Power flow through a lossless line

Consider a line represented by series reactance X, neglecting resistance and shunt charging. Let V_S and V_R be sending- and receiving-end voltage magnitudes, with angle difference δ. Using per-phase RMS quantities, the receiving-end powers are:

P_R = (V_SV_R/X) sin δ

Q_R = (V_SV_R cos δ − V_R²)/X

Positive Q_R 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:

Q_R ≈ (V_SV_R − V_R²)/X

Step 2: Factor out the receiving voltage. Define the voltage-magnitude drop as ΔV = V_S − V_R:

Q_R ≈ (V_R/X)(V_S − V_R)

Q_R ≈ (V_R/X)ΔV

At fixed V_R 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.

Quick Trick

Real power follows angle; reactive power follows voltage.

Why Other Options Are Wrong

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.

Power flow through a lossless line

Consider a line represented by series reactance X, neglecting resistance and shunt charging. Let V_S and V_R be sending- and receiving-end voltage magnitudes, with angle difference δ. Using per-phase RMS quantities, the receiving-end powers are:

P_R = (V_SV_R/X) sin δ

Q_R = (V_SV_R cos δ − V_R²)/X

Positive Q_R 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:

Q_R ≈ (V_SV_R − V_R²)/X

Step 2: Factor out the receiving voltage. Define the voltage-magnitude drop as ΔV = V_S − V_R:

Q_R ≈ (V_R/X)(V_S − V_R)

Q_R ≈ (V_R/X)ΔV

At fixed V_R 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.

Quick Trick

Real power follows angle; reactive power follows voltage.

Why Other Options Are Wrong

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.

Power flow through a lossless line

Consider a line represented by series reactance X, neglecting resistance and shunt charging. Let V_S and V_R be sending- and receiving-end voltage magnitudes, with angle difference δ. Using per-phase RMS quantities, the receiving-end powers are:

P_R = (V_SV_R/X) sin δ

Q_R = (V_SV_R cos δ − V_R²)/X

Positive Q_R 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:

Q_R ≈ (V_SV_R − V_R²)/X

Step 2: Factor out the receiving voltage. Define the voltage-magnitude drop as ΔV = V_S − V_R:

Q_R ≈ (V_R/X)(V_S − V_R)

Q_R ≈ (V_R/X)ΔV

At fixed V_R 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.

Quick Trick

Real power follows angle; reactive power follows voltage.

Why Other Options Are Wrong

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.