Chapter 9, Problem 9.1P

Given a 2.0-cm length of AWG20 copper wire, (a) calculate R_dC, (b) calculate R_ac at 800 MHz, (c) estimate L.

To determine

The DC resistance $R_{\text{dc}}$ for $\text{AWG20}$ copper wire.

Answer to Problem 9.1P

The DC resistance $R_{\text{dc}}$ for $\text{AWG20}$ copper wire is $670\text{ μ}\Omega$ .

Explanation of Solution

Given:

The length of a $\text{AWG20}$ copper wire is $2\text{ cm}$ .

Concept used:

The expression for DC resistance is shown below.

  $R_{\text{dc}}=\frac{l}{\text{σπ}{a}^2}$     ........... (1)

Calculation:

The diameter of $\text{AWG20}$ copper wire is $31.96\text{ mil}$ .

The radius of $\text{AWG20}$ copper wire is calculated as,

  $\begin{array}{c}a=\frac{31.96\text{ mil}}{2}\left(25.4\times {10}^{-6}\frac{\text{m}}{\text{mil}}\right)\\ =405.892\times {10}^{-6}\text{ m}\end{array}$

The conductivity of copper is $5.8\times {10}^7\text{S}/\text{m}$ .

Substitute $2\times {10}^{-2}$ for $l$ , $5.8\times {10}^7$ for $\text{σ}$ and $405.892\times {10}^{-6}$ for $a$ in equation (1).

  $\begin{array}{c}{R}_{\text{dc}}=\frac{2\times {10}^{-2}}{5.8\times {10}^7\text{π}{\left(405.892\times {10}^{-6}\right)}^2}\\ \approx 670\times {10}^{-6}\end{array}$

Therefore, the DC resistance is $670\text{ μ}\Omega$ .

Conclusion:

Thus, the DC resistance $R_{\text{dc}}$ for $\text{AWG20}$ copper wire is $670\text{ μ}\Omega$ .

To determine

The AC resistance at $800\text{ MHz}$ .

Answer to Problem 9.1P

The AC resistance at $800\text{ MHz}$ is $58\text{ m}\Omega$ .

Explanation of Solution

Concept used:

The expression for Ac resistance is shown below.

  $R_{\text{ac}}=\frac{l}{\text{σ}2\text{π}a\text{δ}}$   ........ (1)

Calculation:

The skin depth for the copper wire is calculated as,

  $\text{δ}=\frac{1}{\sqrt{\text{π}f\text{μσ}}}$

Substitute $800\times {10}^6$ for $f$ , $4\text{π}\times {10}^{-7}$ for $\text{μ}$ and $5.8\times {10}^7$ for $\text{σ}$ in above equation.

  $\begin{array}{c}\text{δ}=\frac{1}{\sqrt{\text{π}\left(800\times {10}^6\right)\left(4\text{π}\times {10}^{-7}\right)\left(5.8\times {10}^7\right)}}\\ =2.336\times {10}^{-6}\text{ m}\end{array}$

Substitute $2\times {10}^{-2}$ for $l$ , $5.8\times {10}^7$ for $\text{σ}$ , $405.892\times {10}^{-6}$ for $a$ and $2.336\times {10}^{-6}$ for $\text{δ}$ in equation (1).

  $\begin{array}{c}{R}_{\text{ac}}=\frac{2\times {10}^{-2}}{5.8\times {10}^7\cdot 2\text{π}\left(405.892\times {10}^{-6}\right)\left(2.336\times {10}^{-6}\right)}\\ \approx 58\times {10}^{-3}\end{array}$

Therefore, the AC resistance is $58\text{ m}\Omega$ .

Conclusion:

Thus, the AC resistance at $800\text{ MHz}$ is $58\text{ m}\Omega$ .

To determine

The external inductance for copper wire.

Answer to Problem 9.1P

The external inductance for copper wire is $14\text{ nH}$ .

Explanation of Solution

Concept used:

The expression for external inductance for wire segment is shown below.

  $L=2\times {10}^{-7}l\left[\ln \left(\frac{2l}{a}\right)-1\right]$   ........ (3)

Calculation:

Substitute $2\times {10}^{-2}$ for $l$ and $405.892$ for $a$ in equation (3).

  $\begin{array}{c}L=2\times {10}^{-7}\left(2\times {10}^{-2}\right)\left[\ln \left(\frac{2\cdot 2\times {10}^{-2}}{405.892\times {10}^{-6}}\right)-1\right]\\ \approx 14\times {10}^{-9}\end{array}$

Therefore, the inductance is $14\text{ nH}$ .

Conclusion:

Thus, the external inductance for copper wire is $14\text{ nH}$ .