7.2 Electromagnetics & Passive Components

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7.2 Electromagnetics & Passive Components

Key Takeaways Explain This Topic AI

Faraday's Law of Electromagnetic Induction states induced EMF is proportional to time rate of magnetic flux change ( e = − N d Φ d t e = -N \frac{d\Phi}{dt} e = − N d t d Φ ​ ), with Lenz's Law determining the opposing polarity.

Magnetic flux density ( B = μ H = Φ A B = \mu H = \frac{\Phi}{A} B = μ H = A Φ ​ ) and magnetomotive force ( F = N I = Φ R \mathcal{F} = N I = \Phi \mathcal{R} F = N I = Φ R ) describe magnetic circuits analogous to Ohm's law in electric circuits.

Passive component behaviors (R, L, C) are governed by fundamental field relations, temperature coefficients, dielectric permeability, and core saturation characteristics.

RC and RL first-order transient responses follow exponential charging/discharging curves governed by time constants τ = R C \tau = RC τ = RC and τ = L / R \tau = L/R τ = L / R , taking 5 τ 5\tau 5 τ to reach ≈ 99.3 % \approx 99.3\% ≈ 99.3% of final steady state.

Second-order RLC transient responses are categorized into overdamped ( α > ω 0 \alpha > \omega_0 α > ω 0 ​ ), critically damped ( α = ω 0 \alpha = \omega_0 α = ω 0 ​ ), and underdamped ( α < ω 0 \alpha < \omega_0 α < ω 0 ​ ) regimes based on the damping factor α = R / ( 2 L ) \alpha = R/(2L) α = R / ( 2 L ) relative to natural frequency ω 0 = 1 / L C \omega_0 = 1/\sqrt{LC} ω 0 ​ = 1/ L C ​ .

Last updated: July 2026

7.2 Electromagnetics & Passive Components

Quick Answer: Faraday's Law ( e = − N d Φ d t e = -N \frac{d\Phi}{dt} e = − N d t d Φ ​ ) and Lenz's Law define electromagnetic induction. Magnetic circuits operate via Ohm's Law analogue: Magnetomotive Force F = N I = Φ R \mathcal{F} = NI = \Phi \mathcal{R} F = N I = Φ R , where reluctance R = l μ A \mathcal{R} = \frac{l}{\mu A} R = μ A l ​ . Passive components store energy in electric fields ( E C = 1 2 C V 2 E_C = \frac{1}{2} C V^2 E C ​ = 2 1 ​ C V 2 ) and magnetic fields ( E L = 1 2 L I 2 E_L = \frac{1}{2} L I^2 E L ​ = 2 1 ​ L I 2 ). First-order transient response is determined by time constants τ = R C \tau = RC τ = RC (for RC circuits) and τ = L R \tau = \frac{L}{R} τ = R L ​ (for RL circuits), reaching steady state in 5 τ 5\tau 5 τ . Second-order RLC transients are underdamped, critically damped, or overdamped based on damping factor α = R 2 L \alpha = \frac{R}{2L} α = 2 L R ​ vs natural frequency ω 0 = 1 L C \omega_0 = \frac{1}{\sqrt{LC}} ω 0 ​ = L C ​ 1 ​ .

Fundamentals of Electromagnetics & Magnetic Circuits

Electromagnetism connects electric currents with magnetic fields, governing the behavior of inductors, transformers, relays, and electric motors.

1. Magnetic Field Quantities & Relations

Magnetic Flux ( Φ \Phi Φ ): Total magnetic lines of force, measured in Webers (Wb).

Magnetic Flux Density ( B B B ): Flux per unit area perpendicular to the magnetic field, measured in Tesla (T) or Wb/m 2 ^2 2 : B = Φ A B = \frac{\Phi}{A} B = A Φ ​

Magnetic Field Intensity ( H H H ): Magnetizing force per unit length, measured in A-turns/m: H = N I l H = \frac{N I}{l} H = l N I ​

Permeability ( μ \mu μ ): Ability of a medium to conduct magnetic flux: μ = μ 0 μ r \mu = \mu_0 \mu_r μ = μ 0 ​ μ r ​ , where μ 0 = 4 π × 10 − 7 H/m \mu_0 = 4\pi \times 10^{-7}\text{ H/m} μ 0 ​ = 4 π × 1 0 − 7 H/m is free space permeability and μ r \mu_r μ r ​ is relative permeability. B = μ H = μ 0 μ r H B = \mu H = \mu_0 \mu_r H B = μ H = μ 0 ​ μ r ​ H

2. Electric vs. Magnetic Circuit Analogy

Magnetic circuits can be analyzed using an equivalent of Ohm's Law:

Parameter Electric Circuit Magnetic Circuit Analogy Formula

Driving Force Electromotive Force ( E E E or V V V , Volts) Magnetomotive Force ( F = N I \mathcal{F} = N I F = N I , Ampere-turns) F = N I \mathcal{F} = N I F = N I

Flow Quantity Electric Current ( I I I , Amperes) Magnetic Flux ( Φ \Phi Φ , Webers) Φ = F R \Phi = \frac{\mathcal{F}}{\mathcal{R}} Φ = R F ​

Opposition Resistance ( R = ρ l A R = \rho \frac{l}{A} R = ρ A l ​ , Ohms) Reluctance ( R = l μ A \mathcal{R} = \frac{l}{\mu A} R = μ A l ​ , A-t/Wb) R = l μ 0 μ r A \mathcal{R} = \frac{l}{\mu_0 \mu_r A} R = μ 0 ​ μ r ​ A l ​

Conductance/Permeance Conductance ( G = 1 R G = \frac{1}{R} G = R 1 ​ , Siemens) Permeance ( P = 1 R \mathcal{P} = \frac{1}{\mathcal{R}} P = R 1 ​ , Wb/A-t) P = μ A l \mathcal{P} = \frac{\mu A}{l} P = l μ A ​

Constitutive Law Ohm's Law: I = V R I = \frac{V}{R} I = R V ​ Hopkinson's Law: Φ = F R \Phi = \frac{\mathcal{F}}{\mathcal{R}} Φ = R F ​ F = Φ R \mathcal{F} = \Phi \mathcal{R} F = Φ R

3. Faraday's & Lenz's Laws of Induction

Faraday's Law: An electromotive force (EMF) is induced in a conductor or coil whenever the magnetic flux linking it changes over time: e = − N d Φ d t e = -N \frac{d\Phi}{dt} e = − N d t d Φ ​

Lenz's Law: The minus sign represents Lenz's Law: the induced EMF generates a current whose magnetic field opposes the original change in magnetic flux that created it.

Self-Inductance ( L L L ) & Mutual Inductance ( M M M ): e L = − L d i d t ⟹ L = N Φ I = N 2 μ A l e_L = -L \frac{di}{dt} \implies L = \frac{N \Phi}{I} = \frac{N^2 \mu A}{l} e L ​ = − L d t d i ​ ⟹ L = I N Φ ​ = l N 2 μ A ​ M = k L 1 L 2 ( 0 ≤ k ≤ 1 , magnetic coupling coefficient ) M = k \sqrt{L_1 L_2} \quad (0 \le k \le 1, \text{ magnetic coupling coefficient}) M = k L 1 ​ L 2 ​ ​ ( 0 ≤ k ≤ 1 , magnetic coupling coefficient )

Passive Components: Energy Storage & Properties

Resistors, capacitors, and inductors form the foundation of electronic circuits.

1. Capacitors & Energy Storage

A parallel-plate capacitor stores energy in an electric field between its dielectric-separated plates: C = ϵ A d = ϵ 0 ϵ r A d C = \frac{\epsilon A}{d} = \frac{\epsilon_0 \epsilon_r A}{d} C = d ϵ A ​ = d ϵ 0 ​ ϵ r ​ A ​

where ϵ 0 = 8.854 × 10 − 12 F/m \epsilon_0 = 8.854 \times 10^{-12}\text{ F/m} ϵ 0 ​ = 8.854 × 1 0 − 12 F/m .

Instantaneous Current-Voltage Relation: i ( t ) = C d v ( t ) d t i(t) = C \frac{dv(t)}{dt} i ( t ) = C d t d v ( t ) ​

Stored Energy ( E C E_C E C ​ ): E C = 1 2 C V 2 E_C = \frac{1}{2} C V^2 E C ​ = 2 1 ​ C V 2 (Joules)

Key Property: Voltage across a capacitor cannot change instantaneously ( v ( 0 + ) = v ( 0 − ) v(0^+) = v(0^-) v ( 0 + ) = v ( 0 − ) ).

2. Inductors & Energy Storage

An inductor stores energy in a magnetic field established by current flow through its turns:

Instantaneous Current-Voltage Relation: v ( t ) = L d i ( t ) d t v(t) = L \frac{di(t)}{dt} v ( t ) = L d t d i ( t ) ​

Stored Energy ( E L E_L E L ​ ): E L = 1 2 L I 2 E_L = \frac{1}{2} L I^2 E L ​ = 2 1 ​ L I 2 (Joules)

Key Property: Current through an inductor cannot change instantaneously ( i ( 0 + ) = i ( 0 − ) i(0^+) = i(0^-) i ( 0 + ) = i ( 0 − ) ).

• First-Order & Second-Order Transient Responses
• 1. First-Order RC & RL Transient Response

When a step DC voltage V S V_S V S ​ is applied at t = 0 t=0 t = 0 :

Charging RC Circuit: v C ( t ) = V S ( 1 − e − t / τ ) , i C ( t ) = V S R e − t / τ ( where τ = R C ) v_C(t) = V_S \left(1 - e^{-t/\tau}\right), \quad i_C(t) = \frac{V_S}{R} e^{-t/\tau} \quad (\text{where } \tau = RC) v C ​ ( t ) = V S ​ ( 1 − e − t / τ ) , i C ​ ( t ) = R V S ​ ​ e − t / τ ( where τ = RC )

Discharging RC Circuit: v C ( t ) = V 0 e − t / τ , i C ( t ) = − V 0 R e − t / τ v_C(t) = V_0 e^{-t/\tau}, \quad i_C(t) = -\frac{V_0}{R} e^{-t/\tau} v C ​ ( t ) = V 0 ​ e − t / τ , i C ​ ( t ) = − R V 0 ​ ​ e − t / τ

Energizing RL Circuit: i L ( t ) = V S R ( 1 − e − t / τ ) , v L ( t ) = V S e − t / τ ( where τ = L R ) i_L(t) = \frac{V_S}{R} \left(1 - e^{-t/\tau}\right), \quad v_L(t) = V_S e^{-t/\tau} \quad \left(\text{where } \tau = \frac{L}{R}\right) i L ​ ( t ) = R V S ​ ​ ( 1 − e − t / τ ) , v L ​ ( t ) = V S ​ e − t / τ ( where τ = R L ​ )

• Time Elapsed Percentage of Final Steady-State Value Charging Expression Factor
• 1 τ 1\tau 1 τ 63.2 % 63.2\% 63.2% 1 − e − 1 ≈ 0.6321 1 - e^{-1} \approx 0.6321 1 − e − 1 ≈ 0.6321
• 2 τ 2\tau 2 τ 86.5 % 86.5\% 86.5% 1 − e − 2 ≈ 0.8647 1 - e^{-2} \approx 0.8647 1 − e − 2 ≈ 0.8647
• 3 τ 3\tau 3 τ 95.0 % 95.0\% 95.0% 1 − e − 3 ≈ 0.9502 1 - e^{-3} \approx 0.9502 1 − e − 3 ≈ 0.9502
• 4 τ 4\tau 4 τ 98.2 % 98.2\% 98.2% 1 − e − 4 ≈ 0.9817 1 - e^{-4} \approx 0.9817 1 − e − 4 ≈ 0.9817

5 τ 5\tau 5 τ 99.3 % 99.3\% 99.3% (considered full steady-state) 1 − e − 5 ≈ 0.9933 1 - e^{-5} \approx 0.9933 1 − e − 5 ≈ 0.9933

2. Second-Order RLC Transient Response

The transient differential equation for a series RLC circuit driven by a step voltage is: d 2 i d t 2 + R L d i d t + 1 L C i = 0 \frac{d^2 i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{1}{LC} i = 0 d t 2 d 2 i ​ + L R ​ d t d i ​ + L C 1 ​ i = 0

The characteristic equation roots s 1 , 2 s_{1,2} s 1 , 2 ​ are: s 1 , 2 = − α ± α 2 − ω 0 2 s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} s 1 , 2 ​ = − α ± α 2 − ω 0 2 ​ ​

where α = R 2 L \alpha = \frac{R}{2L} α = 2 L R ​ is the neper damping factor and ω 0 = 1 L C \omega_0 = \frac{1}{\sqrt{LC}} ω 0 ​ = L C ​ 1 ​ is the undamped resonant frequency.

Depending on the relationship between α \alpha α and ω 0 \omega_0 ω 0 ​ , three transient response regimes exist:

Overdamped ( α > ω 0 ⟹ R > 2 L C \alpha > \omega_0 \implies R > 2\sqrt{\frac{L}{C}} α > ω 0 ​ ⟹ R > 2 C L ​ ​ ): Real, distinct roots ( s 1 , s 2 s_1, s_2 s 1 ​ , s 2 ​ ). Response returns to equilibrium slowly without oscillation.

Critically Damped ( α = ω 0 ⟹ R = 2 L C \alpha = \omega_0 \implies R = 2\sqrt{\frac{L}{C}} α = ω 0 ​ ⟹ R = 2 C L ​ ​ ): Real, equal roots ( s 1 = s 2 = − α s_1 = s_2 = -\alpha s 1 ​ = s 2 ​ = − α ). Response returns to equilibrium in the shortest possible time without overshoot.

Underdamped ( α < ω 0 ⟹ R < 2 L C \alpha < \omega_0 \implies R < 2\sqrt{\frac{L}{C}} α < ω 0 ​ ⟹ R < 2 C L ​ ​ ): Complex conjugate roots ( s 1 , 2 = − α ± j ω d s_{1,2} = -\alpha \pm j\omega_d s 1 , 2 ​ = − α ± j ω d ​ ). Response oscillates with damped natural frequency ω d = ω 0 2 − α 2 \omega_d = \sqrt{\omega_0^2 - \alpha^2} ω d ​ = ω 0 2 ​ − α 2 ​ while exponentially decaying.

• Step-by-Step Worked Examples
• Example 1: Magnetic Reluctance & Flux Calculation

Problem: An iron magnetic core has a mean path length l = 0.5 m l = 0.5\text{ m} l = 0.5 m , cross-sectional area A = 10 cm 2 A = 10\text{ cm}^2 A = 10 cm 2 ( 10 − 3 m 2 10^{-3}\text{ m}^2 1 0 − 3 m 2 ), relative permeability μ r = 2000 \mu_r = 2000 μ r ​ = 2000 , and a coil of N = 500 turns N = 500\text{ turns} N = 500 turns . Calculate: (a) Reluctance R \mathcal{R} R , and (b) Magnetic flux Φ \Phi Φ when coil current I = 2 A I = 2\text{ A} I = 2 A .

Solution:

Calculate Reluctance R \mathcal{R} R : R = l μ 0 μ r A = 0.5 ( 4 π × 10 − 7 ) × ( 2000 ) × ( 10 − 3 ) = 0.5 8 π × 10 − 7 = 0.5 2.5133 × 10 − 6 ≈ 198 , 944 A-t/Wb \mathcal{R} = \frac{l}{\mu_0 \mu_r A} = \frac{0.5}{(4\pi \times 10^{-7}) \times (2000) \times (10^{-3})} = \frac{0.5}{8\pi \times 10^{-7}} = \frac{0.5}{2.5133 \times 10^{-6}} \approx 198,944\text{ A-t/Wb} R = μ 0 ​ μ r ​ A l ​ = ( 4 π × 1 0 − 7 ) × ( 2000 ) × ( 1 0 − 3 ) 0.5 ​ = 8 π × 1 0 − 7 0.5 ​ = 2.5133 × 1 0 − 6 0.5 ​ ≈ 198 , 944 A-t/Wb

Calculate MMF ( F \mathcal{F} F ): F = N I = 500 × 2 = 1000 A-turns \mathcal{F} = N I = 500 \times 2 = 1000\text{ A-turns} F = N I = 500 × 2 = 1000 A-turns

Calculate Flux ( Φ \Phi Φ ): Φ = F R = 1000 198 , 944 ≈ 5.027 × 10 − 3 Wb = 5.027 mWb \Phi = \frac{\mathcal{F}}{\mathcal{R}} = \frac{1000}{198,944} \approx 5.027 \times 10^{-3}\text{ Wb} = 5.027\text{ mWb} Φ = R F ​ = 198 , 944 1000 ​ ≈ 5.027 × 1 0 − 3 Wb = 5.027 mWb

Example 2: RC Transient Capacitor Voltage Calculation

Problem: A 100 μ F 100\ \mu\text{F} 100 μ F capacitor is charged through a 50 k Ω 50\text{ k}\Omega 50 k Ω resistor from a 12 V 12\text{ V} 12 V DC source. Determine: (a) Time constant τ \tau τ , and (b) Capacitor voltage at t = 10 seconds t = 10\text{ seconds} t = 10 seconds .

Solution:

Calculate Time Constant τ \tau τ : τ = R C = ( 50 × 10 3 Ω ) × ( 100 × 10 − 6 F ) = 5.0 seconds \tau = R C = (50 \times 10^3\ \Omega) \times (100 \times 10^{-6}\text{ F}) = 5.0\text{ seconds} τ = RC = ( 50 × 1 0 3 Ω ) × ( 100 × 1 0 − 6 F ) = 5.0 seconds

Calculate Voltage at t = 10 s t = 10\text{ s} t = 10 s ( t = 2 τ t = 2\tau t = 2 τ ): v C ( 10 ) = 12 ( 1 − e − 10 / 5 ) = 12 ( 1 − e − 2 ) = 12 ( 1 − 0.1353 ) = 12 × 0.8647 ≈ 10.38 V v_C(10) = 12 \left(1 - e^{-10/5}\right) = 12 \left(1 - e^{-2}\right) = 12 (1 - 0.1353) = 12 \times 0.8647 \approx 10.38\text{ V} v C ​ ( 10 ) = 12 ( 1 − e − 10/5 ) = 12 ( 1 − e − 2 ) = 12 ( 1 − 0.1353 ) = 12 × 0.8647 ≈ 10.38 V