Phasors and Complex Impedance

Summary

This chapter introduces phasors - rotating vectors that elegantly represent sinusoidal signals - and shows how they transform differential equations into algebraic ones. Students will learn about impedance, the AC equivalent of resistance, and how capacitors and inductors create frequency-dependent reactance. The chapter covers the impedance triangle, admittance, and the phasor domain approach to AC circuit analysis. Mastering phasor techniques is essential for efficient analysis of AC circuits.

Concepts Covered

This chapter covers the following 21 concepts from the learning graph:

Phasor
Phasor Diagram
Phasor Addition
Impedance
Reactance
Capacitive Reactance
Inductive Reactance
Admittance
Susceptance
AC Resistance
Impedance Triangle
Complex Impedance
AC Circuit Analysis
Phasor Domain
Resonance
Series Resonance
Parallel Resonance
Selectivity
Bandwidth
Passband
Stopband

Prerequisites

This chapter builds on concepts from:

title: Phasors and Complex Impedance description: Transform AC circuit analysis from differential equations to simple algebra using phasors and impedance generated_by: chapter-content-generator v0.03 date: 2026-01-30

Introduction: Freezing the Dance

In the last chapter, you learned that AC signals are constantly in motion—sinusoids that rise and fall, shift and dance through time. Analyzing circuits with these moving signals seems like it should be complicated. And with differential equations, it is complicated.

But here's the secret that electrical engineers discovered over a century ago: if all your signals are sinusoids at the same frequency (which is true for most AC power and signal analysis), you can take a snapshot of the dance. You can freeze the rotating signals into stationary arrows called phasors, and suddenly all those intimidating differential equations become simple algebra that you already know how to do.

This chapter introduces the phasor transformation—one of the most powerful tools in circuit analysis. You'll learn that capacitors and inductors have a frequency-dependent "resistance" called impedance, and that Ohm's law works perfectly in the phasor domain. By the end, you'll be analyzing AC circuits with the same ease as DC circuits, just with complex numbers instead of real ones.

Phasors: Rotating Vectors Frozen in Time

A phasor is a complex number that represents the amplitude and phase of a sinusoidal signal. It captures the "personality" of a sinusoid—how big and where in its cycle—while stripping away the time variation.

The transformation:

For a sinusoidal voltage: [v(t) = V_m \cos(\omega t + \phi)]

The corresponding phasor is: [\mathbf{V} = V_m \angle \phi = V_m e^{j\phi}]

Key insight: The phasor contains all the information about the sinusoid except the frequency, which we track separately. Since all signals in a typical AC analysis share the same frequency, we can factor it out!

To recover the time-domain signal from a phasor: [v(t) = \text{Re}{\mathbf{V} e^{j\omega t}} = \text{Re}{V_m e^{j(\omega t + \phi)}}]

Notation conventions:

Notation	Meaning
\(v(t)\)	Time-domain signal (lowercase, function of t)
\(\mathbf{V}\)	Phasor (bold uppercase, no time dependence)
\(V_m\) or (	\mathbf{V}
\(\angle\phi\)	Phase angle

Why Cosine as Reference?

Engineers typically use cosine (not sine) as the reference for phasor analysis because it aligns naturally with the real axis on the complex plane. A phasor at angle 0° corresponds to a cosine wave. Converting from sine is easy: \(\sin(\omega t) = \cos(\omega t - 90°)\).

Diagram: Phasor Transformation Visualizer

Phasor Transformation Visualizer

Type: microsim Bloom Level: Understand (L2) Bloom Verb: explain Learning Objective: Students will explain the relationship between a time-domain sinusoid and its phasor representation by observing both simultaneously. Visual elements: - Left panel: Time-domain waveform v(t) = Vm cos(ωt + φ) - Right panel: Complex plane with phasor arrow - Animated rotating phasor (optional) showing e^(jωt) rotation - Real-axis projection tracking the time waveform - Current time marker on waveform - Phase angle arc on phasor diagram Interactive controls: - Slider: Magnitude Vm (1 to 10) - Slider: Phase angle φ (-180° to 180°) - Toggle: Show rotating version vs frozen phasor - Slider: Animation speed (when rotating) - Button: Freeze at current angle - Display: Phasor in polar and rectangular form Step-through mode: - Stage 1: Show time-domain sinusoid - Stage 2: Show rotating phasor and how projection creates sinusoid - Stage 3: "Freeze" the phasor at t=0 to get the static phasor Annotations: - "Magnitude = peak amplitude" - "Phase = starting angle at t=0" - Formula showing transformation Default parameters: - Vm = 5 - φ = 45° - ω = 2π rad/s (1 Hz for visible animation) Canvas layout: - Dual panel: 300 × 350 pixels each - Control area: 100 pixels below Implementation: p5.js

Phasor Diagrams: Seeing Phase Relationships

A phasor diagram displays multiple phasors on the same complex plane, making phase relationships visually obvious. The lengths represent magnitudes, and the angles show relative phases.

Reading a phasor diagram:

Phasors pointing in the same direction are in phase
A phasor rotated counterclockwise from another leads it in phase
A phasor rotated clockwise lags
Perpendicular phasors are 90° out of phase

Common reference conventions:

In circuit analysis, we often choose one signal as the "reference" and assign it zero phase. Then all other phases are measured relative to this reference.

Reference Choice	Common in
Source voltage	Voltage divider analysis
Source current	Current divider analysis
Total current	Series circuit analysis
Total voltage	Parallel circuit analysis

Phasor Addition

Phasor addition is how we add sinusoidal signals of the same frequency. Instead of wrestling with trigonometric identities, we simply add the complex numbers!

In rectangular form: [\mathbf{V}_1 + \mathbf{V}_2 = (a_1 + a_2) + j(b_1 + b_2)]

Graphically: Place phasors head-to-tail like vector addition.

Example: Add \(v_1(t) = 10\cos(\omega t)\) and \(v_2(t) = 10\cos(\omega t + 90°)\)

Phasors: [\mathbf{V}_1 = 10\angle 0° = 10 + j0] [\mathbf{V}_2 = 10\angle 90° = 0 + j10]

Sum: [\mathbf{V}_{total} = 10 + j10 = 14.14\angle 45°]

Time domain result: [v_{total}(t) = 14.14\cos(\omega t + 45°)]

Try doing this with trig identities—phasors are much easier!

Diagram: Phasor Addition Visualizer

Phasor Addition Visualizer

Type: microsim Bloom Level: Apply (L3) Bloom Verb: calculate Learning Objective: Students will calculate the sum of two sinusoidal signals using phasor addition and verify the result graphically. Visual elements: - Left panel: Phasor diagram showing V₁, V₂, and V_total - V₁ and V₂ shown in different colors (blue, red) - V_total shown as resultant (green) - Head-to-tail construction visible - Right panel: Time-domain plot showing all three waveforms - Verification that sum waveform matches vector sum Interactive controls: - Slider: V₁ magnitude (1 to 10) - Slider: V₁ phase (-180° to 180°) - Slider: V₂ magnitude (1 to 10) - Slider: V₂ phase (-180° to 180°) - Toggle: Show/hide construction lines - Display: All three phasors in polar form Calculations displayed: - V₁ + V₂ in rectangular: (a₁+a₂) + j(b₁+b₂) - Conversion to polar: |V_total|∠θ - Time-domain expression of sum Default parameters: - V₁ = 10∠0° - V₂ = 10∠90° - V_total = 14.14∠45° Canvas layout: - Phasor diagram: 300 × 350 pixels - Time-domain plot: 300 × 350 pixels - Control area: 100 pixels below Implementation: p5.js

Impedance: AC's Generalized Resistance

Impedance (\(Z\)) is the complex ratio of voltage phasor to current phasor:

\[Z = \frac{\mathbf{V}}{\mathbf{I}}\]

Impedance plays the same role in AC circuits that resistance plays in DC circuits. In fact, Ohm's law works perfectly with impedance:

\[\mathbf{V} = \mathbf{I} \cdot Z\]

But unlike resistance (which is just a real number), impedance is complex:

\[Z = R + jX\]

Where:

\(R\) = Resistance (real part) - dissipates power
\(X\) = Reactance (imaginary part) - stores and returns power

Units: Impedance, resistance, and reactance are all measured in ohms (Ω).

Component	Impedance	Notes
Resistor	\(Z_R = R\)	Purely real
Capacitor	\(Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}\)	Purely imaginary, negative
Inductor	\(Z_L = j\omega L\)	Purely imaginary, positive

Reactance: The Imaginary Part

Reactance (\(X\)) is the imaginary component of impedance. It represents opposition to current that doesn't dissipate power—instead, it stores energy temporarily and returns it to the circuit.

Inductive Reactance (\(X_L\)): [X_L = \omega L = 2\pi f L]

Positive reactance (current lags voltage by 90°)
Increases with frequency (inductors "resist" fast changes)
At DC (\(f = 0\)): \(X_L = 0\) (short circuit)

Capacitive Reactance (\(X_C\)): [X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}]

By convention, written as negative in impedance: \(Z_C = -jX_C\)
Current leads voltage by 90°
Decreases with frequency (capacitors pass high frequencies more easily)
At DC (\(f = 0\)): \(X_C \to \infty\) (open circuit)

Component	Low Frequency Behavior	High Frequency Behavior
Inductor	Small X_L, acts like short	Large X_L, acts like open
Capacitor	Large X_C, acts like open	Small X_C, acts like short

Why Reactance Doesn't Dissipate Power

When current and voltage are 90° out of phase (as with pure reactance), energy flows into the component for half the cycle and flows back out for the other half. The average power over a complete cycle is zero. Power is only dissipated when there's a resistance component.

Diagram: Reactance vs Frequency

Reactance vs Frequency

Type: microsim Bloom Level: Analyze (L4) Bloom Verb: examine Learning Objective: Students will examine how inductive and capacitive reactance vary with frequency and identify the crossover point where they are equal. Visual elements: - Log-log plot of reactance vs frequency - X_L curve (positive, increasing) in red - X_C curve (positive magnitude, decreasing) in blue - Crossover point where X_L = X_C marked (resonant frequency) - Current frequency marker (vertical line) - Display showing X_L and X_C at current frequency Interactive controls: - Slider: Inductance L (0.1mH to 100mH, log scale) - Slider: Capacitance C (0.1nF to 100μF, log scale) - Slider: Frequency f (1Hz to 1MHz, log scale) - Toggle: Linear vs Log scale - Display: f₀ = 1/(2π√LC) resonant frequency Annotations: - "X_L = ωL = 2πfL" on inductor curve - "X_C = 1/(ωC) = 1/(2πfC)" on capacitor curve - "Resonance: X_L = X_C" at crossover Default parameters: - L = 10mH - C = 1μF - f₀ ≈ 1592 Hz Canvas layout: - Plot area: 600 × 350 pixels - Control area: 100 pixels below Implementation: p5.js

Complex Impedance and the Impedance Triangle

When a circuit contains both resistance and reactance, the complex impedance is:

\[Z = R + jX\]

In polar form:

\[Z = |Z| \angle \theta\]

Where:

Magnitude: \(|Z| = \sqrt{R^2 + X^2}\)
Phase angle: \(\theta = \tan^{-1}(X/R)\)

The impedance triangle is a right triangle showing the relationship:

        |Z|
         /|
        / |
       /  | X (reactance)
      /   |
     /θ___|
       R (resistance)

Phase angle interpretation:

Condition	Phase Angle	Circuit Behavior
X = 0	θ = 0°	Purely resistive
X > 0 (inductive)	θ > 0°	Current lags voltage
X < 0 (capacitive)	θ < 0°	Current leads voltage
R = 0	θ = ±90°	Purely reactive

Example: Series R-L circuit with R = 30Ω and X_L = 40Ω

\[Z = 30 + j40 = 50\angle 53.13°\]

The magnitude is 50Ω, and current lags voltage by 53.13°.

Diagram: Impedance Triangle Explorer

Impedance Triangle Explorer

Type: microsim Bloom Level: Apply (L3) Bloom Verb: calculate Learning Objective: Students will calculate impedance magnitude and phase angle from resistance and reactance using the impedance triangle. Visual elements: - Impedance triangle with R on horizontal, X on vertical, |Z| as hypotenuse - Phase angle θ marked with arc - Labels showing R, X, |Z|, and θ values - Color coding: R (blue), X (red for inductive, green for capacitive), |Z| (purple) - Phasor diagram showing V and I relationship Interactive controls: - Slider: Resistance R (0 to 100Ω) - Slider: Reactance X (-100 to +100Ω) - Toggle: Show as R-L or R-C circuit - Display: Z in rectangular and polar form - Display: Phase relationship (leads/lags) Calculations displayed: - |Z| = √(R² + X²) with values - θ = tan⁻¹(X/R) with values - Power factor = cos(θ) Visual feedback: - Triangle updates in real-time - Positive X shows above horizontal (inductive) - Negative X shows below horizontal (capacitive) Default parameters: - R = 30Ω - X = 40Ω (inductive) - |Z| = 50Ω, θ = 53.13° Canvas layout: - Triangle/phasor area: 450 × 350 pixels - Control area: 100 pixels below Implementation: p5.js

Admittance and Susceptance

Admittance (\(Y\)) is the reciprocal of impedance—it measures how easily current flows:

\[Y = \frac{1}{Z} = \frac{\mathbf{I}}{\mathbf{V}}\]

Units: Siemens (S), formerly called mhos (℧)

\[Y = G + jB\]

Where:

\(G\) = Conductance (real part)
\(B\) = Susceptance (imaginary part)

Susceptance (\(B\)) is the imaginary part of admittance:

Capacitive susceptance: \(B_C = \omega C\) (positive)

Inductive susceptance: \(B_L = -\frac{1}{\omega L}\) (negative)

Note the sign flip compared to reactance!

Quantity	Symbol	Units	Formula
Impedance	Z	Ω	R + jX
Admittance	Y	S	G + jB
Resistance	R	Ω	Re{Z}
Conductance	G	S	Re{Y}
Reactance	X	Ω	Im{Z}
Susceptance	B	S	Im{Y}

When to use admittance:

Parallel circuits (admittances add directly)
High-frequency analysis
When measuring with siemens makes more sense

Converting between Z and Y:

If \(Z = R + jX\), then: [Y = \frac{1}{R + jX} = \frac{R - jX}{R^2 + X^2}]

So: \(G = \frac{R}{R^2 + X^2}\) and \(B = \frac{-X}{R^2 + X^2}\)

Note: \(G \neq 1/R\) unless \(X = 0\)!

AC Resistance and Power Factor

In AC circuits, AC resistance (also called effective resistance) includes all power-dissipating effects, which may differ from DC resistance due to:

Skin effect: Current crowds toward conductor surface at high frequencies
Proximity effect: Nearby conductors affect current distribution
Core losses: In inductors with magnetic cores

The power factor relates real power to apparent power:

\[\text{Power Factor} = \cos\theta = \frac{R}{|Z|}\]

Power Factor	θ	Circuit Type
1.0	0°	Purely resistive
0	±90°	Purely reactive
0.8 lagging	36.87°	Inductive load
0.8 leading	-36.87°	Capacitive load

We'll explore power factor in depth in the next chapter on AC power analysis.

Phasor Domain Analysis: The Method

The phasor domain (or frequency domain) approach transforms time-domain differential equations into algebraic equations. Here's the systematic method:

Step 1: Transform to Phasor Domain

Replace time-domain sources \(v(t)\) with phasors \(\mathbf{V}\)
Replace R, L, C with their impedances \(Z_R\), \(Z_L\), \(Z_C\)

Step 2: Analyze Using DC Techniques

Apply Ohm's law: \(\mathbf{V} = \mathbf{I}Z\)
Apply KVL around loops (with phasor voltages)
Apply KCL at nodes (with phasor currents)
Use voltage/current dividers
Apply Thevenin/Norton equivalents

Step 3: Transform Back to Time Domain

Convert phasor results to time-domain expressions
\(\mathbf{V} = V_m \angle \phi \rightarrow v(t) = V_m \cos(\omega t + \phi)\)

The magic: Differentiation becomes multiplication by \(j\omega\), integration becomes division by \(j\omega\). All the calculus disappears!

Time Domain	Phasor Domain
\(v(t) = V_m\cos(\omega t + \phi)\)	\(\mathbf{V} = V_m\angle\phi\)
\(\frac{dv}{dt}\)	\(j\omega \mathbf{V}\)
\(\int v \, dt\)	\(\frac{\mathbf{V}}{j\omega}\)
Differential equation	Algebraic equation

Worked Example: Series RLC Analysis

Problem: Find the steady-state current in a series RLC circuit with: - \(v_s(t) = 100\cos(1000t)\) V - R = 10Ω, L = 20mH, C = 50μF

Solution:

Step 1: Calculate impedances at ω = 1000 rad/s

[Z_R = 10 \text{ Ω}] [Z_L = j\omega L = j(1000)(0.02) = j20 \text{ Ω}] [Z_C = \frac{1}{j\omega C} = \frac{1}{j(1000)(50 \times 10^{-6})} = \frac{1}{j0.05} = -j20 \text{ Ω}]

Step 2: Find total impedance

\[Z_{total} = Z_R + Z_L + Z_C = 10 + j20 - j20 = 10 + j0 = 10 \text{ Ω}\]

Notice that \(Z_L\) and \(Z_C\) cancel! This is resonance.

Step 3: Find current phasor

[\mathbf{V}s = 100\angle 0°] [\mathbf{I} = \frac{\mathbf{V}_s}{Z]}} = \frac{100\angle 0°}{10\angle 0°} = 10\angle 0° \text{ A

Step 4: Convert to time domain

\[i(t) = 10\cos(1000t) \text{ A}\]

At resonance, the current is in phase with the voltage and limited only by the resistance!

Diagram: Phasor Domain Circuit Solver

Phasor Domain Circuit Solver

Type: microsim Bloom Level: Apply (L3) Bloom Verb: solve Learning Objective: Students will solve for current and voltage in series RLC circuits using phasor domain analysis. Visual elements: - Circuit schematic showing series RLC with AC source - Impedance calculation panel showing Z_R, Z_L, Z_C, Z_total - Phasor diagram showing V_s, V_R, V_L, V_C, I - KVL verification: V_s = V_R + V_L + V_C shown graphically - Time-domain waveform showing voltage and current Step-through mode: - Step 1: "Transform source to phasor" - Step 2: "Calculate component impedances" - Step 3: "Sum impedances for Z_total" - Step 4: "Calculate current phasor I = V/Z" - Step 5: "Calculate component voltage phasors" - Step 6: "Transform back to time domain" Interactive controls: - Slider: Frequency f or ω - Slider: R (1 to 100Ω) - Slider: L (1mH to 100mH) - Slider: C (0.1μF to 100μF) - Source voltage amplitude (fixed or adjustable) - Button: Step through / Auto-solve Displays: - Z_total in rectangular and polar form - I in polar form and time domain - V_R, V_L, V_C phasors - Resonant frequency indicator Default parameters: - f = 1000/(2π) ≈ 159 Hz (ω = 1000 rad/s) - R = 10Ω, L = 20mH, C = 50μF - At resonance, X_L = X_C = 20Ω Canvas layout: - Circuit diagram: 250 × 400 pixels (left) - Phasor diagram: 350 × 400 pixels (right) - Control area: 100 pixels below Implementation: p5.js

Resonance in AC Circuits

Resonance occurs when the inductive and capacitive reactances are equal, causing them to cancel. The resonant frequency is:

\[f_0 = \frac{1}{2\pi\sqrt{LC}} \text{ Hz}\]

Or in angular frequency:

\[\omega_0 = \frac{1}{\sqrt{LC}} \text{ rad/s}\]

At resonance, the circuit behavior depends on whether components are in series or parallel.

Series Resonance

In a series resonant circuit:

\(X_L = X_C\), so \(Z = R\) (minimum impedance)
Current is maximum: \(I = V_s/R\)
Current is in phase with voltage
Voltage across L and C can exceed source voltage!

Voltage magnification: [Q = \frac{V_L}{V_s} = \frac{V_C}{V_s} = \frac{X_L}{R} = \frac{\omega_0 L}{R}]

At resonance with Q = 100, the voltage across L or C is 100 times the source voltage! This is useful for selecting frequencies but requires care to avoid component damage.

Parallel Resonance

In a parallel resonant circuit (also called a "tank" circuit):

Admittances cancel, so impedance is maximum
Current from source is minimum
Large circulating current flows between L and C
Acts as a high-impedance "trap" at the resonant frequency

Property	Series Resonance	Parallel Resonance
Impedance at f₀	Minimum (= R)	Maximum
Current at f₀	Maximum	Minimum
Use	Bandpass (select f₀)	Bandstop (reject f₀)
Q effect	Voltage magnification	Current magnification

Diagram: Series vs Parallel Resonance

Series vs Parallel Resonance

Type: microsim Bloom Level: Analyze (L4) Bloom Verb: compare Learning Objective: Students will compare series and parallel resonance by observing impedance and current behavior as frequency sweeps through resonance. Visual elements: - Split display: Series RLC (left) and Parallel RLC (right) - Top plots: |Z| vs frequency for each configuration - Bottom plots: |I| vs frequency for each configuration - Resonant frequency marked on all plots - Current frequency marker (sweepable) - Energy flow animations showing circulating current at resonance Interactive controls: - Slider: Frequency sweep (0.1× to 10× f₀) - Slider: Q factor (affects bandwidth) - Button: Auto-sweep animation - Toggle: Show series only / parallel only / both - Display: Current impedance and current values Annotations: - Series at f₀: "Z minimum, I maximum" - Parallel at f₀: "Z maximum, I minimum" - "Same resonant frequency!" Default parameters: - L = 10mH, C = 1μF - f₀ ≈ 1592 Hz - Q = 10 (for visible bandwidth) Canvas layout: - Dual configuration display: 600 × 400 pixels - Control area: 100 pixels below Implementation: p5.js

Bandwidth, Selectivity, Passband, and Stopband

Bandwidth (BW) is the range of frequencies over which a circuit or filter passes signals effectively. For a resonant circuit:

\[BW = \frac{f_0}{Q}\]

Selectivity describes how well a circuit discriminates between desired and undesired frequencies. Higher Q means better selectivity (narrower bandwidth).

Term	Definition
Passband	Frequency range where signals pass through with little attenuation
Stopband	Frequency range where signals are significantly attenuated
Cutoff frequency	Boundary between passband and stopband (usually -3dB point)
Transition band	Region between passband and stopband

The -3dB points:

The bandwidth is typically measured between the frequencies where the response drops to 70.7% of its peak value (half power, or -3dB):

[f_1 = f_0\sqrt{1 + \frac{1}{4Q^2}} - \frac{f_0}{2Q} \approx f_0 - \frac{BW}{2}] [f_2 = f_0\sqrt{1 + \frac{1}{4Q^2}} + \frac{f_0}{2Q} \approx f_0 + \frac{BW}{2}]

Quality factor and bandwidth relationship:

Q	Bandwidth (% of f₀)	Selectivity
1	100%	Poor
10	10%	Moderate
100	1%	Good
1000	0.1%	Excellent

Diagram: Bandwidth and Selectivity

Bandwidth and Selectivity

Type: microsim Bloom Level: Understand (L2) Bloom Verb: explain Learning Objective: Students will explain how Q factor affects bandwidth and selectivity by observing frequency response curves at different Q values. Visual elements: - Frequency response plot showing |H(f)| vs frequency - Multiple curves for different Q values displayed simultaneously - -3dB line (0.707) marked horizontally - Bandwidth markers showing f₁ and f₂ for selected curve - Passband region shaded - Stopband regions indicated Interactive controls: - Slider: Q factor (1 to 100) - Slider: Resonant frequency f₀ - Toggle: Show multiple Q curves (5, 10, 20, 50) simultaneously - Toggle: Linear vs dB scale - Display: Calculated bandwidth BW = f₀/Q Annotations: - "Higher Q = narrower bandwidth" - "Better selectivity = sharper peak" - Passband and stopband labels Default parameters: - Q = 10 - f₀ = 1000 Hz - BW = 100 Hz Canvas layout: - Plot area: 600 × 350 pixels - Control area: 100 pixels below Implementation: p5.js

Component Impedances Summary

Here's a comprehensive reference for working with component impedances:

Component	Time Domain	Impedance	Admittance
Resistor	\(v = iR\)	\(Z = R\)	\(Y = G = 1/R\)
Capacitor	\(i = C\frac{dv}{dt}\)	\(Z = \frac{1}{j\omega C}\)	\(Y = j\omega C\)
Inductor	\(v = L\frac{di}{dt}\)	\(Z = j\omega L\)	\(Y = \frac{1}{j\omega L}\)

Series combinations: Add impedances [Z_{series} = Z_1 + Z_2 + Z_3 + ...]

Parallel combinations: Add admittances (or use product/sum) [Y_{parallel} = Y_1 + Y_2 + Y_3 + ...] [Z_{parallel} = \frac{Z_1 Z_2}{Z_1 + Z_2}] (for two impedances)

Voltage divider (series): [\mathbf{V}_1 = \mathbf{V}_s \cdot \frac{Z_1}{Z_1 + Z_2}]

Current divider (parallel): [\mathbf{I}_1 = \mathbf{I}_s \cdot \frac{Z_2}{Z_1 + Z_2}]

Self-Check Questions

1. A sinusoidal voltage v(t) = 20cos(500t + 30°) V is applied to a series RL circuit with R = 40Ω and L = 60mH. Find the impedance, current phasor, and time-domain current expression.

Find impedance:

[Z_R = 40 \text{ Ω}] [Z_L = j\omega L = j(500)(0.06) = j30 \text{ Ω}] [Z_{total} = 40 + j30 = 50\angle 36.87° \text{ Ω}]

Find current phasor:

[\mathbf{V} = 20\angle 30°] [\mathbf{I} = \frac{\mathbf{V}}{Z} = \frac{20\angle 30°}{50\angle 36.87°} = 0.4\angle (30° - 36.87°) = 0.4\angle -6.87° \text{ A}]

Time-domain expression:

\[i(t) = 0.4\cos(500t - 6.87°) \text{ A}\]

The current lags the voltage by 36.87° because the circuit is inductive.

2. At what frequency will a 100mH inductor have the same reactance magnitude as a 10μF capacitor?

Set \(X_L = X_C\):

[\omega L = \frac{1}{\omega C}] [\omega^2 = \frac{1}{LC} = \frac{1}{(0.1)(10 \times 10^{-6})} = 10^6] [\omega = 1000 \text{ rad/s}]

Converting to Hz: [f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = 159.2 \text{ Hz}]

At this frequency, both have reactance: [X_L = X_C = \omega L = (1000)(0.1) = 100 \text{ Ω}]

This is the resonant frequency of the L-C combination.

3. A series RLC circuit has R = 20Ω, and at resonance the Q factor is 50. If the resonant frequency is 10 kHz, what are L and C?

From Q factor:

[Q = \frac{\omega_0 L}{R} = \frac{X_L}{R}] [50 = \frac{X_L}{20}] [X_L = 1000 \text{ Ω}]

Find L:

[\omega_0 = 2\pi f_0 = 2\pi(10000) = 62832 \text{ rad/s}] [X_L = \omega_0 L] [L = \frac{X_L}{\omega_0} = \frac{1000}{62832} = 15.92 \text{ mH}]

Find C (at resonance X_C = X_L):

[X_C = \frac{1}{\omega_0 C} = 1000 \text{ Ω}] [C = \frac{1}{\omega_0 X_C} = \frac{1}{62832 \times 1000} = 15.92 \text{ nF}]

Verify: \(\omega_0 = 1/\sqrt{LC} = 1/\sqrt{(15.92 \times 10^{-3})(15.92 \times 10^{-9})} = 62832\) rad/s ✓

4. What is the bandwidth of the circuit in Question 3, and what are the lower and upper cutoff frequencies?

Bandwidth:

\[BW = \frac{f_0}{Q} = \frac{10000}{50} = 200 \text{ Hz}\]

Cutoff frequencies (approximately):

[f_1 \approx f_0 - \frac{BW}{2} = 10000 - 100 = 9900 \text{ Hz}] [f_2 \approx f_0 + \frac{BW}{2} = 10000 + 100 = 10100 \text{ Hz}]

The passband extends from 9900 Hz to 10100 Hz (just 200 Hz wide). This high-Q circuit is very selective, passing only frequencies very close to 10 kHz.

Summary

This chapter equipped you with the essential tools for AC circuit analysis:

Phasors freeze rotating signals - Complex numbers represent sinusoid amplitude and phase, allowing algebraic analysis
Phasor diagrams reveal phase relationships - Visualizing multiple phasors shows leading/lagging relationships at a glance
Impedance extends Ohm's law to AC - Z = R + jX includes both dissipative (R) and reactive (X) components
Reactance is frequency-dependent - Inductors impede high frequencies; capacitors impede low frequencies
The impedance triangle relates R, X, and Z - Magnitude, angle, and power factor are geometrically linked
Admittance simplifies parallel circuits - Y = G + jB is the reciprocal of impedance
Resonance cancels reactance - At ω₀ = 1/√(LC), inductive and capacitive effects cancel
Series resonance minimizes impedance - Used for bandpass filters that select specific frequencies
Parallel resonance maximizes impedance - Used for bandstop filters or "traps"
Q determines selectivity - Higher Q means narrower bandwidth and sharper frequency selection

With phasors and impedance in your toolkit, you can now analyze any AC circuit using the same familiar techniques you learned for DC. Kirchhoff's laws, voltage dividers, and Thevenin/Norton equivalents all work perfectly—just with complex numbers. The next chapter will show you how to analyze power in AC circuits, where the phase relationships you've learned become crucial for understanding real, reactive, and apparent power.