ELECTENG291 · Fundamentals of Electrical Engineering
AC Steady-State: Phasors & Impedance
Module 3 of University of Auckland ELECTENG 291 analyses circuits driven by sinusoids without solving differential equations, by working in the frequency domain. It covers sinusoids (amplitude, phase, average and RMS), the phasor representation, and complex impedance Z = R + jX, which turns Ohm's law into V = Z·I so that resistive-circuit techniques carry straight over into complex arithmetic. Phasor/impedance solving underpins the Module 3 assignment, the AC tutorials and the final exam.
What this chapter covers
- 01Sinusoid (cosine convention): x(t) = A cos(ωt + φ), with ω = 2πf = 2π/T
- 02Average value X = (1/T)∫_T x dt and RMS value; for a sinusoid of amplitude A_p, X_rms = A_p/√2
- 03Phasor (peak convention): v(t) = V_p cos(ωt + θ_v) ↔ V = V_p∠θ_v
- 04Frequency-domain Ohm's law V = Z·I (bold = phasors)
- 05Element impedances: Z_R = R, Z_L = jωL (X_L = ωL), Z_C = 1/(jωC) = −j/(ωC) (X_C = −1/(ωC))
- 06General impedance Z = R + jX Ω, with |Z| = √(R² + X²) and arg Z = tan⁻¹(X/R)
- 07Method: map the circuit to the phasor domain, solve with series/parallel and node/mesh in complex arithmetic, convert back to time
Series R-L impedance and the current phasor
- +1Inductor impedance: Z_L = jωL = j(1000)(0.040) = j40 Ω. The resistor is Z_R = 30 Ω.
- +1Total series impedance: Z = R + jX = 30 + j40 Ω (rectangular).
- +1Convert to polar: |Z| = √(30² + 40²) = √(900 + 1600) = √2500 = 50 Ω, and arg Z = tan⁻¹(40/30) = 53.13°, so Z = 50∠53.13° Ω.
- +1Current phasor by Ohm's law: I = V/Z = (100∠0°)/(50∠53.13°) = 2∠−53.13° A (peak). The current angle is negative, so the current lags the voltage by 53.13° — as expected for an inductive branch.
Key terms
- Sinusoid
- A signal of the form x(t) = A cos(ωt + φ), with amplitude A, angular frequency ω = 2πf = 2π/T and phase φ. The cosine convention is used throughout Module 3.
- RMS value
- The root-mean-square value X_rms = √[(1/T)∫_T x² dt], the effective DC-equivalent magnitude. For a sinusoid of peak amplitude A_p it is A_p/√2 — the quantity used in AC power.
- Phasor
- A complex number encoding a sinusoid's amplitude and phase at a fixed frequency: v(t) = V_p cos(ωt + θ_v) ↔ V = V_p∠θ_v (peak convention). Phasors turn calculus on sinusoids into algebra.
- Impedance (Z)
- The AC generalisation of resistance, Z = R + jX Ω, relating voltage and current phasors by V = Z·I. R is resistance, X is reactance; |Z| = √(R²+X²) and arg Z = tan⁻¹(X/R).
- Reactance (X)
- The imaginary part of impedance: inductive X_L = ωL (positive) and capacitive X_C = −1/(ωC) (negative). Its sign tells you whether the current lags (inductive) or leads (capacitive) the voltage.
- Element impedances
- The phasor-domain models of the basic elements: resistor Z_R = R, inductor Z_L = jωL, capacitor Z_C = 1/(jωC) = −j/(ωC). Obtained from the s-domain impedances by setting s = jω.
AC Steady-State: Phasors & Impedance FAQ
What is a phasor and why use one?
A phasor is a complex number V = V_p∠θ_v that captures a sinusoid's amplitude and phase at a fixed frequency, so v(t) = V_p cos(ωt + θ_v). Because differentiation and integration of sinusoids become simple multiplications by jω in the phasor domain, AC circuits reduce to algebra: you solve them with the same series/parallel and nodal/mesh techniques as resistive circuits, just in complex numbers, then convert the answer back to a time-domain sinusoid.
What is impedance and how do I combine impedances?
Impedance Z = R + jX is the AC version of resistance, defined so that V = Z·I for phasors. The element impedances are Z_R = R, Z_L = jωL and Z_C = 1/(jωC) = −j/(ωC). They combine exactly like resistances — impedances in series add, and in parallel add as reciprocals — but the arithmetic is complex, so track both magnitude and angle throughout.
What is the difference between inductive and capacitive reactance?
Inductive reactance X_L = ωL is positive and grows with frequency, and it makes the current lag the voltage (Z_L = jωL). Capacitive reactance X_C = −1/(ωC) is negative and shrinks with frequency, and it makes the current lead the voltage (Z_C = −j/(ωC)). The sign of the total reactance X in Z = R + jX therefore tells you whether a load is net inductive or net capacitive.
Can Sia help me with phasors and impedance in ELECTENG 291?
Yes, as a study aid. Sia can convert sinusoids to phasors, build element impedances, combine them, solve V = Z·I in complex arithmetic and convert answers back to the time domain, checking your rectangular↔polar conversions on the way. 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
Build fluency with complex arithmetic first — rectangular↔polar conversion, and multiply/divide phasors in polar (magnitudes multiply/divide, angles add/subtract) — because every AC solve rides on it. Memorise the three element impedances (Z_R = R, Z_L = jωL, Z_C = −j/(ωC)) since they are not all on the provided formula page, and burn in the capacitor's reciprocal-and-minus sign, the classic error. Then practise the standard workflow: map the circuit to the phasor domain, solve with the same series/parallel and nodal/mesh tools you already know, and convert back to a time-domain sinusoid. Always read arg Z to sanity-check whether the current should lead (capacitive) or lag (inductive). Confirm assessment details on Canvas.
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