Question

A transmission line has series impedance $Z=(20+j40)\Omega$, shunt admittance $Y=j4\times {10}^{-4}\mho$, the line is modeled as receiving end capacitor model. The value of shunt inductive reactance to be placed at the end of line to avoid ferranti effect is

• A. $1600\Omega$
• B. $2000\Omega$
• C. $2500\Omega$
• D. $2400\Omega$

Answer 1

The correct option is C $2500\Omega$

Given transmission line model can be drawn,

Equivalent T -matrix will be
$\left[\begin{array}{cc}{A}_1& B_1\\ C_1& D_1\end{array}\right]=\left[\begin{array}{cc}1+YZ& Z\\ Y& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1/j{X}_r& 1\end{array}\right]$

For no ferranti effect,

$V_s=V_R\to A_1=1$
$1+YZ+\frac{Z}{j{X}_r}=1$

$Simplifying,YZ=j\frac{Z}{X_r},$
$or|X_r|=\frac{Z}{YZ}=\frac{1}{Y}=\frac{1}{4\times {10}^{-4}}=2500\Omega$