Series R, X and shunt admittance from ABCD

Given: $A = D = 0.9955\angle 0.086^\circ$, $B = 31.552\angle 71.608^\circ\ \Omega$, $C = 3\times10^{-4}\angle 90^\circ$ S.

Since $|A|$ is close to 1 and $C$ is purely capacitive, this is a medium line. The nominal-Π model is assumed, for which $B = Z$, $A = 1 + ZY/2$ and $C = Y(1 + ZY/4)$.

Series impedance

$$\begin{aligned} Z &= B = 31.552\angle 71.608^\circ \\ R &= 31.552\cos 71.608^\circ = 9.955\ \Omega \\ X_L &= 31.552\sin 71.608^\circ = 29.94\ \Omega \end{aligned}$$

Shunt admittance

From $A = 1 + \frac{ZY}{2}$:

$$\begin{aligned} A - 1 &= (0.995499 + j0.001494) - 1 = -0.004501 + j0.001494 = 4.743\times10^{-3}\angle 161.63^\circ \\ Y &= \frac{2(A - 1)}{Z} = \frac{2\times4.743\times10^{-3}\angle 161.63^\circ}{31.552\angle 71.608^\circ} \\ &= 3.006\times10^{-4}\angle 90.03^\circ \approx j3.0\times10^{-4}\ \text{S} \end{aligned}$$

Check with the given $C$: $Y(1 + ZY/4) = 2.999\times10^{-4}\angle 90.07^\circ$ S $\approx 3\times10^{-4}\angle 90^\circ$ S, which agrees. (A nominal-T reading, $C = Y$, gives nearly the same numbers: $Z = 9.96 + j30.01\ \Omega$.)

Answer: Series resistance $R \approx 9.96\ \Omega$, series inductive reactance $X_L \approx 29.94\ \Omega$, shunt admittance $Y \approx j3.0\times10^{-4}$ S (all per phase).