13.3 First-Order & Second-Order Circuit Transients

13.3 First-Order & Second-Order Circuit Transients

Circuit transients represent the temporary transition state when an electrical network shifts from one steady-state condition to another due to switching operations, short-circuit faults, or sudden load changes. For the PRC Registered Electrical Engineer (REE) Licensure Examination, mastering time constants, step response formulas, second-order $RLC$ damping classifications, and Laplace transform $s$-domain methods is vital.

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1. Fundamentals of Transients & Initial Energy Continuity

Transients arise because energy stored in reactive elements cannot change instantaneously:

• Inductor Magnetic Energy: $W_L = \frac{1}{2} L i_L^2 \implies$ Inductor current cannot change instantaneously: $i_L(0^+) = i_L(0^-)$
• Capacitor Electric Energy: $W_C = \frac{1}{2} C v_C^2 \implies$ Capacitor voltage cannot change instantaneously: $v_C(0^+) = v_C(0^-)$

where $t = 0^-$ represents the instant immediately prior to switching, and $t = 0^+$ represents the instant immediately after switching.

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2. First-Order $RC$ and $RL$ Transients

First-order circuits contain a single energy storage element (or equivalent $L$ or $C$) alongside resistive networks.

Time Constant (\tau) Equations

• $RC$ Circuit Time Constant: $\tau = R_{\text{eq}} C \quad (\text{seconds})$
• $RL$ Circuit Time Constant: $\tau = \frac{L}{R_{\text{eq}}} \quad (\text{seconds})$

where $R_{\text{eq}}$ is the Thevenin equivalent resistance seen from the terminals of the storage element.

General First-Order Step Response Formula

For any first-order variable $x(t)$ (representing capacitor voltage $v_C(t)$ or inductor current $i_L(t)$):

$x(t) = x(\infty) + \left[ x(0^+) - x(\infty) \right] e^{-t/\tau} \quad \text{for } t \ge 0$

where:

• $x(0^+)$ is the initial value immediately after switching;
• $x(\infty)$ is the final steady-state value as $t \to \infty$;
• $\tau$ is the circuit time constant.

Time Constant Progression & Response Characteristics

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3. Second-Order $RLC$ Transients & Damping Regimes

Second-order circuits contain two uncoupled energy storage elements ($L$ and $C$), yielding a second-order linear differential equation.

Governing Differential Equations

• Series $RLC$ Circuit: $L \frac{d^2 i(t)}{dt^2} + R \frac{di(t)}{dt} + \frac{1}{C} i(t) = 0 \implies \frac{d^2 i}{dt^2} + 2\alpha \frac{di}{dt} + \omega_0^2 i = 0$
• Parallel $RLC$ Circuit: $C \frac{d^2 v(t)}{dt^2} + \frac{1}{R} \frac{dv(t)}{dt} + \frac{1}{L} v(t) = 0 \implies \frac{d^2 v}{dt^2} + 2\alpha \frac{dv}{dt} + \omega_0^2 v = 0$

Key Parameters

1. Undamped Natural Frequency (\omega_0): $\omega_0 = \frac{1}{\sqrt{LC}} \quad (\text{rad/s})$
2. Damping Attenuation Factor (\alpha): $\text{Series } RLC: \alpha = \frac{R}{2L}, \qquad \text{Parallel } RLC: \alpha = \frac{1}{2RC}$
3. Damping Ratio (\zeta): $\zeta = \frac{\alpha}{\omega_0}$

Characteristic Equation & Roots

Substituting $x(t) = A e^{st}$ yields the characteristic equation:

$s^2 + 2\alpha s + \omega_0^2 = 0$

Roots are given by the quadratic formula:

$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$

Classification of Damping Regimes

$\text{Damped Natural Frequency: } \omega_d = \sqrt{\omega_0^2 - \alpha^2} \quad (\text{rad/s}), \qquad f_d = \frac{\omega_d}{2\pi} \quad (\text{Hz})$

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4. Laplace Transform ($s$-Domain) Methods for Transients

Laplace transform converts differential equations into algebraic equations in the complex frequency domain ($s = \sigma + j\omega$).

Key Laplace Transform Pairs

• $\mathcal{L}{ 1 } = \frac{1}{s}$
• $\mathcal{L}{ e^{-at} } = \frac{1}{s + a}$
• $\mathcal{L}{ \sin(\omega t) } = \frac{\omega}{s^2 + \omega^2}$
• $\mathcal{L}{ \cos(\omega t) } = \frac{s}{s^2 + \omega^2}$
• $\mathcal{L}\left{ \frac{df(t)}{dt} \right} = s F(s) - f(0^+)$
• $\mathcal{L}\left{ \frac{d^2f(t)}{dt^2} \right} = s^2 F(s) - s f(0^+) - f'(0^+)$

Circuit Element Equivalent $s$-Domain Models

• Resistor ($R$): $Z(s) = R$
• Inductor ($L$): Series combination of impedance $sL$ and voltage source $L i_L(0^+)$ (or parallel current source $\frac{i_L(0^+)}{s}$)
• Capacitor ($C$): Series combination of impedance $\frac{1}{sC}$ and voltage source $\frac{v_C(0^+)}{s}$

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5. Power System Switching Transients & Protective Measures

In high-voltage power networks, transient phenomena can trigger destructive voltage spikes and insulation breakdown:

1. Transient Recovery Voltage (TRV): When a circuit breaker interrupts an inductive fault current, high frequency oscillations occur across the opening contacts as energy transfers between grid inductance and stray capacitance.
2. Inductive Kickback ($v_L = L \frac{di}{dt}$): Opening an inductive circuit (such as a motor field winding or transformer) abruptly produces severe overvoltage surges. Protection requires RC snubber circuits, flyback diodes, or metal-oxide varistors (MOVs).
3. Capacitor Bank Inrush Current: Energizing distribution capacitor banks creates high-magnitude, high-frequency current transients. Damping reactors or pre-insertion resistors are installed to limit inrush currents.

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Solved Board Exam Numerical Problems

Problem 1: First-Order $RL$ Transient Step Response

Question: A series $RL$ circuit containing a resistor $R = 10\ \Omega$ and an inductor $L = 2.0\ \text{H}$ is connected to a constant $100\ \text{V}$ DC source at $t = 0$. Assuming zero initial current ($i_L(0^-) = 0$), determine: (a) the circuit time constant $\tau$, (b) the current at $t = 0.10\ \text{s}$, (c) the steady-state current as $t \to \infty$, and (d) the time required for current to reach $90%$ of its final value.

Solution:

1. Calculate time constant $\tau$: $\tau = \frac{L}{R} = \frac{2.0\ \text{H}}{10\ \Omega} = 0.20\ \text{seconds}$

2. Determine steady-state current $i(\infty)$: $i(\infty) = \frac{V}{R} = \frac{100\ \text{V}}{10\ \Omega} = 10\ \text{Amperes}$

3. Apply first-order step response equation: $i(t) = i(\infty) + [i(0^+) - i(\infty)] e^{-t/\tau} = 10 + [0 - 10] e^{-t/0.20} = 10 \left( 1 - e^{-5t} \right)\ \text{A}$

4. Compute current at $t = 0.10\ \text{s}$: $i(0.10) = 10 \left( 1 - e^{-5(0.10)} \right) = 10 \left( 1 - e^{-0.5} \right) = 10 (1 - 0.6065) = 3.935\ \text{Amperes}$

5. Calculate time $t$ to reach $90%$ ($9.0\ \text{A}$): $9.0 = 10 \left( 1 - e^{-t/0.20} \right) \implies 0.90 = 1 - e^{-t/0.20} \implies e^{-t/0.20} = 0.10$ $-\frac{t}{0.20} = \ln(0.10) = -2.3026 \implies t = 0.20 \times 2.3026 = 0.4605\ \text{seconds}$

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Problem 2: Series $RLC$ Transient Response Classification & Frequency

Question: A series $RLC$ circuit has parameters $R = 40\ \Omega$, $L = 100\ \text{mH}$ ($0.10\ \text{H}$), and $C = 25\ \mu\text{F}$ ($25 \times 10^{-6}\ \text{F}$). Determine: (a) the attenuation factor $\alpha$, (b) the undamped natural frequency $\omega_0$, (c) the damping regime classification, and (d) the damped natural frequency $\omega_d$ and cyclic frequency $f_d$.

Solution:

1. Calculate attenuation factor $\alpha$ for series $RLC$: $\alpha = \frac{R}{2L} = \frac{40}{2 \times 0.10} = \frac{40}{0.20} = 200\ \text{rad/s}$

2. Calculate undamped natural frequency $\omega_0$: $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.10 \times 25 \times 10^{-6}}} = \frac{1}{\sqrt{2.5 \times 10^{-6}}} = \frac{1}{0.0015811} = 632.46\ \text{rad/s}$

3. Classify damping regime by comparing $\alpha$ and $\omega_0$: $\text{Since } \alpha = 200\ \text{rad/s} < \omega_0 = 632.46\ \text{rad/s}, \text{ the system is } \mathbf{UNDERDAMPED} \text{ (oscillatory).}$

4. Calculate damped natural frequency $\omega_d$ and cyclic frequency $f_d$: $\omega_d = \sqrt{\omega_0^2 - \alpha^2} = \sqrt{(632.46)^2 - (200)^2} = \sqrt{400,000 - 40,000} = \sqrt{360,000} = 600\ \text{rad/s}$ $f_d = \frac{\omega_d}{2\pi} = \frac{600}{2\pi} = 95.49\ \text{Hz}$