Correct Answer - Option 2 : |V
S| - |V
R|
The active power transfer over a line is given by,
\(P = \frac{{{V_S}{V_R}}}{{{X_S}}}\sin \delta \)
The reactive power transfer over a line is given by
\(Q = \frac{{{V_S}{V_R}}}{{{X_S}}}\cos \delta - \frac{{V_R^2}}{{{X_S}}}\)
Where VS is the sending end voltage
VR is the receiving end voltage
XS is the line reactance
δ is the power angle
Active power is mainly depends on the power angle whereas the reactive power transfer over a line mainly depends on the difference between sending end voltage and receiving end voltage i.e. |V
S| - |V
R|.
