ELECTENG291 · Fundamentals of Electrical Engineering
AC Power & Phasor Circuit Analysis
The final Module 3 chapter of University of Auckland ELECTENG 291 characterises power in AC circuits via the frequency domain: instantaneous, average (real), reactive and apparent power, the power factor and the power triangle, and analysing a complete AC circuit with phasors and impedances. Because none of the AC-power formulas are on the provided exam formula page, this material — and especially the sign of reactive power — is a high-value, high-risk part of the tests and the final exam.
What this chapter covers
- 01Instantaneous power p(t) = v(t)·i(t) [W]
- 02Average (real) power P = ½ V_p I_p cos(θ_v − θ_i) = V_rms I_rms cos(θ_v − θ_i) [W]
- 03Reactive power Q = ½ V_p I_p sin(θ_v − θ_i) = V_rms I_rms sin(θ_v − θ_i) [VAR]
- 04Apparent power |S| = V_rms I_rms = √(P² + Q²) [VA] and the power triangle (P, Q, |S|)
- 05Power factor pf = cos(θ_v − θ_i) = P/|S|; leading (capacitive, θ < 0) vs lagging (inductive, θ > 0)
- 06Complex power S = ½ V I* = V_rms (I_rms)* with P = Re[S], Q = Im[S], pf angle = arg[S] (S is NOT a phasor)
- 07Complex power via impedance: S = |I_rms|² Z, so P = |I_rms|² R and Q = |I_rms|² X
Power flow from source through a line to a load
- +1Total series impedance: Z = Z_line + Z_load = (1 + j4) + (39 + j26) = 40 + j30 Ω, so |Z| = √(40² + 30²) = √2500 = 50 Ω.
- +1RMS line current magnitude: I_rms = V_rms/|Z| = 250/50 = 5 A (its phase is −arg Z = −tan⁻¹(30/40) = −36.87°, but only |I_rms| = 5 A is needed for the powers).
- +1Power to the load using S = |I_rms|² Z_load: S_load = 25 × (39 + j26) = 975 + j650, so P_load = 975 W, Q_load = 650 VAR.
- +1Power absorbed by the line: S_line = |I_rms|² Z_line = 25 × (1 + j4) = 25 + j100, so P_line = 25 W, Q_line = 100 VAR.
- +1Power supplied by the source: S_src = |I_rms|² Z = 25 × (40 + j30) = 1000 + j750, so P_src = 1000 W, Q_src = 750 VAR.
- +1Conservation check: real power 975 + 25 = 1000 W ✓ and reactive power 650 + 100 = 750 VAR ✓ — load plus line equals source, as they must.
Key terms
- Instantaneous power
- p(t) = v(t)·i(t) in watts [W], the actual power at each moment. Its time average over a cycle is the real power P; the rest sloshes back and forth as reactive power.
- Average (real) power (P)
- The power actually converted to work or heat, P = ½ V_p I_p cos(θ_v − θ_i) = V_rms I_rms cos(θ_v − θ_i) [W]. It equals Re[S] and is dissipated only in the resistive part: P = |I_rms|² R.
- Reactive power (Q)
- The power exchanged with the reactive elements, Q = ½ V_p I_p sin(θ_v − θ_i) = V_rms I_rms sin(θ_v − θ_i) [VAR]. It equals Im[S]; Q > 0 for an inductive load and Q < 0 for a capacitive load.
- Apparent power (|S|)
- The product of RMS voltage and current, |S| = V_rms I_rms = √(P² + Q²) [VA]. It is the hypotenuse of the power triangle whose legs are P and Q.
- Power factor (pf)
- pf = cos(θ_v − θ_i) = P/|S|, the fraction of apparent power that is real. It is 'leading' for a capacitive load (θ < 0) and 'lagging' for an inductive load (θ > 0).
- Complex power (S)
- S = ½ V I* = V_rms (I_rms)* [VA], a single complex number with P = Re[S], Q = Im[S] and pf angle arg[S]. It is a constant, not a phasor, and can be found from impedance as S = |I_rms|² Z.
AC Power & Phasor Circuit Analysis FAQ
What is the difference between real, reactive and apparent power?
Real power P (watts) is the power actually dissipated or converted to work, P = V_rms I_rms cos θ. Reactive power Q (VAR) is the power that shuttles back and forth with inductors and capacitors without net consumption, Q = V_rms I_rms sin θ. Apparent power |S| (VA) is their combined magnitude, |S| = V_rms I_rms = √(P²+Q²) — the hypotenuse of the power triangle. The power factor cos θ = P/|S| ties them together.
How do I tell if a load is leading or lagging?
Look at the sign of the angle θ = θ_v − θ_i (equivalently arg Z or arg S). If θ > 0 the load is inductive with a lagging power factor and Q > 0; if θ < 0 the load is capacitive with a leading power factor and Q < 0. This sign is the single highest-risk step in AC power, so double-check whether you took θ_v − θ_i or its negative.
Why is complex power S = ½ V I* and not ½ V I?
Because using the conjugate of the current makes the angle of S equal to θ_v − θ_i, the power-factor angle, so that Re[S] gives the real power and Im[S] gives the reactive power with the correct signs. If you used ½ V I instead, the angle would be θ_v + θ_i and the physics would be wrong. Note S is a complex constant, not a phasor — it has no time-domain waveform.
Can Sia help me with AC power in ELECTENG 291?
Yes, as a study aid. Sia can compute P, Q, |S| and the power factor, build complex power from phasors or from S = |I_rms|² Z, trace power flow through a line to a load, and check the all-important sign of Q. It explains and drills you on fresh numbers; it does not do graded assessment for you, and University of Auckland academic-integrity rules apply — confirm what is permitted on Canvas.
Exam move
Treat this chapter as the payoff of Module 3 and drill the formulas cold, because none of them are on the provided exam formula page: P = ½V_pI_p cos θ = V_rms I_rms cos θ, Q with sin θ, |S| = V_rms I_rms = √(P²+Q²), and complex power S = ½ V I* = |I_rms|² Z. Fix the highest-risk step first — the sign of θ = θ_v − θ_i, which sets whether Q is positive (inductive, lagging) or negative (capacitive, leading). Keep RMS and peak separate so you never double-count the ½ factor. Practise the two routes to complex power (from phasors via ½ V I*, and from impedance via |I_rms|² Z) and use conservation of P and Q as a built-in check on multi-element power-flow problems. Put the whole AC-power block on your A4 sheet. Confirm assessment details on Canvas.
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