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Electrical Circuits and Sensors · Module 5 of 10

AC Circuits and Phasors

Sinusoids are everywhere: mains power, vibration signals, rotating machinery. The phasor turns the calculus of a sine wave into the algebra of a complex number, and impedance lets Ohm's law work in the frequency domain.

01

Readiness check

This module moves into the frequency domain. Tick only what you can do closed-notes.

  • Recall the form A cos(ωt + φ) of a sinusoid.
  • Add and multiply complex numbers in rectangular and polar form.
  • Convert between a + jb and magnitude-angle.
  • Recall ω = 2πf.
  • Apply Ohm's law and series impedance addition.
0 or 1 weak itemsContinue with this module.
2 weak itemsRefresh complex arithmetic in Mathematics: Complex Numbers.
3 or more weak itemsRevisit capacitors and inductors in Module 4.
02

The core idea

A phasor represents a sinusoid by its amplitude and phase. Impedance generalises resistance to complex numbers, so that Ohm's law, V = IZ, solves any AC circuit with algebra instead of differential equations.

ZR = R, ZL = jωL, ZC = 1/(jωC)V = I ZZ = R + jX, |Z| = √(R2 + X2)

At a single frequency, every voltage and current in a linear circuit is a sinusoid of that same frequency, differing only in amplitude and phase. A phasor captures exactly those two numbers as a complex quantity, dropping the common ωt. In the phasor domain a resistor still has impedance R, but an inductor has jωL and a capacitor has 1/(jωC): both store energy and shift phase by 90 degrees, the inductor making current lag and the capacitor making it lead. Impedance Z = R + jX combines resistance and reactance, and Ohm's law becomes V = IZ with complex quantities. The magnitude of Z scales the amplitudes; its angle sets the phase between voltage and current.

The skill works when: you build each element's impedance at the given ω and combine them like resistors, in complex form.
The skill breaks down when: magnitudes are added without their angles, or reactance is treated as a real resistance.
The concept. Impedance is a complex number: resistance along the real axis, reactance along the imaginary. Its length scales the amplitude and its angle sets the phase between voltage and current.
03

The skills, taught in order

Five skills move from the sinusoid to the phasor, to impedance, and to solving a circuit in the frequency domain.

5.1 Sinusoidal sources

An AC source is described by v(t) = Vm cos(ωt + φ): an amplitude Vm, an angular frequency ω = 2πf, and a phase φ. Frequency is fixed by the source; the circuit changes only amplitude and phase, which is what makes the phasor method work.

5.2 The phasor

A phasor V = Vm∠φ encodes the amplitude and phase of a sinusoid as a complex number, omitting the common ejωt. Adding sinusoids of the same frequency becomes adding complex numbers, and differentiation becomes multiplication by jω.

5.3 Impedance of R, L, and C

In the phasor domain each element has an impedance: ZR = R is real; ZL = jωL grows with frequency and makes current lag; ZC = 1/(jωC) shrinks with frequency and makes current lead. Reactance X is the imaginary part, in ohms.

ElementImpedanceReactancePhase of current
ResistorR0in phase
InductorjωL+ωLlags by 90°
Capacitor1/(jωC)−1/(ωC)leads by 90°

The three element impedances. Inductive reactance is positive, capacitive reactance negative, and they partly cancel in series.

5.4 Series and parallel impedance

Impedances combine exactly like resistances, but as complex numbers: in series they add, in parallel their reciprocals add. The total impedance Z = R + jX has a magnitude that scales amplitude and an angle that gives the phase between voltage and current.

5.5 Solving an AC circuit

Convert the source to a phasor, replace each element by its impedance at ω, then apply Ohm's law and the divider rules in complex arithmetic. The phasor result, magnitude and angle, converts back to a time-domain sinusoid.

Engineering connection: the frequency response of a sensor filter, in Module 9, is the magnitude of an impedance ratio plotted against ω.

04

Worked example 1: impedance of a series RL circuit

A series circuit has R = 30 Ω and an inductive reactance XL = 40 Ω at the source frequency. A sinusoidal source of amplitude 50 V drives it. Find the impedance magnitude and angle, and the current amplitude and phase.

Figure 1. Resistance and inductive reactance form the legs of a right triangle whose hypotenuse is the impedance. The 30-40-50 triangle gives a phase angle of 53.1 degrees, by which the current lags.
  1. ProblemFind |Z|, the impedance angle, and the current for the RL circuit in Figure 1.
  2. Given / findR = 30 Ω, XL = 40 Ω, Vm = 50 V. Find |Z|, φ, and the current amplitude and phase.
  3. AssumptionsSingle-frequency steady state; the source phase is the reference (0°).
  4. ModelZ = R + jXL; magnitude and angle by Pythagoras and arctangent; I = V/Z.
  5. Equations|Z| = √(R2 + XL2)φ = arctan(XL/R)I = Vm/|Z|
  6. Solve|Z| = √(302 + 402) = √2500 = 50 Ω. φ = arctan(40/30) = 53.1°. I = 50/50 = 1 A, lagging the voltage by 53.1°.
  7. CheckThe 30-40-50 right triangle confirms |Z| = 50. The positive reactance makes φ positive, so current lags, as an inductive circuit requires.
  8. ConclusionThe current is 1 A in amplitude and lags the source by 53.1 degrees, fully describing the phasor V = IZ.
Result. |Z| = 50 Ω at 53.1°, I = 1 A lagging by 53.1°.
05

Worked example 2: a series RLC impedance

A series RLC branch has R = 20 Ω, XL = 60 Ω, and XC = 45 Ω at the source frequency. A 100 V amplitude source drives it. Find the net reactance, the impedance, and the current.

Figure 2. The inductive and capacitive reactances partly cancel, leaving a net inductive reactance of 15 ohms. The 20-15-25 triangle then sets the impedance and a 36.9 degree lag.
  1. ProblemFind the net reactance, |Z|, the angle, and the current for the RLC branch in Figure 2.
  2. Given / findR = 20 Ω, XL = 60 Ω, XC = 45 Ω, Vm = 100 V. Find X, |Z|, φ, I.
  3. AssumptionsSeries connection at a single frequency; source phase is the reference.
  4. ModelNet reactance X = XL − XC; Z = R + jX; I = V/|Z|.
  5. EquationsX = XL − XC|Z| = √(R2 + X2)I = Vm/|Z|
  6. SolveX = 60 − 45 = 15 Ω (net inductive). |Z| = √(202 + 152) = √625 = 25 Ω. φ = arctan(15/20) = 36.9°. I = 100/25 = 4 A, lagging by 36.9°.
  7. CheckThe 20-15-25 triangle (a scaled 4-3-5) confirms |Z| = 25. The reactance is positive, so the branch is net inductive and the current lags.
  8. ConclusionInductive and capacitive reactances subtract; only the net reactance enters the impedance, which here leaves a 4 A current lagging by 36.9 degrees.
Result. X = 15 Ω, |Z| = 25 Ω at 36.9°, I = 4 A lagging.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Adding magnitudes only|Z| taken as R + X"Did I combine them as complex numbers?"Use |Z| = √(R2 + X2), not R + X.
Sign of reactanceCapacitor treated as +jX"Which way does this element shift phase?"Inductive reactance is positive, capacitive negative.
Forgetting the phaseCurrent amplitude reported with no angle"What is the phase of the current?"A phasor needs both magnitude and angle.
Using R for L or CReactance left out of Z"Did I evaluate jωL and 1/jωC?"Build each element's impedance at the actual ω.
07

Practice ladder

Level 1 · Direct skill

Find the inductive reactance of a 50 mH inductor at ω = 400 rad/s.

Show answer

XL = ωL = 400 × 0.05 = 20 Ω.

Level 2 · Mixed concept

A series circuit has R = 6 Ω and XC = 8 Ω. Find the impedance magnitude and angle.

Show answer

|Z| = √(62 + 82) = 10 Ω. φ = arctan(−8/6) = −53.1°, so the current leads (capacitive).

Level 3 · Independent problem

A 120 V amplitude source drives R = 8 Ω in series with XL = 6 Ω. Find the current amplitude and phase.

Show answer

|Z| = √(82 + 62) = 10 Ω, φ = arctan(6/8) = 36.9°. I = 120/10 = 12 A, lagging by 36.9°.

Transfer task | Real engineering

A capacitor's reactance falls as frequency rises. Explain why a capacitor in series with a sensor signal blocks slow drift but passes fast variation.

What good work looks like

At low frequency XC = 1/(ωC) is large and blocks the signal; at high frequency it is small and passes it. That high-pass behaviour removes DC drift while letting the dynamic signal through.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I gave the impedance both a magnitude and an angle."
"Give me three RLC branches; I will find the net reactance of each."
"Solve this AC circuit." Building the impedances at ω yourself is the skill.
"Does the current lead or lag?" Reading the sign of reactance is the point.

Portfolio task

Take a series RL or RC circuit, compute its impedance and the current phase at one frequency, and confirm the phase against the expected lead or lag.

Must include: the element impedances, a complex Z, and a current with magnitude and phase.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What does a phasor represent?

The amplitude and phase of a sinusoid, as a complex number, at a fixed frequency.

2. Give the impedance of R, L, and C.

ZR = R, ZL = jωL, ZC = 1/(jωC).

3. How do you find |Z| from R and X?

|Z| = √(R2 + X2), with angle arctan(X/R).

4. Which element makes current lag, and which lead?

An inductor makes current lag; a capacitor makes it lead.

5. Write Ohm's law in the phasor domain.

V = IZ, with V, I, and Z all complex.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the two impedance triangles from a blank page.
+3 daysSolve three new RLC impedance problems.
+7 daysCarry impedance into AC power and resonance, Module 6.
+30 daysReuse phasors when reading a sensor frequency response.
10

Textbook mapping

This module follows Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition. Use these references to read further.

Topic in this moduleWhere to read more
Sinusoids and phasorsAlexander & Sadiku, Chapter 9
Impedance and admittanceAlexander & Sadiku, Chapter 9
Sinusoidal steady-state analysisAlexander & Sadiku, Chapter 10

Chapter numbers refer to the 4th edition. The phasor method is standard, so any recent edition will align closely.