Question

A 3-phase transmission line is shown in figure:



Voltage drop across the transmission line is given by the following equation:

$\left[\begin{array}{c}\Delta V_a\\ \Delta V_b\\ \Delta V_c\end{array}\right]=\left[\begin{array}{ccc}{Z}_s& Z_m& Z_m\\ Z_m& Z_s& Z_m\\ Z_m& Z_m& Z_s\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

Shunt capacitance of the line can be neglected. If the line has positive sequence impedance of 15$\Omega$ and zero sequence impedance of 48$\Omega$, then the values of Z${}_s$ and Z${}_m$ will be

• A. Z${}_s$ = 31.5$\Omega$; Z${}_m$ = 16.5$\Omega$
• B. Z${}_s$ = 26$\Omega$; Z${}_m$ = 11$\Omega$
• C. Z${}_s$ = 16.5$\Omega$; Z${}_m$ = 31.5$\Omega$
• D. Z${}_s$ = 11$\Omega$; Z${}_m$ = 26$\Omega$

Answer 1

The correct option is B Z${}_s$ = 26$\Omega$; Z${}_m$ = 11$\Omega$

Positive sequence impedance = Negative sequence impedance

= Z${}_s$ - Z${}_m$ = 15 ...(i)

Zero-sequence impedance

=Z${}_s$ + 2Z${}_m$ = 48 .....(ii)

Solving equation (i) and (ii), we get

Z${}_s$ = 26 $\Omega$ and Z${}_m$ = 11 $\Omega$