Correct answer: C — Line voltage drop; δ
Power flow through a lossless line
Consider a line represented by series reactance X, neglecting resistance and shunt charging. Let VS and VR be sending- and receiving-end voltage magnitudes, with angle difference δ. Using per-phase RMS quantities, the receiving-end powers are:
PR = (VSVR/X) sin δ
QR = (VSVR cos δ − VR²)/X
Positive QR represents lagging reactive power delivered at the receiving end. These equations show that active power depends strongly on angle, while reactive power depends strongly on voltage magnitude.
Step 1: Apply the small-angle approximation. For small δ measured in radians, cos δ ≈ 1. Hence:
Step 2: Factor out the receiving voltage. Define the voltage-magnitude drop as ΔV = VS − VR:
QR ≈ (VR/X)(VS − VR)
QR ≈ (VR/X)ΔV
At fixed VR and X, lagging reactive power is approximately proportional to voltage-magnitude drop, and δ does not appear in this approximation. Thus option C is intended. The independence is not exact: at larger angles, the cos δ term matters, and resistance also couples voltage, angle, active power and reactive power.
