AC Fundamentals & Phasors: Sine Waves, RMS, Phase & Phasor Math
A complete beginner-to-advanced path through alternating current. Start with the shape and language of a sine wave, move to RMS values and the complex-number toolkit (rectangular, polar, Euler), and finish by doing real AC circuit analysis with phasors — impedance, phase angle, and power. No prior AC experience assumed.
Interactive40 lessons3 levels
Lesson 1 / 40
Live AC playground — a phasor tracing a sine wave
The arrow on the left is a phasor spinning at the signal's frequency; the curve on the right is the same signal "unrolled" in time. Drag the controls and watch amplitude, frequency, phase and lead/lag change. This one picture is the heart of the whole module.
ω = 6.28 rad/sT = 1.00 sWave 2 leads/lags by 90°
Level 1 · Beginner
Alternating current is the current that the power grid actually delivers, and it changes direction many times every second. Because its value is never still, we need a small vocabulary — amplitude, frequency, period, phase — to describe it precisely, and a couple of clever tricks (RMS values and phasors) to calculate with it. This section builds that vocabulary from nothing, one idea at a time.
The single most useful picture is the one in the playground above: a rotating arrow that projects out a sine wave. Keep it in mind as you go, because every later lesson is just a more careful version of that same idea. By the end of this level you'll be able to look at any sine wave and state its size, speed, and timing.
1 DC vs AC: Why Alternating Current?
Beginner
Direct current (DC) flows one way and, in a steady circuit, stays at a constant value — think of a battery. Alternating current (AC) reverses direction periodically and is usually sinusoidal, swinging positive and negative. AC won out for the grid because its voltage can be raised and lowered cheaply with transformers, and high voltage transmits power over long distances with far less loss in the wires.
The price of AC is that its voltage and current are always changing, so the simple "V = IR" picture from DC needs to be generalized. That's why this module introduces frequency, phase, RMS and phasors — the tools that let us treat a changing quantity with the same algebraic calm we use for a steady one.
2 Anatomy of a Sine Wave
Beginner
A sine wave oscillates smoothly and symmetrically between a maximum (+peak) and a minimum (−peak). One full up-and-down swing is called a cycle. The wave repeats itself forever (in an ideal source), so if you understand one cycle you understand all of them — which is exactly why the playground's spinning arrow and the wave to its right are the same thing viewed two ways.
Four numbers completely specify a sinusoid: its amplitude (how tall), its frequency (how fast it repeats), its phase (where it starts), and its shape (sine). The next few lessons unpack each of these, and by lesson 9 we'll write them all down as a single equation.
3 Amplitude: The Peak Value
Beginner
The amplitude (or peak value) is the maximum value the wave reaches above zero, written as Vmax for voltage or Imax for current. A wave with amplitude 10 V swings between +10 V and −10 V. It's the single most important "size" number because it tells you the extremes a component will actually see.
Peak values matter for ratings: insulation must withstand the peak voltage, a wire must survive the peak current, and a meter's overload limit is a peak concern. Don't confuse peak with the RMS value you'll meet soon — they differ by a factor of √2 for a sine wave.
A voltage wave oscillates between +240 V and −240 V. Its peak value Vmax is…
Peak = the maximum value above zero = 240 V. (480 V would be the peak-to-peak value, from +240 to −240.)
4 Peak-to-Peak and Other Ways to Measure Size
Beginner
Peak-to-peak (Vpp) is the total swing from the positive peak to the negative peak, so for a symmetric wave Vpp = 2 × Vmax. It's the number an oscilloscope often makes you read off directly (top of the trace minus bottom). A 240 V peak wave therefore has a 480 V peak-to-peak value.
Notice we now have two "sizes" for the same wave — peak and peak-to-peak — and there are more to come (RMS, average). The reason so many exist is that each answers a different practical question: peak for stress and ratings, peak-to-peak for measuring on a scope, RMS for power and heating. We'll learn to convert between them in the Intermediate section.
5 Period: One Complete Cycle
Beginner
The period (T) refers to the duration needed to complete one full cycle, and it is measured in seconds. For example, if a wave repeats every 0.02 seconds, its period is 0.02 seconds, or 20 milliseconds. A longer period indicates a slower, more relaxed wave, while a shorter period signifies a faster wave.
In simpler terms, the period tells us the length of one complete oscillation. Imagine being in a playground; the period would be the time it takes for a phasor arrow to go all the way around. Typically, when measuring the period using an oscilloscope, you time the interval between two similar points, like from one positive peak to the next one.
6 Frequency: Cycles per Second
Beginner
Frequency (f) is the number of complete cycles per second, measured in hertz (Hz), where 1 Hz = 1 cycle/second. High frequency means many cycles each second. The unit honors Heinrich Hertz, who first measured electromagnetic waves.
Frequency ranges are all around us: mains power is 50 or 60 Hz, the human hearing range is about 20 Hz to 20 kHz, radio and Wi-Fi operate in the MHz to GHz range. The larger prefixes (k, M, G) are just 10³, 10⁶, 10⁹ — so 2 kHz = 2000 Hz.
A signal completes 120 cycles in 2 seconds. Its frequency is…
f = cycles ÷ time = 120 ÷ 2 = 60 Hz. Frequency is cycles per second, so divide the count by the seconds.
7 Frequency & Period: f = 1/T
Beginner
Frequency and period are reciprocals — one tells you cycles per second, the other seconds per cycle. So f = 1/T and T = 1/f. A 60 Hz signal has a period of 1/60 ≈ 16.67 ms; a 1 kHz signal repeats every 1 ms.
f = 1 / T | T = 1 / f
Use the calculator to flip between them and see the angular frequency (the next lesson) at the same time. Remember: a high frequency is a short period, and vice versa.
Try it — frequency, period & angular frequency
Edit either field — the other updates automatically.
8 Angular Frequency ω
Beginner
Angular frequency (ω, "omega") measures how fast the phase angle advances, in radians per second. Because a sine is a function of angle, and one full cycle corresponds to 2π radians, ω is just 2π times the ordinary frequency: ω = 2πf = 2π/T. For 60 Hz, ω = 2π × 60 ≈ 377 rad/s.
ω = 2πf = 2π / T (radians per second)
You'll see ω inside the sinusoid's equation and everywhere phasors appear, because a rotating phasor's spin rate is exactly ω. Think of it as frequency expressed in "radians per second" instead of "cycles per second."
9 The General Sinusoidal Equation
Beginner
Every sinusoid can be written as v(t) = Vmax sin(ωt + φ). Read it piece by piece: Vmax sets the height, ω sets how fast it oscillates, and φ (phi) sets where in its cycle the wave is at t = 0. The argument of the sine, ωt + φ, is the phase angle at time t.
v(t) = Vmax · sin(ωt + φ)
This one line is the master description. Everything in this module — RMS, phasors, impedance — is built by manipulating this equation or a version of it. Learn to read it fluently and the rest of AC analysis stops feeling like magic.
10 Phase and the Phase Angle φ
Beginner
The phase angle φ shifts the whole wave left or right along the time axis. With φ = 0, the sine starts at zero and rising. A positive φ shifts the wave to the left (it "starts ahead"), a negative φ shifts it right. Phase is measured in degrees (0–360) or radians (0–2π); 360° = 2π rad = one full cycle.
Phase answers "where in its cycle is this wave right now?" Two identical-frequency waves can have the same height and speed yet be perfectly out of step — that difference is phase. It's the property that makes lead/lag and, later, the whole of AC circuit behavior possible.
11 Phase Difference Between Two Waves
Beginner
When two waves have the same frequency, their phase difference Δφ is constant and fully describes their timing relationship. If they peak at the same instant they are in phase (Δφ = 0°); if one peaks exactly when the other troughs they are out of phase or in antiphase (Δφ = 180°); and a 90° difference is called quadrature.
Common phase relationships
Phase difference
Name
What it looks like
0°
In phase
Peaks and troughs line up exactly
90°
Quadrature
One peaks as the other crosses zero
180°
Antiphase / out of phase
One's peak is the other's trough
Turn on the second wave in the playground and set its phase difference to 0°, 90° and 180° to see each relationship directly.
12 Leading and Lagging
Beginner
If wave A reaches its peak before wave B, we say A leads B (and B lags A) by the phase difference. "Lead" means "gets there first." The amount is the same Δφ either way — only the labels flip depending on which wave you're describing.
Lead/lag is a relative, time-based statement and it's central to AC: in a resistor voltage and current are in phase; in an inductor current lags voltage; in a capacitor current leads voltage. Enable the second wave in the playground and drag its phase to see which one leads as you change Δφ.
Wave A peaks at t = 0 ms and wave B peaks at t = 2 ms (same frequency, T = 8 ms). Which is true?
A peaks first, so A leads. The offset is 2 ms = T/4 = 360°/4 = 90°. So A leads B by 90° (B lags A by 90°).
13 Other Waveforms and Why Sine Is Special
Beginner
Not everything alternates as a sine. Square waves jump between two levels, triangle waves ramp linearly up and down, and sawtooth waves rise then snap back. Each has the same ideas of period and frequency, but different amplitude statistics and different harmonic content.
Common waveform characteristics
Waveform
Shape
RMS (for amplitude A)
Typical use
Sine
Smooth oscillation
A/√2 = 0.707 A
Power, RF, audio
Square
Two-level jumps
A
Digital logic, clocks
Triangle
Linear ramps
A/√3 = 0.577 A
Test signals, ramps
Sawtooth
Rise + snap-back
A/√3 = 0.577 A
Synth, scanning, sweep
The sine is special because it's the natural output of resonant circuits and transformers, it contains a single frequency (no harmonics), and it's the building block (via Fourier's theorem) of every other periodic waveform. That's why AC power analysis is built on the sine.
14 AC Terminology & Key Terms
Beginner
Here's the Beginner vocabulary in one place. Internalize these and the Intermediate and Advanced sections will read like a continuation rather than a new subject.
Key AC terms
Term
Symbol
Meaning
Unit
Amplitude / peak
Vmax, Imax
Maximum value above zero
V, A
Peak-to-peak
Vpp
Swing from +peak to −peak (= 2·peak)
V, A
Period
T
Time for one cycle
s
Frequency
f
Cycles per second (f = 1/T)
Hz
Angular frequency
ω
Phase rate (ω = 2πf)
rad/s
Phase angle
φ
Time-axis shift of the wave
°, rad
Phase difference
Δφ
Timing offset between two same-f waves
°, rad
Cycle
—
One complete oscillation (360°)
—
Quick self-check: can you name the amplitude, period, frequency and phase of a wave from its equation v(t) = Vmax sin(ωt + φ)? If yes, you're ready for RMS.
Level 2 · Intermediate
Now that you understand how to describe a sine wave, this section addresses the practical question of "how big is it, really?" using RMS values. It then provides you with the mathematical tools that make AC analysis easy, including complex numbers, both rectangular and polar forms, Euler's identity, and the concept of a phasor. The shift from recognizing "a changing value" to seeing "a fixed number with an angle" is the most crucial change in all of AC.
The benefit is that adding, multiplying, and dividing sinusoids becomes just basic arithmetic with phasors, without needing calculus. By the end of this section, you will be able to transform a time-domain equation into a phasor and back again. You'll also grasp why the length of the phasor represents an RMS value, while its angle indicates the phase.
15 RMS: Why We Use It
Intermediate
An AC voltage is never a single number, so when we say "the mains is 240 V" we need a fair, single figure. The RMS (Root Mean Square) value is defined as "the DC voltage that would produce the same heating in a resistor." Because heating power is proportional to V², the correct way to average a squaring-then-rooting process is: square the waveform, take the average (mean), then take the square root — hence "root-mean-square."
RMS is the value that meters read, that appears on appliance ratings, and that you use in P = VI for AC power. It's not the peak and not the average — it's the effective, power-equivalent size. Getting this straight now saves confusion later, because peak and RMS differ by √2 for a sine.
16 RMS of a Sine Wave
Intermediate
For a pure sine wave the RMS value is the peak divided by √2: Vrms = Vmax/√2 = 0.707 · Vmax. Equivalently Vmax = √2 · Vrms = 1.414 · Vrms. The factor √2 ≈ 1.414 comes from averaging sin² over a cycle, which equals ½.
Vrms = Vmax / √2 = 0.707 Vmax Irms = Imax / √2
Real-world anchor: a 240 V RMS supply has a peak of 240 × √2 ≈ 340 V. So the "240 V" on the wall is the RMS; the insulation is really seeing ±340 V. (US 120 V RMS ≈ 170 V peak.)
Worked example — mains peak
Given: Vrms = 240 V (sinusoidal).
Peak: Vmax = 240 × 1.414 = 339.4 V.
Peak-to-peak: Vpp = 2 × 339.4 = 678.8 V.
A sine wave has a peak of 340 V. Its RMS value is about…
V_rms = V_max/√2 = 340/1.414 ≈ 240 V. This is exactly the European mains rating — 340 V peak, 240 V RMS.
17 Peak, Peak-to-Peak & RMS — Converting
Intermediate
For a sine wave the three sizes are locked together by two factors: peak-to-peak is 2× peak, and RMS is peak/√2 (≈0.707× peak). Two handy ratios follow: the form factor (RMS ÷ peak) = 0.707, and the peak factor (peak ÷ RMS) = 1.414.
Amplitude measures for a sine wave
Quantity
In terms of peak (Vmax)
Example (Vmax = 340 V)
Peak
1.000 × Vmax
340.0 V
Peak-to-peak
2.000 × Vmax
680.0 V
RMS
0.707 × Vmax
240.4 V
Use the converter: enter any one of the three and pick which it is, and it computes the rest with a visual comparison.
Try it — peak / RMS / peak-to-peak converter
18 RMS of Non-Sinusoidal Waves
Intermediate
The 0.707 factor is only true for a pure sine. The general definition — square, average, root — still applies to any waveform, but the number you get is different. A symmetric square wave that swings ±A has RMS = A (it's at full amplitude all the time), and a triangle wave has RMS = A/√3 ≈ 0.577 A.
This matters for real electronics: switching power supplies, PWM motor drives and audio amplifiers produce non-sine currents, and their RMS (hence their heating and their meter reading) is not 0.707× the peak. Always check the waveform before assuming the sine factor.
19 Instantaneous, Peak & RMS — When to Use Which
Intermediate
The instantaneous value v(t) is the voltage at one specific moment — it's what the sine equation gives you and what you integrate to find energy. The peak is the maximum of those instantaneous values, used for insulation, breakdown and overload ratings. The RMS is the effective size, used for power, heating and what meters display.
A good rule of thumb: analyze with instantaneous values, rate components with peak values, and compute power and read meters with RMS. Confusing the three is one of the most common AC mistakes, so keep the "what question am I answering" check habit.
20 Sinusoids as Equations
Intermediate
Writing a real sinusoid as an equation just means plugging numbers into v(t) = Vmax sin(ωt + φ). The trap most learners hit: datasheets and problems usually give the RMS value, but the equation needs the peak. So first convert Vrms → Vmax = √2·Vrms, then set ω = 2πf, then write it.
Worked Write v(t) for a 240 V RMS, 60 Hz source that leads by 30°.
Vmax = 240 × √2 = 339.4 V. ω = 2π × 60 = 377 rad/s. φ = +30°. So v(t) = 339.4 sin(377t + 30°) V.
Note the equation uses the peak (339.4 V), not the 240 V RMS, while the phase angle φ is used directly.
The same applies to current: i(t) = Imax sin(ωt + θ). Once you can write both v(t) and i(t), you can compare their phases — which is the setup for the phasor method.
21 Phase Relationships in Real Circuits
Intermediate
In a purely resistive load, voltage and current rise and fall together — they are in phase, and all the power is real. In a reactive load (inductor or capacitor) the current and voltage get out of step: an inductor makes current lag voltage, a capacitor makes current lead voltage. The angle between them is the circuit's phase angle θ.
This is why phase is not just a curiosity: the phase angle between v and i determines how much of the apparent power is "useful" real power versus "sloshing back and forth" reactive power. We'll quantify that in the Advanced section, but the idea starts here — and the playground's lead/lag control is a perfect way to feel it.
22 Complex Numbers: a + jb
Intermediate
A complex number has a real part and an imaginary part, written a + jb. Engineers use j (not the mathematicians' i) for the imaginary unit so it doesn't collide with current i, where j = √(−1) and j² = −1. A complex number is most usefully thought of as a point, or an arrow, on a 2-D plane: a is the horizontal coordinate, b the vertical one.
Why introduce this for AC? Because a sinusoid's two essential features — its magnitude and its phase — map perfectly onto a 2-D arrow. Once voltage and current become arrows (phasors), circuit math becomes geometry: adding arrows, rotating them, scaling them. Complex numbers are simply the algebra for that geometry.
23 Rectangular vs Polar Forms
Intermediate
The same complex number can be written two ways. Rectangular (Cartesian) form uses the coordinates: a + jb. Polar form uses the arrow's length and angle: M ∠ θ, where M is the magnitude and θ the angle from the positive real axis.
Rectangular vs polar
Rectangular
Polar
Form
a + jb
M ∠ θ
Best for
Addition & subtraction
Multiplication & division
Example
3 + j4
5 ∠ 53.13°
Rule of thumb: to add or subtract phasors, work in rectangular; to multiply or divide, work in polar. The converter in the next lesson shows both forms of the same number updating live.
Which form is most convenient for ADDING two phasors?
Addition is component-wise (add reals, add imaginaries), so rectangular is easiest. Polar is easiest for multiplication/division (scale magnitudes, add/subtract angles).
24 Converting Between Rectangular & Polar
Intermediate
From rectangular to polar: M = √(a² + b²) and θ = atan2(b, a). From polar to rectangular: a = M cos θ and b = M sin θ. The atan2 function (not plain arctan) is important because it picks the correct quadrant from the signs of a and b.
M = √(a² + b²) θ = atan2(b, a) | a = M·cos θ b = M·sin θ
Example: 3 + j4 → M = √(9+16) = 5, θ = atan2(4,3) = 53.13°, so 3 + j4 = 5 ∠ 53.13°. Try the converter — edit either side and watch the other (and the vector) update.
Try it — rectangular ↔ polar converter
25 Euler's Identity
Intermediate
Euler's formula ties rotation to complex numbers: e^(jθ) = cos θ + j sin θ. The left side is a point on the unit circle at angle θ; the right side is its rectangular coordinates. Multiply by a magnitude M and you get a phasor: M·e^(jθ) = M(cos θ + j sin θ) = M ∠ θ.
e^(jθ) = cos θ + j·sin θ ⇒ M∠θ = M·e^(jθ)
This is the bridge that lets a rotating phasor (a time-varying picture) be represented as a fixed complex number (a static picture). The rotation is "absorbed" into the e^(jωt) factor, which cancels out in linear circuit equations — that's the entire reason the phasor method works.
26 Phasors: Freezing the Sine Wave
Intermediate
A phasor is a rotating vector of fixed length that represents a sinusoid. As it spins at angular speed ω, its vertical projection is the sine wave. Instead of tracking the rotation in time, we freeze the phasor (conventionally at t = 0) and do algebra on the still vector — the "ωt" part is understood and dropped.
The arrow in the playground is a live phasor. When you pause it, you're looking at the frozen phasor that all the algebra in this and the next level operates on. The length is the (RMS) magnitude and the angle is the phase — exactly the two numbers that define the wave.
27 Phasor Form of a Sinusoid
Intermediate
The standard convention: a sinusoid v(t) = Vmax sin(ωt + φ) is represented by the phasor V = Vrms ∠ φ. Note the two conventions baked in — the magnitude is the RMS value (not the peak), and the angle is the phase φ. This keeps power calculations consistent, since power uses RMS.
v(t) = Vmax sin(ωt + φ) ⇔ phasor V = Vrms ∠ φ
Worked Convert 339.4 sin(377t + 30°) V to phasor form.
Vrms = 339.4 / √2 = 240 V. Phase φ = 30°. So the phasor is V = 240 ∠ 30° V.
The 377 rad/s (ω) is dropped — all phasors in a circuit share the same ω, so it's implied.
Going the other way, V = 240 ∠ 30° V means v(t) = (240√2) sin(ωt + 30°) = 339.4 sin(ωt + 30°) V, with ω supplied by the problem. Master this two-way conversion and phasor analysis opens up.
Level 3 · Advanced
With phasors in hand, this level is where AC analysis becomes mechanical and powerful. You'll add, subtract, multiply and divide phasors, meet the j operator that encodes 90° rotation, and bring in impedance — resistance plus reactance — for capacitors and inductors. Then you'll analyze real series and parallel AC circuits, find phase angles, and close with the full power picture: real, reactive and apparent power, power factor, and complex power.
The through-line is that every element (R, L, C) becomes a complex impedance, every circuit becomes an algebra problem in phasors, and every power question resolves into the power triangle. The final lesson puts the entire toolkit together on one complete circuit so you can see the whole method at a glance.
28 Adding & Subtracting Phasors
Advanced
To add phasors, convert each to rectangular (a + jb), add the real parts and the imaginary parts separately, then convert the result back to polar. You can't just add the magnitudes or the angles — phasors are vectors, so they add head-to-tail geometrically.
(a₁ + jb₁) + (a₂ + jb₂) = (a₁+a₂) + j(b₁+b₂)
Example: 5∠0° + 5∠90° = (5 + j0) + (0 + j5) = 5 + j5 = 7.07 ∠ 45°. The two equal perpendicular phasors add to a longer one at 45°. Use the interactive adder to see the vector sum change as you move the two inputs.
Try it — add two phasors
29 Multiplying & Dividing Phasors
Advanced
Multiplication and division are where polar form shines. To multiply, multiply the magnitudes and add the angles. To divide, divide the magnitudes and subtract the angles. This is a direct consequence of Euler's formula: e^(jθ₁)·e^(jθ₂) = e^(j(θ₁+θ₂)).
Example: (10∠30°)(2∠−10°) = 20 ∠ 20°. And (10∠30°) ÷ (2∠−10°) = 5 ∠ 40°. This is exactly how Ohm's law works for AC: I = V/Z is a division of two phasors, giving a current whose magnitude and angle fall out directly.
30 The j Operator & the 90° Rotation
Advanced
Remember j = 1∠90°. So multiplying any phasor by j rotates it 90° counter-clockwise without changing its size. And since j² = (1∠90°)² = 1∠180° = −1, multiplying by j twice gives a 180° turn (a sign flip). The powers of j cycle: j, −1, −j, 1, j, …
j = 1∠90° j² = −1 j³ = −j j⁴ = 1
This is the algebraic form of rotation, and it's the seed of reactance: an inductor's voltage leads its current by 90°, which is exactly "multiply the current phasor by j" (times a magnitude). Hold that thought — the next two lessons make it concrete.
Multiplying a phasor by j rotates it…
j = 1∠90°, so it's a pure 90° CCW rotation (magnitude 1). j² = −1 (180°), j³ = −j (270° CCW / 90° CW), j⁴ = 1.
31 Impedance: Resistance Meets Reactance
Advanced
Impedance Z is the AC generalization of resistance — it measures how much a component opposes sinusoidal current, and it carries a phase angle. Like a complex number, Z = R + jX, where R is the resistive part (in phase with current) and X is the reactive part (90° out of phase). Its magnitude |Z| = √(R² + X²) is the "effective opposition" in ohms.
Ohm's law becomes V = I·Z in phasor form. A resistor has Z = R (purely real, angle 0°). Inductors and capacitors have purely imaginary impedance — and that imaginary part is the reactance X, which the next lesson defines in terms of frequency.
A capacitor's impedance is…
Z_C = 1/(jωC) = −j/(ωC) = −jX_C, a negative imaginary number (current leads voltage). Inductors are +jX_L.
32 Capacitive & Inductive Reactance
Advanced
Both reactances are in ohms and both depend on frequency. Inductive reactance is XL = 2πfL = ωL (grows with frequency — an inductor blocks high frequencies). Capacitive reactance is XC = 1/(2πfC) = 1/(ωC) (shrinks with frequency — a capacitor passes high frequencies). Their signs encode the phase: XL is +j (current lags), XC is −j (current leads).
XL = ωL = 2πfL | XC = 1/(ωC) = 1/(2πfC)
Worked example — inductive reactance
Given: L = 10 mH = 0.01 H, f = 60 Hz.
Compute: XL = 2π × 60 × 0.01 = 3.77 Ω.
Impedance: Z = j3.77 Ω = 3.77 ∠ 90° Ω (current lags V by 90°).
At DC (f = 0) an inductor is a short (XL = 0) and a capacitor is an open (XC → ∞) — the reason "capacitor blocks DC, inductor passes DC."
33 Impedance in Series: Z = R + jX
Advanced
Impedances in series simply add, just like resistors: Z = R + j(XL − XC). The net reactance is X = XL − XC (inductive minus capacitive). From there, |Z| = √(R² + X²) and the phase angle θ = atan2(X, R). A positive X (net inductive) means current lags; a negative X (net capacitive) means current leads.
Z = R + j(XL − XC) | |Z| = √(R² + X²) | θ = atan2(X, R)
Use the calculator below: enter R, f, L and C and it computes XL, XC, the net X, the impedance Z, its magnitude and its angle, with a live impedance triangle.
Try it — series impedance calculator
Worked Series R = 4 Ω with XL = 3 Ω. Find |Z| and θ.
Z = 4 + j3. |Z| = √(16+9) = 5 Ω. θ = atan2(3,4) = 36.87°. Inductive (X>0), so current lags voltage by 36.87°.
This is the classic 3-4-5 impedance triangle; cos θ = R/|Z| = 0.8 is the power factor (see lesson 38).
34 Impedance in Parallel
Advanced
Parallel impedances combine with the reciprocal rule, exactly like parallel resistors but with complex numbers: 1/Z = 1/Z₁ + 1/Z₂ + …, so for two, Z = (Z₁·Z₂)/(Z₁+Z₂). In practice you usually work in admittance Y = 1/Z (in siemens), because admittances in parallel just add: Y = Y₁ + Y₂.
Example: a 10 Ω resistor in parallel with a −j10 Ω capacitor. In admittance: Y = 1/10 + 1/(−j10) = 0.1 + j0.1 S. Then Z = 1/Y = 1/(0.1 + j0.1) = 1/(0.1414∠45°) = 7.07 ∠ −45° Ω. Parallel analysis is where keeping rectangular vs polar straight pays off.
35 Series RL & RC Circuits and Phase Angle
Advanced
The phase angle of the total impedance is the phase angle between the source voltage and the circuit current. For a series RL circuit, X is positive so θ > 0 and current lags voltage. For a series RC circuit, X is negative so θ < 0 and current leads voltage. The bigger the reactance relative to R, the closer θ gets to 90°.
Worked Series R = 5 Ω, XC = 5 Ω (capacitive). Find θ and the lead/lag.
Z = 5 + j(0 − 5) = 5 − j5. |Z| = √(25+25) = 7.07 Ω. θ = atan2(−5,5) = −45°. Negative θ → current leads voltage by 45°.
The negative sign is the capacitive fingerprint: current leads. Magnitude of the angle (45°) comes from equal R and X.
This is the whole phase story for first-order series circuits: compute Z, read θ, and the sign tells you lead vs lag. The playground's lead/lag control is a nice way to visualize the resulting current-vs-voltage timing.
36 Phase-Angle Calculations
Advanced
The phase angle follows directly from the impedance components: tan θ = X/R, so θ = arctan(X/R) (use atan2 for the correct quadrant). Because cos θ = R/|Z| and sin θ = X/|Z|, the phase angle also gives you the power factor and the real/reactive split for free.
θ = atan2(X, R) cos θ = R / |Z| sin θ = X / |Z|
Worked: R = 4 Ω, X = 3 Ω → θ = atan(3/4) = 36.87°, cos θ = 4/5 = 0.8, sin θ = 3/5 = 0.6. That 0.8 power factor and 0.6 reactive fraction feed straight into the power calculations in the next lessons. The sign of X (and thus of θ) is your lead/lag indicator.
For R = 6 Ω and X = 8 Ω, the phase angle θ is…
θ = atan(8/6) = atan(1.333) = 53.13°. X is positive (inductive) so it's positive and current lags. (6-8-10 triangle; cos θ = 0.6.)
37 AC Power: Real, Reactive & Apparent
Advanced
With a phase angle θ between V and I, power splits into three related quantities. Real power P (watts) is the power actually consumed: P = Vrms·Irms·cos θ. Reactive power Q (VAR) is the power that sloshes back and forth with the reactive elements: Q = Vrms·Irms·sin θ. Apparent power S (VA) is the product the source must supply: S = Vrms·Irms.
P = VI·cos θ (W) Q = VI·sin θ (VAR) S = VI (VA)
These three form a right triangle (the power triangle) with P horizontal, Q vertical and S as the hypotenuse, so S = √(P² + Q²). The angle of that triangle is the same θ you've been computing all along — the power triangle is the impedance triangle in power clothing.
38 The Power Factor
Advanced
The power factor (pf) is cos θ = P/S, a number between 0 and 1 that says what fraction of the apparent power is actually doing useful work. A pf of 1 (purely resistive) means all the apparent power is real. A low pf means a lot of the current is just supporting reactive power, which wastes capacity in wires and generators without delivering energy.
power factor = cos θ = P / S
That's why industry cares: a low power factor forces larger currents (and bigger, costlier equipment) for the same real power. Utilities may charge penalties for low pf, and capacitor banks are added to "correct" it by cancelling reactive power. The 0.8 pf from the 3-4-5 example means 80% of the apparent power is real.
A load has P = 800 W and S = 1000 VA. Its power factor is…
pf = P/S = 800/1000 = 0.8. (We can't say leading or lagging from P and S alone — that needs the sign of Q/θ.)
39 Complex Power S = P + jQ
Advanced
Just as impedance is a complex number, so is power. The complex power is S = P + jQ, with magnitude |S| = √(P² + Q²) (the apparent power in VA) and angle θ (so the power factor is cos θ). A positive Q means inductive (lagging); a negative Q means capacitive (leading).
S = P + jQ |S| = √(P² + Q²) θ = atan2(Q, P) pf = P/|S|
Try it — power triangle & complex power
Worked From the 3-4-5 circuit (V = 120 V, |Z| = 5 Ω), find P, Q and S.
I = V/|Z| = 120/5 = 24 A. S = V·I = 120 × 24 = 2880 VA. pf = cos θ = 0.8, so P = 2880 × 0.8 = 2304 W. Q = 2880 × sin θ = 2880 × 0.6 = 1728 VAR. Check: √(2304² + 1728²) = 2880 VA. ✓
Complex power S = 2304 + j1728 VA = 2880 ∠ 36.87° VA. Inductive (positive Q), pf 0.8 lagging.
40 Putting It All Together: A Complete AC Circuit Analysis
Advanced
This is the whole toolkit on one problem. Given a series RLC circuit with a source, a frequency, and R, L and C, the method is: (1) find XL and XC, (2) build Z = R + j(XL − XC), (3) find I = V/Z, (4) read the phase θ from Z, (5) compute S, P, Q and the power factor. Every earlier lesson is a step in this list.
Worked — full analysis V = 120 V (RMS), f = 60 Hz, R = 10 Ω, L = 0.05 H, C = 0.0001 F. Find Z, I, θ, P, Q, S, pf.
I = V/|Z| = 120/12.61 = 9.52 A. S = V·I = 120 × 9.52 = 1142 VA. pf = cos θ = 0.792. P = S·pf = 1142 × 0.792 = 905 W. Q = S·sin θ = 1142 × (−0.610) = −696 VAR (capacitive, leading).
Complex power S = 905 − j696 VA = 1142 ∠ −37.6° VA. The negative Q confirms the net-capacitive, current-leading result from the sign of X.
You now have the complete phasor method: describe the wave (Beginner), measure it with RMS and convert to a phasor (Intermediate), and analyze circuits and power with complex algebra (Advanced). That's AC Fundamentals & Phasors, end to end.