Reactive power proportional to voltage drop

Consider a short line with negligible resistance, $Z \approx jX$, sending voltage $V_S\angle\delta$ and receiving voltage $V_R\angle0^\circ$.

Derivation

$$I = \frac{V_S\angle\delta - V_R}{jX}$$ $$\begin{aligned} S_R &= V_R I^* = V_R\,\frac{V_S\angle-\delta - V_R}{-jX} \\ &= \frac{V_SV_R\sin\delta}{X} + j\,\frac{V_SV_R\cos\delta - V_R^2}{X} \end{aligned}$$

So

$$P_R = \frac{V_SV_R}{X}\sin\delta,\qquad Q_R = \frac{V_R}{X}\left(V_S\cos\delta - V_R\right)$$

Reactive power and voltage drop

In practice the power angle $\delta$ is small (normally below about $10^\circ$–$15^\circ$), so $\cos\delta \approx 1$:

$$Q_R \approx \frac{V_R\,(V_S - V_R)}{X} = \frac{V_R}{X}\,\Delta V$$

where $\Delta V = |V_S| - |V_R|$ is the magnitude drop along the line.

Hence:

• $Q_R \propto \Delta V$: reactive power transfer is set by the difference in voltage magnitudes.
• $Q_R$ is (nearly) independent of $\delta$, since $\cos\delta \approx 1$ changes very little.
• Conversely, $P_R \propto \sin\delta$ depends mainly on the angle, not on the magnitude difference.

Example

$V_S = 134$ kV, $V_R = 132$ kV, $X = 50\ \Omega$:

For small $\delta$ the value stays close to $V_R\Delta V/X$, but it would double if $\Delta V$ doubled.

This is why voltage magnitude is controlled by reactive power (Q–V coupling), and frequency/angle by active power (P–δ coupling).