Answer to Question #226223 in Electrical Engineering for Masande

Question

2.1 The per-phase parameters for a 60 Hz, 200 km long transmission line are

Answer 1

The total resistance $= 2 \pi fL= 2\pi*60*310.8= 117.16 \Omega\\$

Capacitance resistance $X_c= \frac{1}{jwC}=- \frac{j}{2 \pi fC}=- \frac{j}{2 \pi *60*1.477 \mu F}$

Therefore admittance $Y= \frac{1}{X_c}=2 \pi fC=2 \pi *60*1.477*10^{-6}= 0.0005565$