Phasor Diagrams and Phasor Algebra

1. Definition and Representation of Phasors

Definition and Representation of Phasors

Mathematical Foundation of Phasors

A phasor is a complex number representation of a sinusoidal signal, enabling simplified analysis of linear time-invariant (LTI) systems. For a time-domain signal x(t) with amplitude A, angular frequency ω, and phase φ:

$$ x(t) = A \cos(\omega t + \phi) $$

This can be expressed as the real part of a rotating complex exponential:

$$ x(t) = \Re\{A e^{j(\omega t + \phi)}\} = \Re\{A e^{j\phi} e^{j\omega t}\} $$

The time-invariant component A ejϕ is the phasor, denoted as X:

$$ X = A e^{j\phi} = A \angle \phi $$

Phasor Representation in Polar and Rectangular Forms

Phasors can be represented in:

Conversion between forms uses Euler’s formula:

$$ A e^{j\phi} = A \cos \phi + j A \sin \phi $$

Visualization: Phasor Diagrams

A phasor diagram plots phasors as vectors in the complex plane, where:

Phasor (A∠ϕ) Re Im

Applications in AC Circuit Analysis

Phasors simplify steady-state AC analysis by:

Historical Context

Phasor notation was popularized by Charles Proteus Steinmetz in the late 19th century to simplify AC power system calculations, replacing laborious differential equation solutions.

Phasor Representation in Complex Plane A vector diagram showing a phasor in the complex plane with labeled real (Re) and imaginary (Im) axes, magnitude (A), and phase angle (ϕ). Re Im ϕ A∠ϕ
Diagram Description: The diagram would physically show the phasor as a vector in the complex plane with labeled real (Re) and imaginary (Im) axes, illustrating its magnitude and phase angle.

Sinusoidal Signals and Phasor Conversion

Mathematical Representation of Sinusoidal Signals

A sinusoidal signal in the time domain is represented as:

$$ v(t) = V_m \cos(\omega t + \phi) $$

where:

This can also be expressed in terms of sine:

$$ v(t) = V_m \sin(\omega t + \phi') $$

where \( \phi' = \phi + \frac{\pi}{2} \). The choice between sine and cosine is arbitrary but must remain consistent in analysis.

Phasor Representation

A phasor is a complex number representation of a sinusoidal signal, capturing its amplitude and phase while omitting the time-varying component. Using Euler's formula:

$$ V_m e^{j(\omega t + \phi)} = V_m \cos(\omega t + \phi) + j V_m \sin(\omega t + \phi) $$

The phasor form extracts the time-independent part:

$$ \mathbf{V} = V_m e^{j\phi} = V_m \angle \phi $$

This simplifies AC circuit analysis by converting differential equations into algebraic ones.

Conversion Between Time Domain and Phasor Domain

To convert a sinusoidal signal to its phasor equivalent:

  1. Express the signal in cosine form \( V_m \cos(\omega t + \phi) \).
  2. Extract the amplitude \( V_m \) and phase \( \phi \).
  3. Represent it as \( \mathbf{V} = V_m \angle \phi \).

For example, given \( v(t) = 10 \cos(100t + 30^\circ) \), the phasor is:

$$ \mathbf{V} = 10 \angle 30^\circ $$

To revert to the time domain, multiply the phasor by \( e^{j\omega t} \) and take the real part:

$$ v(t) = \Re \{ \mathbf{V} e^{j\omega t} \} $$

Phasor Algebra Operations

Phasors follow complex arithmetic rules:

Example: Given \( \mathbf{V}_1 = 5 \angle 20^\circ \) and \( \mathbf{V}_2 = 3 \angle -30^\circ \), their product is:

$$ \mathbf{V}_1 \times \mathbf{V}_2 = 15 \angle (-10^\circ) $$

Practical Applications

Phasors are indispensable in:

Visualizing Phasors

A phasor diagram plots vectors in the complex plane, where:

\( \mathbf{V} = V_m \angle \phi \) \( \phi \)

The angle \( \phi \) indicates phase shift relative to a reference phasor (typically the voltage in power systems).

Phasor Representation in Complex Plane A vector diagram showing a phasor in the complex plane with labeled real and imaginary axes, phase angle, and magnitude. Re Im φ Vₘ ∠ φ
Diagram Description: The diagram would physically show the phasor representation in the complex plane, illustrating the relationship between the real and imaginary components and the phase angle.

1.3 Phasor Notation and Complex Numbers

Phasor notation leverages complex numbers to simplify the analysis of sinusoidal steady-state circuits. A phasor V representing a sinusoidal voltage v(t) = Vmcos(ωt + φ) is expressed in complex form as:

$$ \mathbf{V} = V_m e^{j\phi} = V_m \angle \phi $$

This compact representation encodes both amplitude Vm and phase angle φ while eliminating the explicit time dependence. The transformation from time domain to phasor domain follows from Euler's formula:

$$ V_m \cos(\omega t + \phi) = \text{Re}\{V_m e^{j(\omega t + \phi)}\} = \text{Re}\{\mathbf{V}e^{j\omega t}\} $$

Complex Number Representations

Phasors utilize three equivalent complex number forms:

The conversions between these forms are fundamental:

$$ a = r \cos \theta \quad b = r \sin \theta $$ $$ r = \sqrt{a^2 + b^2} \quad \theta = \tan^{-1}\left(\frac{b}{a}\right) $$

Phasor Arithmetic Operations

Phasor mathematics follows complex number algebra:

For two phasors V1 = a + jb = r1∠θ1 and V2 = c + jd = r2∠θ2:

$$ \mathbf{V}_1 + \mathbf{V}_2 = (a + c) + j(b + d) $$ $$ \mathbf{V}_1 \times \mathbf{V}_2 = r_1 r_2 \angle (\theta_1 + \theta_2) $$ $$ \frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{r_1}{r_2} \angle (\theta_1 - \theta_2) $$

Impedance as Complex Phasor

Circuit elements exhibit frequency-dependent complex impedance:

Ohm's Law extends to phasors:

$$ \mathbf{V} = \mathbf{I} \times \mathbf{Z} $$

This formulation enables analysis of AC circuits using DC-like methods with complex numbers.

Practical Computation Considerations

Modern engineering practice utilizes:

The complex conjugate (a - jb) proves particularly useful when calculating power in AC circuits:

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

where P is real power and Q is reactive power.

Complex Number Forms and Phasor Operations A diagram illustrating the geometric relationship between rectangular and polar forms of complex numbers, and how phasor arithmetic operations transform these representations. Re Im r∠θ (r e^{jθ}) θ a + jb a = r·cosθ b = r·sinθ + - × ÷ ∠ Resultant
Diagram Description: The diagram would show the geometric relationship between rectangular and polar forms of complex numbers, and how phasor arithmetic operations transform these representations.

2. Constructing Phasor Diagrams

2.1 Constructing Phasor Diagrams

Fundamentals of Phasor Representation

A phasor is a complex number representing a sinusoidal function whose amplitude (A), angular frequency (ω), and phase (φ) are time-invariant. The general form of a sinusoidal signal is:

$$ v(t) = A \cos(\omega t + \phi) $$

This can be expressed as the real part of a rotating complex vector:

$$ v(t) = \Re\{A e^{j(\omega t + \phi)}\} = \Re\{A e^{j\phi} e^{j\omega t}\} $$

The time-independent part, A ejφ, is the phasor representation. It simplifies AC circuit analysis by converting differential equations into algebraic ones.

Graphical Construction Steps

To construct a phasor diagram:

Example: RLC Series Circuit

Consider a series RLC circuit with:

$$ V_s = 100\angle 0^\circ \text{ V}, R = 5\Omega, X_L = 10\Omega, X_C = 3\Omega $$

The current phasor I is calculated first:

$$ I = \frac{V_s}{Z} = \frac{100\angle 0^\circ}{5 + j(10 - 3)} = \frac{100}{5 + j7} = 11.2\angle -54.5^\circ \text{ A} $$

Voltage phasors across each component:

$$ V_R = I \times R = 56\angle -54.5^\circ \text{ V} $$ $$ V_L = I \times jX_L = 112\angle 35.5^\circ \text{ V} $$ $$ V_C = I \times (-jX_C) = 33.6\angle -144.5^\circ \text{ V} $$

The phasor diagram shows:

Advanced Techniques

Impedance Phasor Diagrams

For networks with multiple impedances, construct an impedance phasor diagram by plotting:

$$ Z = R + jX $$

where X is the net reactance (XL - XC). The angle of Z determines the phase difference between voltage and current.

Power Phasors

Power in AC circuits is represented using:

$$ S = P + jQ = V_{rms} I_{rms}^* $$

where S (apparent power), P (real power), and Q (reactive power) form a right triangle in the phasor diagram.

Practical Applications

Phasor diagrams are critical in:

Phasor Diagram for RLC Series Circuit A vector diagram showing the voltage and current phasors in an RLC series circuit, including their magnitudes and phase angles. Real Imaginary I (11.2∠-54.5°) V_R (56∠-54.5°) V_L (112∠35.5°) V_C (33.6∠-144.5°) -54.5°
Diagram Description: The diagram would physically show the vector relationships between the voltage and current phasors in the RLC series circuit example, including their magnitudes and phase angles.

Interpreting Phase Relationships

Phase Difference in Phasor Representation

When two sinusoidal signals of the same frequency are represented as phasors, their relative phase difference manifests as the angular separation between the vectors. Consider two voltage phasors:

$$ \tilde{V}_1 = V_1 e^{j\theta_1} $$ $$ \tilde{V}_2 = V_2 e^{j\theta_2} $$

The phase difference φ between them is:

$$ \phi = \theta_2 - \theta_1 $$

This angle determines whether the signals are in-phase (φ = 0), quadrature (φ = ±90°), or out-of-phase (φ = 180°). In power systems, this relationship directly affects real and reactive power flow.

Visualizing Phase Relationships

A phasor diagram clearly shows phase relationships through vector orientation. For example:

V₁ (0°) V₂ (45°) φ=45°

Mathematical Operations on Phasors

Phasor algebra simplifies AC circuit analysis through complex arithmetic:

$$ \tilde{V}_1 + \tilde{V}_2 = (V_1\cos\theta_1 + V_2\cos\theta_2) + j(V_1\sin\theta_1 + V_2\sin\theta_2) $$

The resultant magnitude and phase are:

$$ V_{sum} = \sqrt{(V_1\cos\theta_1 + V_2\cos\theta_2)^2 + (V_1\sin\theta_1 + V_2\sin\theta_2)^2} $$ $$ \theta_{sum} = \tan^{-1}\left(\frac{V_1\sin\theta_1 + V_2\sin\theta_2}{V_1\cos\theta_1 + V_2\cos\theta_2}\right) $$

Practical Applications

Phase Measurement Techniques

Modern instrumentation uses:

$$ \phi = \sin^{-1}\left(\frac{Y_{int}}{Y_{max}}\right) $$

where Yint is the Y-axis intercept and Ymax is the peak amplitude in Lissajous analysis.

Phasor Diagram Showing Phase Difference A vector diagram showing two voltage phasors (V₁ and V₂) with a phase difference of 45 degrees, including reference axes and angle arc. Real Imaginary V₁ (0°) V₂ (45°) φ=45°
Diagram Description: The section visually demonstrates phase relationships between phasors and their angular separation, which is inherently spatial.

Applications in AC Circuit Analysis

Impedance Calculation Using Phasors

The concept of impedance Z in AC circuits generalizes resistance to include both magnitude and phase shift. For a circuit element with voltage V and current I, the impedance is defined as:

$$ Z = \frac{\tilde{V}}{\tilde{I}} = R + jX $$

where R is resistance and X is reactance. In phasor form, inductive reactance XL and capacitive reactance XC are represented as:

$$ X_L = j\omega L, \quad X_C = -\frac{j}{\omega C} $$

The total impedance of series-connected elements is the phasor sum of individual impedances. For a series RLC circuit:

$$ Z_{\text{total}} = R + j\left(\omega L - \frac{1}{\omega C}\right) $$

Power Factor and Phase Angle

The power factor (PF) quantifies the phase difference between voltage and current. It is derived from the cosine of the phase angle θ:

$$ \text{PF} = \cos \theta = \frac{P}{S} $$

where P is real power and S is apparent power. A phasor diagram visually depicts this relationship, with the voltage phasor leading (inductive) or lagging (capacitive) the current phasor. For example, in an inductive load:

$$ \theta = \tan^{-1}\left(\frac{X_L}{R}\right) $$

Kirchhoff’s Laws in Phasor Form

Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) apply to phasors in steady-state AC analysis. For a mesh with phasor voltages V1, V2, ..., Vn:

$$ \sum_{k=1}^{n} \tilde{V}_k = 0 $$

Similarly, KCL for a node with phasor currents I1, I2, ..., In is:

$$ \sum_{k=1}^{n} \tilde{I}_k = 0 $$

These laws simplify AC circuit analysis by converting differential equations into algebraic phasor equations.

Three-Phase Systems and Symmetrical Components

In balanced three-phase systems, phasors are spaced 120° apart. The line-to-neutral voltages for phases A, B, and C are:

$$ \tilde{V}_{AN} = V_p \angle 0°, \quad \tilde{V}_{BN} = V_p \angle -120°, \quad \tilde{V}_{CN} = V_p \angle 120° $$

Unbalanced systems use symmetrical components (positive, negative, zero sequences) to decompose phasors into balanced sets. For example, the positive-sequence component V1 is:

$$ \tilde{V}_1 = \frac{1}{3}\left(\tilde{V}_{AN} + a\tilde{V}_{BN} + a^2\tilde{V}_{CN}\right) $$

where a = 1 ∠120° is the Fortescue operator.

Practical Applications in Filter Design

Phasor analysis is critical in designing frequency-selective circuits. For a second-order RLC bandpass filter, the transfer function H(jω) in phasor form is:

$$ H(j\omega) = \frac{V_{\text{out}}}{V_{\text{in}}} = \frac{j\omega RC}{1 - \omega^2 LC + j\omega RC} $$

The center frequency ω0 and quality factor Q are derived from phasor impedance:

$$ \omega_0 = \frac{1}{\sqrt{LC}}, \quad Q = \frac{\omega_0 L}{R} $$
Phasor diagram for RLC series circuit V_L V_R V_C
Phasor Diagram for Series RLC Circuit A vector diagram showing voltage phasors (V_R, V_L, V_C) and current phasor (I) in a series RLC circuit, with phase angle θ. Real Imaginary I (reference) V_R V_L V_C V_total θ
Diagram Description: The section involves vector relationships (phasor sums) and phase angles that are inherently spatial, which a diagram can clearly depict.

3. Addition and Subtraction of Phasors

3.1 Addition and Subtraction of Phasors

Mathematical Representation of Phasors

Phasors represent sinusoidal signals in the complex plane, where the magnitude corresponds to the amplitude and the angle represents the phase shift. A phasor V can be expressed in rectangular and polar forms:

$$ V = V_m e^{j\phi} = V_m (\cos \phi + j \sin \phi) = a + jb $$

where Vm is the peak amplitude, ϕ is the phase angle, and a and b are the real and imaginary components, respectively.

Addition of Phasors

When adding two phasors V1 = a1 + jb1 and V2 = a2 + jb2, the resultant phasor Vr is obtained by vector addition in the complex plane:

$$ V_r = V_1 + V_2 = (a_1 + a_2) + j(b_1 + b_2) $$

For phasors in polar form, conversion to rectangular form simplifies addition. The magnitude and phase of the resultant phasor are:

$$ |V_r| = \sqrt{(a_1 + a_2)^2 + (b_1 + b_2)^2} $$ $$ \theta_r = \tan^{-1}\left( \frac{b_1 + b_2}{a_1 + a_2} \right) $$

Subtraction of Phasors

Subtraction follows a similar approach but involves negating the second phasor before addition. For Vr = V1 - V2:

$$ V_r = V_1 - V_2 = (a_1 - a_2) + j(b_1 - b_2) $$

The magnitude and phase are derived analogously:

$$ |V_r| = \sqrt{(a_1 - a_2)^2 + (b_1 - b_2)^2} $$ $$ \theta_r = \tan^{-1}\left( \frac{b_1 - b_2}{a_1 - a_2} \right) $$

Graphical Interpretation

Phasor addition and subtraction can be visualized using the parallelogram law. The resultant phasor forms the diagonal of the parallelogram constructed from the original phasors. For subtraction, the direction of the second phasor is reversed before constructing the parallelogram.

V₁ V₂ V₁ + V₂

Practical Applications

Phasor addition is fundamental in AC circuit analysis, particularly when combining voltages or currents in reactive components. For example:

Example Calculation

Consider two phasors V1 = 3 + j4 and V2 = 1 - j2. Their sum is:

$$ V_r = (3 + 1) + j(4 - 2) = 4 + j2 $$

Converted to polar form:

$$ |V_r| = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47 $$ $$ \theta_r = \tan^{-1}(2/4) \approx 26.6^\circ $$
Phasor Addition and Subtraction in Complex Plane Vector diagram showing phasor addition and subtraction in the complex plane, illustrating the parallelogram law and resultant phasor direction. Re Im V₁ V₂ V₁ + V₂ V₁ - V₂ ϕ₁ ϕ₂ θᵣ
Diagram Description: The diagram would physically show the vector addition and subtraction of phasors in the complex plane, illustrating the parallelogram law and resultant phasor direction.

3.2 Multiplication and Division of Phasors

Multiplication of Phasors in Polar Form

Given two phasors in polar form:

$$ \mathbf{A} = A \angle \theta_A $$ $$ \mathbf{B} = B \angle \theta_B $$

The product of 𝐀 and 𝐁 is obtained by multiplying their magnitudes and adding their phase angles:

$$ \mathbf{A} \cdot \mathbf{B} = (A \cdot B) \angle (\theta_A + \theta_B) $$

This follows from Euler’s formula, where:

$$ A e^{j\theta_A} \cdot B e^{j\theta_B} = AB e^{j(\theta_A + \theta_B)} $$

In practical applications, such as AC power calculations, this property simplifies the analysis of complex impedances and power factors.

Division of Phasors in Polar Form

Division of two phasors involves dividing their magnitudes and subtracting their phase angles:

$$ \frac{\mathbf{A}}{\mathbf{B}} = \left( \frac{A}{B} \right) \angle (\theta_A - \theta_B) $$

Again, using Euler’s representation:

$$ \frac{A e^{j\theta_A}}{B e^{j\theta_B}} = \frac{A}{B} e^{j(\theta_A - \theta_B)} $$

This operation is essential in circuit analysis when computing transfer functions or impedance ratios in AC circuits.

Multiplication and Division in Rectangular Form

For phasors in rectangular form (a + jb and c + jd), multiplication and division require algebraic manipulation:

Multiplication

$$ (a + jb)(c + jd) = (ac - bd) + j(ad + bc) $$

This follows from the distributive property of complex numbers and the fact that j² = -1.

Division

Division requires multiplying the numerator and denominator by the complex conjugate of the denominator:

$$ \frac{a + jb}{c + jd} = \frac{(a + jb)(c - jd)}{c^2 + d^2} = \frac{(ac + bd) + j(bc - ad)}{c^2 + d^2} $$

This ensures the denominator becomes real, simplifying the result into standard rectangular form.

Practical Applications

Phasor Multiplication and Division in Polar Form Vector diagram showing two phasors A and B, their product (A·B) with angle addition, and quotient (A/B) with angle subtraction. Re Im A∠θ_A θ_A B∠θ_B θ_B A·B∠(θ_A+θ_B) θ_A+θ_B A/B∠(θ_A-θ_B) θ_A-θ_B Phasor A Phasor B Product (A·B) Quotient (A/B)
Diagram Description: A diagram would visually demonstrate the multiplication and division of phasors in polar form, showing how magnitudes and angles combine or separate.

Impedance and Admittance in Phasor Form

Impedance in Phasor Notation

Impedance (Z) generalizes resistance to AC circuits by incorporating both magnitude and phase shift. In phasor form, it is expressed as a complex number:

$$ \mathbf{Z} = R + jX $$

where R is resistance (real part) and X is reactance (imaginary part). For inductive circuits (XL = ωL), the reactance is positive, while capacitive circuits (XC = -1/ωC) exhibit negative reactance. The polar representation is:

$$ \mathbf{Z} = |Z| \angle \theta_z \quad \text{where} \quad |Z| = \sqrt{R^2 + X^2}, \quad \theta_z = \tan^{-1}\left(\frac{X}{R}\right) $$

This phase angle (θZ) directly relates to the voltage-current phase difference in the circuit.

Admittance: The Dual Concept

Admittance (Y) is the reciprocal of impedance and simplifies parallel AC circuit analysis. It decomposes into conductance (G) and susceptance (B):

$$ \mathbf{Y} = \frac{1}{\mathbf{Z}} = G + jB $$

For parallel RLC components, the total admittance sums individual admittances. In polar form:

$$ \mathbf{Y} = |Y| \angle -\theta_z $$

Note the phase angle’s sign inversion relative to impedance, reflecting the inverse relationship.

Practical Applications

In power systems, impedance phasors model transmission line losses and voltage drops. Admittance matrices are foundational for nodal analysis in circuit simulation tools like SPICE. For example, a capacitor’s admittance (YC = jωC) directly captures its frequency-dependent behavior.

Example: Series RLC Circuit

For a series RLC circuit driven by a voltage phasor V = Vm∠0°, the total impedance is:

$$ \mathbf{Z}_{\text{total}} = R + j\left(\omega L - \frac{1}{\omega C}\right) $$

The current phasor I follows from Ohm’s law in phasor form:

$$ \mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}_{\text{total}}} = \frac{V_m}{|Z|} \angle -\theta_z $$

This reveals how the circuit’s phase response varies with frequency, particularly at resonance (ωL = 1/ωC).

Impedance Phasor Diagram A vector diagram in the complex plane showing the geometric relationship between resistance (R), reactance (X), and impedance (Z) as vectors, including the phase angle θz. Re (R) jX R jX |Z| θz = tan⁻¹(X/R)
Diagram Description: The diagram would show the geometric relationship between resistance (R), reactance (X), and impedance (Z) as vectors in the complex plane, including the phase angle θz.

4. Phasor Analysis in RLC Circuits

4.1 Phasor Analysis in RLC Circuits

Phasor analysis simplifies the study of sinusoidal steady-state behavior in RLC circuits by converting time-domain differential equations into algebraic equations in the frequency domain. The key advantage lies in representing voltages and currents as complex numbers (phasors), allowing impedance-based circuit analysis without solving differential equations directly.

Impedance Representation of RLC Components

The impedance Z of each circuit element is derived from its voltage-current relationship in the frequency domain:

$$ Z_R = R $$
$$ Z_L = j\omega L $$
$$ Z_C = \frac{1}{j\omega C} = -j\frac{1}{\omega C} $$

where ω is the angular frequency of the sinusoidal source. These impedances combine algebraically in series or parallel configurations, analogous to resistances in DC circuits.

Kirchhoff’s Laws in Phasor Form

Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) apply directly to phasors:

For a series RLC circuit driven by a voltage phasor V_s, KVL yields:

$$ V_s = V_R + V_L + V_C = I(R + j\omega L + \frac{1}{j\omega C}) $$

Phasor Diagram Construction

Phasor diagrams visualize phase relationships between circuit quantities. For a series RLC circuit:

The resultant voltage phasor V_s is the vector sum of V_R, V_L, and V_C, forming a right triangle if V_L and V_C are unequal.

Admittance and Parallel RLC Circuits

For parallel RLC circuits, admittance Y (the reciprocal of impedance) simplifies analysis:

$$ Y = \frac{1}{R} + j\omega C + \frac{1}{j\omega L} $$

The total current phasor I_total splits into components through each branch, with the capacitor current leading and inductor current lagging the voltage phasor by 90°.

Resonance and Quality Factor

At resonance (ω_0 = 1/√(LC)), the imaginary parts of impedance cancel:

$$ Z_{series} = R \quad \text{(purely resistive)} $$
$$ Y_{parallel} = \frac{1}{R} \quad \text{(purely conductive)} $$

The quality factor Q quantifies the sharpness of the resonance peak:

$$ Q_{series} = \frac{\omega_0 L}{R} \quad \text{or} \quad Q_{parallel} = R\omega_0 C $$

High-Q circuits exhibit selective frequency response, critical in filter design and RF applications.

Practical Example: Series RLC Bandpass Filter

Consider a series RLC circuit with R = 50 Ω, L = 1 mH, and C = 10 nF. The resonant frequency is:

$$ f_0 = \frac{1}{2\pi\sqrt{LC}} \approx 50.33 \text{ kHz} $$

The phasor analysis reveals maximum current at resonance, with a 3-dB bandwidth Δf = f_0/Q, where Q ≈ 6.33 for this circuit.

Series RLC Circuit Phasor Diagram A vector diagram showing the phasor relationships in a series RLC circuit, with current (I) as the horizontal reference, voltage phasors (V_R, V_L, V_C), and resultant voltage (V_s). I (0°) V_R (0°) V_L (+90°) V_C (-90°) V_s (θ) θ
Diagram Description: The section describes phasor relationships and vector sums in RLC circuits, which are inherently spatial concepts.

4.2 Power Calculations Using Phasors

Instantaneous Power in AC Circuits

The instantaneous power p(t) in an AC circuit is given by the product of the instantaneous voltage v(t) and current i(t). For sinusoidal signals:

$$ v(t) = V_m \cos(\omega t + \phi_V) $$ $$ i(t) = I_m \cos(\omega t + \phi_I) $$

The power waveform becomes:

$$ p(t) = V_m I_m \cos(\omega t + \phi_V) \cos(\omega t + \phi_I) $$

Using the trigonometric identity cos A cos B = ½[cos(A+B) + cos(A-B)], this simplifies to:

$$ p(t) = \frac{V_m I_m}{2} [\cos(2\omega t + \phi_V + \phi_I) + \cos(\phi_V - \phi_I)] $$

Real and Reactive Power

The time-averaged power P (real power) is derived from the constant term:

$$ P = \frac{V_m I_m}{2} \cos(\theta) = V_{rms} I_{rms} \cos(\theta) $$

where θ = ϕV - ϕI is the phase difference. The reactive power Q represents energy oscillating between source and load:

$$ Q = V_{rms} I_{rms} \sin(\theta) $$

Complex Power Representation

Phasor analysis simplifies power calculations through complex power S:

$$ S = V_{rms} I_{rms}^* = P + jQ $$

where Irms* is the complex conjugate of the current phasor. The magnitude of S is the apparent power:

$$ |S| = \sqrt{P^2 + Q^2} $$

Power Factor Correction

In industrial systems, inductive loads cause lagging power factors (cos θ < 1). Capacitors are added to cancel reactive power:

$$ Q_{required} = P (\tan \theta_1 - \tan \theta_2) $$

where θ1 and θ2 are initial and desired phase angles. The required capacitance is:

$$ C = \frac{Q_{required}}{\omega V_{rms}^2} $$

Three-Phase Power Systems

For balanced three-phase systems, total power is the sum of individual phase powers:

$$ P_{total} = 3 V_{phase} I_{phase} \cos(\theta) $$

In line-to-line measurements (VLL, IL):

$$ P_{total} = \sqrt{3} V_{LL} I_{L} \cos(\theta) $$
V I θ
Phasor Diagram for Power Calculations A phasor diagram showing voltage (V), current (I), phase angle (θ), real power (P), reactive power (Q), and complex power (S). Real (P) Imaginary (Q) V I θ S P Q
Diagram Description: The section involves complex relationships between voltage, current, and power in both time and phasor domains, which are inherently visual concepts.

4.3 Phasor Transformations in Three-Phase Systems

Symmetrical Components and Sequence Phasors

Three-phase systems are often analyzed using symmetrical components, which decompose unbalanced phasors into balanced sets of positive, negative, and zero-sequence components. For a three-phase voltage system with phasors Va, Vb, and Vc, the transformation is given by:

$$ \begin{bmatrix} V_0 \\ V_1 \\ V_2 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix} $$

where α = ej120° is the complex phase operator. The inverse transformation reconstructs the original phasors:

$$ \begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha^2 & \alpha \\ 1 & \alpha & \alpha^2 \end{bmatrix} \begin{bmatrix} V_0 \\ V_1 \\ V_2 \end{bmatrix} $$

Clarke (αβ0) Transformation

The Clarke transformation converts three-phase quantities into a stationary two-axis reference frame (αβ0). This simplifies control in power electronics and motor drives. The transformation matrix is:

$$ \begin{bmatrix} V_\alpha \\ V_\beta \\ V_0 \end{bmatrix} = \sqrt{\frac{2}{3}} \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix} $$

This preserves power invariance, ensuring Pabc = Pαβ0.

Park (dq0) Transformation

The Park transformation further converts αβ0 components into a rotating reference frame (dq0), synchronized with the grid frequency. This is essential for synchronous machine analysis and grid-tied inverters:

$$ \begin{bmatrix} V_d \\ V_q \\ V_0 \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} V_\alpha \\ V_\beta \\ V_0 \end{bmatrix} $$

where θ = ωt is the instantaneous phase angle. The inverse Park transformation is:

$$ \begin{bmatrix} V_\alpha \\ V_\beta \\ V_0 \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} V_d \\ V_q \\ V_0 \end{bmatrix} $$

Practical Applications

Visual Representation

A typical phasor diagram for a balanced three-phase system shows three sinusoidal voltages displaced by 120°. Under unbalanced conditions, the symmetrical components reveal the underlying positive, negative, and zero-sequence phasors.

Three-phase phasor diagram with symmetrical components
Three-Phase Phasor Decomposition into Symmetrical Components Diagram showing the decomposition of unbalanced three-phase phasors (Va, Vb, Vc) into their symmetrical components: positive-sequence (V1), negative-sequence (V2), and zero-sequence (V0) phasors. Three-Phase Phasor Decomposition into Symmetrical Components Original Unbalanced Phasors Va Vb Vc Symmetrical Components V0 V1 V2 Decomposition α = e^{j120°} (Phase Shift Operator) 120° 120°
Diagram Description: The diagram would show the spatial relationships between positive, negative, and zero-sequence phasors in a three-phase system, illustrating how symmetrical components decompose unbalanced conditions.

5. Recommended Textbooks

5.1 Recommended Textbooks

5.2 Online Resources and Tutorials

5.3 Advanced Topics in Phasor Analysis