Previous Year Question
Correct answer: C
Explanation
C — Line voltage drop; δ. This is the small-angle, predominantly reactive-line approximation; the exact reactive-power expression still contains cos δ.
Formula
Step-by-step solution
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:
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:
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.
Other options
A reverses the dominant dependencies. B reactance affects the magnitude but reactive power is not independent of power angle in the exact expression simply as stated. D does not represent the standard decoupled relation. C matches the normal small-angle approximation.
Common mistake
Treating the ‘independent of δ’ statement as exact under all operating conditions; it is an approximation associated with a predominantly reactive line and small δ.
Exam tip
P–δ and Q–V is one of the most useful power-system approximation pairs.
Quick method
Real power follows angle; reactive power follows voltage.