Question #209495

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

Expert's answer

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"

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