---
title: "Direction of active and reactive power flow"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:direction-active-reactive-flow"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/direction-active-reactive-flow
site: BCT Engineering
---

# Direction of active and reactive power flow

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2069 Bhadra, 6 marks
- Video: [Does AC power changes its direction or not? | Flow of Active and Reactive Power | Power system](https://www.youtube.com/watch?v=ophotmZogVc) — NO MORE ratta maar
- 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/direction-active-reactive-flow
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

"Direction of active power flow is from leading phase angle to lagging phase angle whereas direction of reactive power flow is from higher voltage magnitude to lower voltage magnitude." Verify the above statement with mathematical explanation and diagrams.

## Short answer

Active power flows from the bus with leading phase angle to the lagging one, as P ∝ sinδ. Reactive power flows from higher voltage magnitude to lower, since Q ∝ (V₁ − V₂) for small angle differences.

## Answer

Consider two buses joined by a short line of reactance $X$ (resistance neglected, as $X \gg R$). Bus 1 voltage $V_1\angle\delta_1$, bus 2 voltage $V_2\angle\delta_2$, and $\delta = \delta_1 - \delta_2$.

```text
  V1/_d1        jX         V2/_d2
   o-----------/\/\/----------o
   |      P, Q  -->           |
  Bus 1                     Bus 2
```

#### Derivation
$$
I = \frac{V_1\angle\delta_1 - V_2\angle\delta_2}{jX}
$$
Power received at bus 2:
$$
S_2 = V_2\angle\delta_2\,I^* = \frac{V_1V_2\sin\delta}{X} + j\,\frac{V_1V_2\cos\delta - V_2^2}{X}
$$
So
$$
P = \frac{V_1V_2}{X}\sin\delta,\qquad Q = \frac{V_2}{X}\left(V_1\cos\delta - V_2\right)
$$

#### Active power: from leading to lagging angle
- $P$ depends on $\sin\delta$. If $\delta_1 > \delta_2$ (bus 1 leads), $\sin\delta > 0$ and $P > 0$: power flows from 1 to 2.
- If bus 2 leads, $P < 0$: power flows from 2 to 1.
- $P$ flows even if $V_1 = V_2$, as long as there is an angle difference.

#### Reactive power: from higher to lower magnitude
In practice $\delta$ is small, so $\cos\delta \approx 1$:
$$
Q \approx \frac{V_2(V_1 - V_2)}{X}
$$
- If $|V_1| > |V_2|$, $Q > 0$: reactive power flows from 1 to 2.
- If $|V_1| < |V_2|$, $Q$ reverses.
- It hardly depends on $\delta$.

#### Phasor illustration
```text
Case A: P flow           Case B: Q flow
  V1 (leads)               V1 (longer)
   /                       ------------>
  / delta                  V2 (shorter, same angle)
 O-------> V2              ------->
 P: 1 -> 2                 Q: 1 -> 2
```

#### Example
$X = 10\ \Omega$, $V_1 = 11\angle5^\circ$ kV, $V_2 = 10.5\angle0^\circ$ kV:
$P = \frac{11\times10.5}{10}\sin5^\circ = 1.01$ MW (1 → 2), $Q = \frac{10.5}{10}(11\cos5^\circ - 10.5) = 0.48$ MVAr (1 → 2).
If instead $V_2 = 11.5\angle0^\circ$ kV, $P$ still flows 1 → 2, but $Q = \frac{11.5}{10}(11\cos5^\circ - 11.5) = -0.62$ MVAr, i.e. Q flows 2 → 1.

Hence the statement is verified: **P follows the phase angle, Q follows the voltage magnitude.**

## Check yourself

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

1. For a lossless line of reactance X, what is the power P from bus 1 to bus 2?
   A. P = V1 V2 cos(delta)/X
   B. P = V2 (V1 - V2)/X
   C. P = V1 V2 sin(delta)/X
   D. P = (V1 - V2)/X
   Answer: C. Taking the real part of V2 I* gives P = V1V2 sin(delta)/X. (Key point: Active power expression)

2. Bus 1 and bus 2 have equal voltage magnitudes, and bus 1 leads by an angle. What happens?
   A. No power flows because magnitudes are equal
   B. P flows from 2 to 1
   C. Only reactive power flows
   D. P flows from 1 to 2
   Answer: D. sin(delta) > 0 so P > 0; P flows from the leading bus to the lagging bus even with equal magnitudes. (Key point: Direction of P)

3. For small delta, Q is approximately which expression?
   A. V2 (V1 - V2)/X
   B. V1 V2 delta/X
   C. V1 V2 sin(delta)/X
   D. V2^2/X
   Answer: A. With cos(delta) near 1, Q = V2(V1 cos delta - V2)/X is about V2(V1 - V2)/X, set by the magnitude difference. (Key point: Reactive power for small angle)

4. In the example (V1 = 11 kV at 5 degrees), V2 is raised from 10.5 kV to 11.5 kV. What happens?
   A. P reverses and Q stays
   B. Both P and Q reverse
   C. P still flows 1 to 2 but Q reverses to flow 2 to 1
   D. Neither changes
   Answer: C. The angle is unchanged so P still flows 1 to 2, but V2 > V1 cos(delta) makes Q negative (-0.63 MVAr). (Key point: Example reversal of Q)

## Related in Power System Analysis I

- [Vs from P+jQ and reactive power flow direction](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Avs-from-pq-reactive-flow) — Power System Analysis I, chapter 6
- [Reactive power flow between machines via j10 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-reactive-j10) — Power System Analysis I, chapter 3
- [Reactive power proportional to voltage drop](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactive-power-voltage-drop) — Power System Analysis I, chapter 6
- [Active power between machines, 200∠0° and 200∠15° V](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-200v-15deg) — Power System Analysis I, chapter 3
- [Two sources via 0 - j5 Ω: power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-0-j5) — Power System Analysis I, chapter 3
- [Sending end active power from ABCD, 132/130.5 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asending-power-132kv-130-5kv) — Power System Analysis I, chapter 6

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