Phasor Representation

Pack a sinusoid's amplitude and phase into one complex number — and turn AC calculus into plain algebra.

Circuit AnalysisElectrical Engineering Year 2Free preview
⏱️ About 18 min

Solving an AC circuit by hand looks like a wall of trig — until one clever step collapses every sinusoid into a single complex number.

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The big idea: Because every source and response in a linear AC circuit shares the same frequency, that frequency can be factored out and dropped, leaving only amplitude and phase — a complex number called a phasor that turns differential equations into algebra.
🎯 By the end, you'll be able to
  • Derive the phasor transform from Euler's formula and define a phasor in both polar and rectangular form
  • Apply the rule that time-domain differentiation d/dt becomes multiplication by jω in the phasor domain
  • Add same-frequency sinusoids by adding their phasors as complex numbers
  • Convert between polar (Vm∠φ) and rectangular (Vm cos φ + j Vm sin φ) phasor forms
📎 Helpful to know first

From trigonometry to complex numbers

Euler's formula states that e^(jθ) = cos θ + j sin θ. A real cosine can therefore be recovered as the real part of a complex exponential: V_m cos(ωt + φ) = Re[V_m e^(j(ωt + φ))] = Re[V_m e^(jφ) · e^(jωt)]. The factor e^(jωt) carries the time dependence and is identical for every source and response in a linear circuit driven at one frequency. We factor it out and keep only the part that distinguishes one sinusoid from another — the complex number V_m e^(jφ). That number is the phasor.

\[ v(t) = V_m \cos(\omega t + \phi) \;\Longleftrightarrow\; \mathbf{V} = V_m\, e^{j\phi} = V_m \angle \phi \]
The phasor transform. The time signal is recovered as Re[V e^(jωt)].

Polar and rectangular forms

A phasor lives in the complex plane, so it has two equivalent notations. In polar form it is written V_m∠φ — magnitude V_m at angle φ. In rectangular form the same number is V_m cos φ + j V_m sin φ, by Euler's formula. Convert polar → rectangular with the cosine (real) and sine (imaginary) parts; convert rectangular → polar with magnitude √(a² + b²) and angle atan2(b, a).

Convention note: here the phasor magnitude is the amplitude (peak) V_m. Some texts use the RMS value V_m/√2 instead; this course uses amplitude-based phasors throughout, and RMS will be revisited in the AC Power module.

\[ \mathbf{V} = V_m \angle \phi = V_m \cos\phi + j\, V_m \sin\phi \]
Polar and rectangular forms of the same phasor.

Calculus becomes algebra

Here is the real payoff. Differentiate the time signal v(t) = Re[V e^(jωt)]: dv/dt = Re[jω V e^(jωt)]. The derivative only pulls out a factor of jω — the phasor V becomes jωV. No trig identities, no chain rule beyond a constant multiplier. This is the reason phasors dominate AC analysis: an integro-differential equation written in the time domain becomes a plain algebraic equation in the phasor domain, where inductors have impedance jωL and capacitors 1/(jωC).

\[ \frac{d}{dt} \;\Longleftrightarrow\; j\omega \]
Time-domain differentiation maps to multiplication by jω in the phasor domain.

Adding sinusoids by adding phasors

Two sinusoids of the same frequency add to another sinusoid of that frequency. Rather than expand trig identities, add the phasors as complex numbers (easiest in rectangular form) and convert the result back to polar. Adding amplitudes directly is wrong — only the complex-number sum captures the combined magnitude and phase.

📝 Worked example: Convert v(t) = 100 cos(500t + 60°) V to phasor form in both polar and rectangular notation.
  1. Polar form is read directly: V = 100 ∠ 60° V.
  2. Rectangular form: V = 100 cos 60° + j · 100 sin 60°.
  3. cos 60° = 0.5 and sin 60° ≈ 0.866, so V = 100(0.5) + j 100(0.866).
  4. Therefore V = 50 + j86.6 V (polar: 100 ∠ 60° V).
✓ Polar: V = 100 ∠ 60° V. Rectangular: V = 50 + j86.6 V.
📝 Worked example: Add v1(t) = 20 cos(ωt) V and v2(t) = 20 cos(ωt + 90°) V using phasors and find the resultant.
  1. Phasors: V1 = 20 ∠ 0° = 20 + j0. V2 = 20 ∠ 90° = 0 + j20.
  2. Add in rectangular form: V = V1 + V2 = 20 + j20.
  3. Magnitude: |V| = √(20² + 20²) = √800 = 20√2.
  4. Angle: φ = atan2(20, 20) = 45°.
  5. Resultant phasor V = 20√2 ∠ 45°, so v(t) = 20√2 cos(ωt + 45°) V.
✓ The sum is a sinusoid of magnitude 20√2 at phase 45°: v(t) = 20√2 cos(ωt + 45°) V.
✏️ Practice: A phasor in rectangular form is V = 8 + j6 V. What is its magnitude |V|? (Round to one decimal place.)
V
Solution
  1. Magnitude: |V| = √(8² + 6²) = √(64 + 36) = √100 = 10 V.
  2. The angle would be atan2(6, 8) ≈ 36.9°, giving polar form 10 ∠ 36.9° V.

Check your understanding

1. In the phasor domain, the time-domain operation dv/dt becomes:
Differentiating Re[V e^(jωt)] pulls out a factor jω, so dv/dt ↔ jωV.
2. Two sinusoids can be added as phasors only if they have:
Phasor addition assumes a shared e^(jωt) factor, which requires identical ω — i.e., the same frequency.
3. The phasor of v(t) = 100 cos(ωt + 60°) V in rectangular form is:
100 cos 60° = 50, and 100 sin 60° ≈ 86.6, so V = 50 + j86.6 V.
4. Adding 20∠0° and 20∠90° gives a phasor of magnitude:
20 + j20 has magnitude √(20² + 20²) = 20√2 at 45°.
🔎 Related lessons
Sinusoidal Sources & Parameters
✅ Key takeaways
  • A phasor packs a sinusoid's amplitude and phase into one complex number: V = Vm ∠ φ = Vm cos φ + j Vm sin φ.
  • Euler's formula justifies the transform: Vm cos(ωt + φ) = Re[V e^(jωt)], and the shared e^(jωt) is dropped.
  • Differentiation in time becomes multiplication by jω — converting differential equations into algebra.
  • Same-frequency sinusoids are added by adding their phasors as complex numbers, then converting back to polar.
➡️ With phasors in hand, the next step is to give resistors, inductors, and capacitors phasor-domain impedances so an entire AC circuit can be solved like a DC resistive network.
Want to test yourself on this? Try the Electrical Aptitude test →