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Phasors and Impedance

Replace calculus with complex arithmetic and every DC technique carries over intact — at the price of steady state, one frequency, and linearity.

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In steady state at one frequency, every signal in a linear circuit is a sinusoid differing only in amplitude and phase — so each can be written as a complex number, d/dt becomes multiplication by jω, and every DC technique carries over to AC unchanged.

The problem

Inductors and capacitors relate voltage and current through derivatives, so KVL around an RLC loop gives a second-order differential equation. Solvable, and thoroughly tedious to do for every circuit.

The observation that saves it

Differentiating a sinusoid scales it by ω and shifts it 90°; the frequency is unchanged. So in steady state every voltage and current in a linear circuit is a sinusoid at the source frequency, differing only in amplitude and phase — two numbers each, instead of a function of time.

A phasor encodes those two numbers as a complex number, dropping ωt because it is common to every signal. That step is legitimate only at a single frequency in steady state — the time dependence has not been solved, it has been factored out.

Impedance

ElementImpedancePhase relationship
ResistorZ = RVoltage and current in phase
InductorZ = jωLVoltage leads current by 90°
CapacitorZ = 1/jωCCurrent leads voltage by 90°

Every element now satisfies V = I·Z — Ohm's law with a complex constant. Impedances combine exactly like resistances: series adds, parallel takes reciprocals, dividers divide. Nodal analysis, mesh analysis and Thévenin all carry over unchanged.

The j is doing real work — it is a 90° rotation, which is precisely the phase shift a reactive element introduces. With exactly 90° between voltage and current, the average power is exactly zero: energy stored and returned rather than consumed.

Resistance and reactance

Z = R + jX separates the dissipative part from the storing part. Resistance converts energy to heat permanently; reactance stores it in a field and returns it twice per cycle. The same distinction reappears as real against reactive power.

Frequency dependence

Inductive reactance rises with frequency and capacitive reactance falls, so every RC and RL network is a filter whether intended or not. Where the two are equal they cancel exactly, leaving pure resistance — which is resonance.

The restrictions

  • Steady state only — the transient after a switch closes is invisible to the method.
  • One frequency only — a square wave needs each harmonic solved separately and summed, which is Fourier plus superposition.
  • Linear only — a diode produces harmonics that were not in the input.

Laplace generalises it: s = σ + jω covers transient and steady state together, and phasor analysis is the special case with σ = 0. Phasors handle the case an engineer meets most often, which is why they are worth the restriction.

The numbers you will be asked for

Phasor form

v(t) = V_m·cos(ωt + φ) → V = V_m∠φ

Impedances

Z_R = R · Z_L = jωL · Z_C = 1/jωC

Ohm's law, AC

V = I · Z

Rectangular and polar

Z = R + jX = |Z|∠θ · |Z| = √(R²+X²) · θ = arctan(X/R)

RMS

V_rms = V_peak / √2

for a sinusoid only

Watch it work

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Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

What makes a sinusoid special enough to justify phasor analysis?
Why is dropping ωt legitimate?
What does multiplying by j accomplish in an impedance?
A square wave drives an RC circuit. Can phasor analysis be used directly?

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4 still unanswered — the dots above jump straight to them.