Question

Consider a lossy transmission line with $V_1$ and $V_2$ as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be

• A. greater than $\left|\frac{V_1{V}_2}{X}\right|$
• B. less than $\left|\frac{V_1{V}_2}{X}\right|$
• C. equal to $\left|\frac{V_1{V}_2}{X}\right|$
• D. equal to $\left|\frac{V_1{V}_2}{X}\right|$

Answer 1

The correct option is B less than $\left|\frac{V_1{V}_2}{X}\right|$

With only x:

$P_{max}=\left|\frac{V_1{V}_2}{X}\right|$With Lossy Tr. Line

$P=\left|\frac{V_1{V}_2}{Z}\right|cos(\beta -\delta )-\left|\frac{A{V}_2^2}{Z}\right|cos(\beta -\alpha )$

$\therefore WithLossyLine{P}_{max}\le \left|\frac{V_1{V}_2}{X}\right|$