Question

8. Calculate the intrinsic impedance of aluminum (? = 3.816 × 107 S/m. ,1-110) and nickel (?-1.449 x 107 S/m) at 200 Hz and ad 25 GHz Note that the relative peraility of nicke is 700 at 200 and lr - 150 at 150 at 25 GHz.

Answer 1

Given that: aluminum: sigma = 3.816 times 10* s/m, mu = mu_0 sigma = 1.449 times 10^7 s/m, mu_r = 700 at 200 Hz, mu_r = 150 at 25 Hz impedance: we know that intrinsic impedance formula that eat = squareroot 900 mu/ sigma + (1) For aluminum where mu = rightarrow mu = mu_0 = mu = mu_ = times 10^-7 H/m, mu_r = 900 e = e_0 e_s e_D = 8.85 times 10^-12, e_r = 16-886 From eqn. (1) eta squareroot 3 times 2 H times 20 times 40 times 40^-7 times 700/3.816 times 10^7 + 9 (2 pi times 200 times 8.85 times 10^-12 times 16.886) eta = squareroot 1.576 times 10^-3 j times 700/3.816 times 10^-7 + 1.877 times 10^-7 g eta = squareroot 2.89 times 10^-8 90 degree Therefore impedance (eta) = 1.7 times 10^-4 95 degree For nickel: we know that mu_r = 150 at 25 G Hz sigma = 1.449 times 10^7