Answer to Question #209495 in Electrical Engineering for Mihri

Question

A parallel conductor transmission line #6 with a distance of 30.48 cm (12 in.)

Made of AWG copper wire (diameter =4.11mm (0.162 inch), sigmax= 58 MS/m). Conductors

Assume that there is air between Neglecting the internal inductance; L, C. G's per meter

their values; Find the dc resistance and the ac resistance at 1 MHz

Answer 1

Inductance (L) =$2*10^{-7} \ln\frac{d}{r'}=2*10^{-7} \ln\frac{30.48*10^{-2}}{1.6*10^{-3}}=10.49* 10^{-7}H/m$

Capacitor (C) =$\frac{2 \pi \epsilon_0}{ \ln (\frac{d}{r})}=\frac{2 \pi *8.85*10^{-12}}{ \ln (\frac{30.48*10^{-2}}{4.11/2 *10^{-3}})}=11.122*10^{-12}F/m$

DC resistance, $R= \rho \frac{l}{A}= \frac{1}{\delta}\frac{l}{A}= \frac{1}{58*10^6}\frac{30.48*10^{-2}}{13.267*10^{-6}}=39.63* 10^{-3} \Omega$

$G= \frac{1}{R}=25.23 \Omega ^{-1}$

Approximate AC resistance

$Z=R_{AC}=R+j wL=39.63*10^{-3}+j2 \pi*10^6*10.49*10^{-7}=39.63*10^{-2}+j6.6$