f}t}\left\{\because \text{ 1 }\mu \Omega ={\text{10}}^{-6}\Omega \right\}\end{array}$

Substitute $156.3\frac{\mu \Omega }{\text{f}t}$ for $R_{L\left(\text{AWG 2}\right)}$ and $50\text{ ft}$ for $L$ in equation (2).

$\begin{array}{c}{R}_{\left(\text{AWG 2}\right)}=\left(50\text{ ft}\times 156.3\frac{\mu \Omega }{\text{f}t}\right)\\ =7815\mu \Omega \\ =7815\times {10}^{-3}\text{ m}\Omega \left\{\because \text{ 1 }\mu \Omega ={\text{10}}^{-3}\text{m}\Omega \right\}\\ =7.815\text{m}\Omega \end{array}$

Substitute $7.815\text{ m}\Omega$ for $R_{\left(\text{AWG 2}\right)}$ in equation (3).

$\begin{array}{c}{G}_{\left(\text{AWG 2}\right)}=\frac{1}{7.815\text{ m}\Omega }\\ =\frac{1}{7.815\times {\text{10}}^{-3}\Omega }\left\{\because \text{ 1 m}\Omega ={\text{10}}^{-3}\Omega \right\}\\ =127.96\text{ S}\end{array}$

Substitute $\text{2}\text{.52 }\Omega$ for $R_{S\left(\text{AWG 14}\right)}$ in equation (4).

$\begin{array}{c}{R}_{L\left(\text{AWG 14}\right)}=\frac{\text{2}\text{.52 }\Omega }{1000\text{ ft}}\\ =2.52\times {10}^{-3}\frac{\Omega }{\text{ft}}\\ =2.52\frac{\text{m}\Omega }{\text{f}t}\left\{\because \text{ 1 }\Omega ={\text{10}}^3\text{m}\Omega \right\}\end{array}$

Substitute $2.52\frac{\text{m}\Omega }{\text{f}t}$ for $R_{L\left(\text{AWG 14}\right)}$ and $50\text{ ft}$ for $L$ in equation (5).

$\begin{array}{c}{R}_{\left(\text{AWG 14}\right)}=\left(50\text{ ft}\times 2.52\frac{\text{m}\Omega }{\text{f}t}\right)\\ =126\text{ m}\Omega \end{array}$

Substitute $126\text{ m}\Omega$ for $R_{\left(\text{AWG 14}\right)}$ in equation (6).

$\begin{array}{c}{G}_{\left(\text{AWG 14}\right)}=\frac{1}{\text{126 m}\Omega }\\ =\frac{1}{\text{126}\times {\text{10}}^{-3}\Omega }\left\{\because \text{ 1 m}\Omega ={\text{10}}^{-3}\Omega \right\}\\ =7.936\text{ S}\end{array}$

Substitute $\text{65}\text{.3 }\Omega$ for $R_{S\left(\text{AWG 28}\right)}$ in equation (7).

$\begin{array}{c}{R}_{L\left(\text{AWG 28}\right)}=\frac{\text{65}\text{.3 }\Omega }{1000\text{ ft}}\\ =65.3\times {10}^{-3}\frac{\Omega }{\text{ft}}\\ =65.3\frac{\text{m}\Omega }{\text{f}t}\left\{\because \text{ 1 }\Omega ={\text{10}}^3\text{m}\Omega \right\}\end{array}$

Substitute $65.3\frac{\text{m}\Omega }{\text{f}t}$ for $R_{L\left(\text{AWG 28}\right)}$ and $50\text{ ft}$ for $L$ in equation (8).

$\begin{array}{c}{R}_{\left(\text{AWG 28}\right)}=\left(50\text{ ft}\times 65.3\frac{\text{m}\Omega }{\text{f}t}\right)\\ =3265\text{ m}\Omega \\ =3265\times {\text{10}}^{-3}\Omega \left\{\because \text{ 1 m}\Omega ={\text{10}}^{-3}\Omega \right\}\\ =3.265\Omega \end{array}$

Substitute $3.265\Omega$ for $R_{\left(\text{AWG 28}\right)}$ in equation (9).

$\begin{array}{c}{G}_{\left(\text{AWG 28}\right)}=\frac{1}{3.265\Omega }\\ =0.306\text{ S}\\ \text{=0}\text{.306}\times {\text{10}}^3\text{}\text{mS }\left\{\because \text{ 1 S}={\text{10}}^3\text{mS}\right\}\\ =306\text{ mS}\end{array}$

Conclusion:

Thus, the resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 2}$ is $7.815\text{ m}\Omega$ and $127.96\text{ S}$ respectively.

The resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 14}$ is $\text{126 m}\Omega$ and $7.936\text{ S}$ respectively.

The resistance and conductance of $50\text{ ft}$ wire with size of $\text{AWG 28}$ is $3.265\Omega$ and $306\text{ mS}$ respectively.