---
title: "Vs from P+jQ and reactive power flow direction"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:vs-from-pq-reactive-flow"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/vs-from-pq-reactive-flow
site: BCT Engineering
---

# Vs from P+jQ and reactive power flow direction

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2073 Magh, 4 marks
- Video: [Master Power Flow Calculations: Deriving the Real (Watts) and Reactive (VAR) Power Formulas (Part 2)](https://www.youtube.com/watch?v=L-aNJafg4wI) — Electrical PE Review
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/vs-from-pq-reactive-flow
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

Derive the expression for sending end voltage (Vs) in terms of receiving end voltage (Vr), line resistance (R) and reactance (X) and complex power (P + jQ) demand at the receiving end and hence justify that reactive power always flows from higher voltage magnitude to lower voltage magnitude.

## Short answer

With V_R as reference, V_S = V_R + (RP+XQ)/V_R + j(XP−RQ)/V_R. For X ≫ R, |V_S|−|V_R| ≈ XQ/V_R, so Q ≈ V_R(|V_S|−|V_R|)/X, thus reactive power always flows from the higher voltage magnitude bus to the lower voltage magnitude bus.

## Answer

#### Derivation
Short line, per phase impedance $Z = R + jX$. Take $V_R = |V_R|\angle0^\circ$ and the receiving-end demand $S_R = P + jQ$ (per phase).

Since $S_R = V_RI^*$,
$$
I = \frac{P - jQ}{V_R}
$$
Then
$$
\begin{aligned}
V_S &= V_R + (R + jX)\frac{P - jQ}{V_R} \\
&= \left(V_R + \frac{RP + XQ}{V_R}\right) + j\left(\frac{XP - RQ}{V_R}\right)
\end{aligned}
$$
So
$$
|V_S| = \sqrt{\left(V_R + \frac{RP + XQ}{V_R}\right)^2 + \left(\frac{XP - RQ}{V_R}\right)^2}
$$
The first bracket is the **in-phase drop** $\Delta V = (RP + XQ)/V_R$ and the second is the **quadrature drop** $\delta V = (XP - RQ)/V_R$.

#### Reactive power flows from higher to lower voltage
For transmission lines $X \gg R$ and the quadrature part changes the magnitude very little. Then
$$
|V_S| - |V_R| \approx \frac{RP + XQ}{V_R} \approx \frac{XQ}{V_R}
$$
$$
\Rightarrow\ Q \approx \frac{V_R\,(|V_S| - |V_R|)}{X}
$$
- If $|V_S| > |V_R|$: $Q > 0$, reactive power flows from sending end to receiving end.
- If $|V_S| < |V_R|$: $Q < 0$, reactive power flows from receiving end to sending end.
- If $|V_S| = |V_R|$: almost no reactive power transfer.

So reactive power always flows from the bus with the **higher voltage magnitude** to the bus with the **lower voltage magnitude**, whatever the phase angles. (Active power, on the other hand, follows the angle: $P \approx V_SV_R\sin\delta/X$.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With V_R as reference and S_R = P + jQ, the line current is:
   A. I = (P + jQ)/V_R
   B. I = (P - jQ)/V_R
   C. I = P/V_R
   D. I = (Q - jP)/V_R
   Answer: B. Since S_R = V_R I*, I* = (P + jQ)/V_R, so I = (P - jQ)/V_R. (Key point: Current from complex power)

2. What are the in-phase and quadrature drops?
   A. (XP + RQ)/V_R and (RP - XQ)/V_R
   B. (RP + XQ)/V_R and (XP - RQ)/V_R
   C. (RQ)/V_R and (XP)/V_R
   D. (P R)/V_R only
   Answer: B. Expanding (R + jX)(P - jQ)/V_R gives real part (RP + XQ)/V_R and imaginary part (XP - RQ)/V_R. (Key point: Voltage drops)

3. Why is |V_S| - |V_R| approximately X Q/V_R?
   A. Because P is zero
   B. Because X >> R and the quadrature part changes magnitude little
   C. Because R is large
   D. Because V_S = V_R
   Answer: B. In transmission lines X >> R, so RP is small and the quadrature drop adds little to magnitude. (Key point: Approximation)

4. What does Q = V_R(|V_S| - |V_R|)/X say?
   A. Q flows from lower to higher voltage
   B. Q flows from higher to lower voltage magnitude, regardless of angle
   C. Q flows with the angle
   D. Q is zero always
   Answer: B. A positive value (V_S > V_R) means Q goes sending to receiving; negative means reverse. (Key point: Direction of Q)

## Related in Power System Analysis I

- [Receiving end complex power and power circle](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-circle) — Power System Analysis I, chapter 6
- [Long line sending end voltage derivation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-sending-voltage-derivation) — Power System Analysis I, chapter 5
- [Receiving end power in ABCD terms and maximum power](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-abcd-max) — Power System Analysis I, chapter 6
- [Direction of active and reactive power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adirection-active-reactive-flow) — Power System Analysis I, chapter 6
- [Receiving and sending end powers from ABCD, 132/134 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-power-132kv-134kv) — Power System Analysis I, chapter 6
- [Sending and receiving power, 140∠10° and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-140kv-132kv-a099) — Power System Analysis I, chapter 6

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- [Justify higher voltage/more circuits for more power](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ajustify-higher-voltage-more-circuits) — Transmission and Distribution System Design, chapter 2
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