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9.1 Sinusoidal Sources and Phasors

Updated March 2026Fiveable Content Team
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⚡Electrical Circuits and Systems I Unit 9 Review

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9.1 Sinusoidal Sources and Phasors

Sinusoidal Sources and Phasors

Sinusoidal sources and phasors are the foundation of AC circuit analysis. Instead of solving differential equations every time you encounter a time-varying signal, phasors let you convert sinusoidal voltages and currents into complex numbers. Once you do that, you can apply the same techniques you already know from DC analysis (Ohm's Law, KVL, KCL, voltage dividers) using complex arithmetic. This is the core idea behind sinusoidal steady-state analysis.

Sinusoidal Sources and Characteristics

Fundamentals of Sinusoidal Sources

A sinusoidal source produces a voltage or current that varies with time according to a sine or cosine function. The general form for voltage is:

v(t)=Vmcos⁡(ωt+θ)v(t) = V_m \cos(\omega t + \theta)

and for current:

i(t)=Imcos⁡(ωt+θ)i(t) = I_m \cos(\omega t + \theta)

Why cosine and not sine? Either works, but cosine is the standard convention because it aligns directly with the real part of a complex exponential (more on that in the phasor section). Sinusoidal waveforms are also fundamental because any periodic signal can be decomposed into a sum of sinusoids through Fourier analysis.

Key Parameters of Sinusoidal Waveforms

Each sinusoidal waveform is fully described by a handful of parameters:

  • Amplitude (VmV_m or ImI_m): The peak value of the waveform, measured from zero to the maximum.
  • Frequency (ff): The number of complete cycles per second, in Hertz (Hz). Household AC in the US runs at f=60 Hzf = 60 \text{ Hz}.
  • Angular frequency (ω\omega): Related to frequency by ω=2πf\omega = 2\pi f, measured in radians per second. For 60 Hz, ω≈377 rad/s\omega \approx 377 \text{ rad/s}.
  • Period (TT): The time for one complete cycle. Calculated as T=1/fT = 1/f or equivalently T=2π/ωT = 2\pi/\omega.
  • Phase angle (θ\theta): The horizontal shift of the waveform relative to a reference, in degrees or radians. A positive θ\theta shifts the waveform to the left (earlier in time), meaning it leads the reference.

Derived Quantities and Measurements

The Root Mean Square (RMS) value is the effective value of a sinusoidal signal. It tells you the equivalent DC value that would deliver the same average power to a resistor.

For sinusoids specifically:

Vrms=Vm2≈0.707 VmV_{rms} = \frac{V_m}{\sqrt{2}} \approx 0.707 \, V_m

Irms=Im2≈0.707 ImI_{rms} = \frac{I_m}{\sqrt{2}} \approx 0.707 \, I_m

When someone says "120 V AC," they mean 120 V RMS. The actual peak voltage is Vm=1202≈170 VV_m = 120\sqrt{2} \approx 170 \text{ V}.

One other useful fact: instantaneous power in an AC circuit varies sinusoidally at twice the source frequency, but average power depends on RMS values.

Phasors for Sinusoidal Signals

Fundamentals of Sinusoidal Sources, Sinusoidal Waveforms - Electronics-Lab.com

Phasor Representation Basics

A phasor is a complex number that captures the amplitude and phase of a sinusoidal signal while dropping the time dependence. The key relationship comes from Euler's formula:

ejθ=cos⁡θ+jsin⁡θe^{j\theta} = \cos\theta + j\sin\theta

Using this, you can write a sinusoidal voltage as:

v(t)=Re{Vmej(ωt+θ)}=Re{Vmejθ⋅ejωt}v(t) = \text{Re}\{V_m e^{j(\omega t + \theta)}\} = \text{Re}\{V_m e^{j\theta} \cdot e^{j\omega t}\}

The phasor V\mathbf{V} is defined as just the part that doesn't depend on time:

V=Vmejθ=Vm∠θ\mathbf{V} = V_m e^{j\theta} = V_m\angle\theta

To recover the time-domain signal, you multiply the phasor by ejωte^{j\omega t} and take the real part.

Phasors can be written in two equivalent forms:

  • Polar form: V=Vm∠θ\mathbf{V} = V_m\angle\theta
  • Rectangular form: V=Vmcos⁡θ+jVmsin⁡θ\mathbf{V} = V_m\cos\theta + jV_m\sin\theta

Phasor Domain Characteristics

There are a few conditions that must hold for phasor analysis to be valid:

  • The circuit must be linear and time-invariant (LTI).
  • The circuit must be in sinusoidal steady state, meaning all transients have died out.
  • All sources must operate at the same frequency. (The phasor strips out ω\omega, so it's implicitly assumed to be the same everywhere.)

The payoff is significant: instead of solving differential equations, you solve algebraic equations with complex numbers.

Phasor Transformations and Applications

To work in the phasor domain, you transform each circuit element into its impedance:

ElementTime DomainPhasor Domain Impedance
Resistorv=Riv = RiZR=RZ_R = R
Inductorv=Ldidtv = L\frac{di}{dt}ZL=jωLZ_L = j\omega L
Capacitorv=1C∫i dtv = \frac{1}{C}\int i \, dtZC=1jωCZ_C = \frac{1}{j\omega C}

Notice what happens: differentiation in the time domain becomes multiplication by jωj\omega, and integration becomes division by jωj\omega. That's why phasors eliminate differential equations.

Once everything is in the phasor domain, Ohm's Law becomes:

V=ZI\mathbf{V} = Z\mathbf{I}

where V\mathbf{V} and I\mathbf{I} are phasors and ZZ is the complex impedance. Admittance is the reciprocal: Y=1/ZY = 1/Z.

KVL and KCL apply directly to phasor quantities, just as they do in DC circuits.

Phasor Analysis of Circuits

Circuit Analysis Techniques

Every DC technique you've learned carries over to the phasor domain, with real resistances replaced by complex impedances:

  • Series impedances add directly: Zeq=Z1+Z2+⋯Z_{eq} = Z_1 + Z_2 + \cdots
  • Parallel impedances combine as: 1Zeq=1Z1+1Z2+⋯\frac{1}{Z_{eq}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \cdots
  • Voltage division: Vk=ZkZtotalVs\mathbf{V}_k = \frac{Z_k}{Z_{total}} \mathbf{V}_s
  • Current division: Ik=ZotherZtotalIs\mathbf{I}_k = \frac{Z_{other}}{Z_{total}} \mathbf{I}_s
  • Source transformations and Thévenin/Norton equivalents work the same way, but with complex values for ZThZ_{Th} and VTh\mathbf{V}_{Th}.

The only difference from DC analysis is that you're doing arithmetic with complex numbers instead of real numbers.

Fundamentals of Sinusoidal Sources, Power in AC Circuits - Electronics-Lab.com

Power Calculations in Phasor Domain

Complex power S\mathbf{S} combines real and reactive power into one quantity:

S=VI∗=P+jQ\mathbf{S} = \mathbf{V}\mathbf{I}^* = P + jQ

where I∗\mathbf{I}^* is the complex conjugate of the current phasor. (You conjugate the current, not the voltage.)

  • Real power: P=Re{S}P = \text{Re}\{\mathbf{S}\}, measured in watts (W). This is the power actually consumed.
  • Reactive power: Q=Im{S}Q = \text{Im}\{\mathbf{S}\}, measured in volt-amperes reactive (VAR). This represents energy stored and released by inductors and capacitors.
  • Apparent power: ∣S∣=P2+Q2|\mathbf{S}| = \sqrt{P^2 + Q^2}, measured in volt-amperes (VA).
  • Power factor: pf=cos⁡θ=P/∣S∣\text{pf} = \cos\theta = P/|\mathbf{S}|, where θ\theta is the angle between the voltage and current phasors. A power factor of 1 means all power is real (purely resistive load).

Advanced Analysis Methods

These techniques extend naturally into the phasor domain:

  • Superposition applies when multiple sinusoidal sources share the same frequency. (If sources have different frequencies, you must analyze each frequency separately and add the time-domain results.)
  • Nodal and mesh analysis work with complex impedances and admittances, producing systems of equations with complex coefficients.
  • Maximum power transfer occurs when the load impedance equals the complex conjugate of the Thévenin impedance: ZL=ZTh∗Z_L = Z_{Th}^*.
  • Balanced three-phase systems can be reduced to a single per-phase equivalent circuit using phasor notation, greatly simplifying analysis.

Phasor Operations

Basic Phasor Arithmetic

Different operations are easier in different forms. Here's the practical rule:

  • Addition/Subtraction: Use rectangular form. Add or subtract the real and imaginary parts separately.

(a1+jb1)+(a2+jb2)=(a1+a2)+j(b1+b2)(a_1 + jb_1) + (a_2 + jb_2) = (a_1 + a_2) + j(b_1 + b_2)

  • Multiplication: Use polar form. Multiply the magnitudes and add the angles.

(A∠θ1)(B∠θ2)=AB∠(θ1+θ2)(A\angle\theta_1)(B\angle\theta_2) = AB\angle(\theta_1 + \theta_2)

  • Division: Use polar form. Divide the magnitudes and subtract the angles.

A∠θ1B∠θ2=AB∠(θ1−θ2)\frac{A\angle\theta_1}{B\angle\theta_2} = \frac{A}{B}\angle(\theta_1 - \theta_2)

The j-Operator and Rotations

The imaginary unit j=−1j = \sqrt{-1} has a geometric meaning: multiplying a phasor by jj rotates it by +90°+90° in the complex plane. Multiplying by −j-j rotates it by −90°-90°. This is why inductor impedance (jωLj\omega L) causes current to lag voltage by 90°, and capacitor impedance (1/jωC=−j/ωC1/j\omega C = -j/\omega C) causes current to lead voltage by 90°.

Phasor diagrams are graphical plots of phasors in the complex plane. They're useful for visualizing the phase relationships between voltages and currents in a circuit.

Coordinate System Conversions

You'll convert between rectangular and polar forms constantly, so these should become second nature.

Rectangular to Polar:

  1. Calculate the magnitude: r=a2+b2r = \sqrt{a^2 + b^2}
  2. Calculate the phase angle: θ=tan⁡−1(b/a)\theta = \tan^{-1}(b/a)
  3. Check the quadrant. If a<0a < 0, you need to add or subtract 180° to the angle from your calculator, since tan⁡−1\tan^{-1} only returns values between −90°-90° and +90°+90°.

Polar to Rectangular:

  1. Calculate the real part: a=rcos⁡θa = r\cos\theta
  2. Calculate the imaginary part: b=rsin⁡θb = r\sin\theta

The quadrant check in step 3 of the rectangular-to-polar conversion is a common source of errors on exams. If you have −3+j4-3 + j4, your calculator gives tan⁡−1(4/(−3))=−53.1°\tan^{-1}(4/(-3)) = -53.1°, but the correct angle is 180°−53.1°=126.9°180° - 53.1° = 126.9° since the phasor is in the second quadrant.

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