3.1 Differential Equations & Circuit Transients

PRC Electronics Engineering Licensure Examination Study Guide Exam

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1 Chapter 1: Introduction & Board Exam Strategy

1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy

2 Chapter 2: Mathematics Foundations

2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications

3 Chapter 3: Advanced Mathematics for ECE

3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability

4 Chapter 4: GEAS - Engineering Sciences & Economics

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5 Chapter 5: GEAS - Management, Materials, Environment & CAD

5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics

6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts

6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)

7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes

7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators

8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits

8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design

9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems

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10 Chapter 10: EST - Communications Fundamentals & Transmission Lines

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11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems

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3.1 Differential Equations & Circuit Transients

Key Takeaways Explain This Topic AI

First-order differential equations model RL and RC transient circuits, with time constants τ = R C \tau = RC τ = RC and τ = L / R \tau = L/R τ = L / R governing exponential charging/discharging rates.

Second-order linear ODEs model RLC circuits, exhibiting three distinct transient responses based on the damping factor α \alpha α and natural frequency ω 0 \omega_0 ω 0 ​ : overdamped, critically damped, and underdamped.

The complete transient response of a non-homogeneous ODE consists of the natural response y h ( t ) y_h(t) y h ​ ( t ) plus the particular (forced) response y p ( t ) y_p(t) y p ​ ( t ) .

Physical continuity of capacitor voltage v C ( 0 − ) = v C ( 0 + ) v_C(0^-) = v_C(0^+) v C ​ ( 0 − ) = v C ​ ( 0 + ) and inductor current i L ( 0 − ) = i L ( 0 + ) i_L(0^-) = i_L(0^+) i L ​ ( 0 − ) = i L ​ ( 0 + ) provides essential initial conditions for evaluating integration constants.

Last updated: July 2026

3.1 Differential Equations & Circuit Transients

Transient analysis is a core competency evaluated in the PRC Electronics Engineering Licensure Examination. When switches operate in electrical networks, voltages and currents transition from an initial state to a final steady state. Ordinary Differential Equations (ODEs) provide the fundamental mathematical framework for describing these time-dependent physical phenomena in circuits containing energy storage elements (capacitors and inductors).

1. Fundamentals of Ordinary Differential Equations in ECE

An Ordinary Differential Equation (ODE) relates an unknown function y ( t ) y(t) y ( t ) to its derivatives with respect to a single independent variable, typically time t t t . The order of a differential equation is defined by the highest derivative present in the equation.

In circuit theory, first-order differential equations govern networks with a single energy storage element (or an equivalent single capacitor/inductor), whereas second-order differential equations govern networks with two non-combinable energy storage elements (such as RLC circuits).

1.1 First-Order Separable & Linear Differential Equations

A first-order ordinary differential equation is termed separable if it can be written in the form: d y d t = g ( t ) h ( y ) ⟹ 1 h ( y ) d y = g ( t ) d t \frac{dy}{dt} = g(t) h(y) \implies \frac{1}{h(y)} dy = g(t) dt d t d y ​ = g ( t ) h ( y ) ⟹ h ( y ) 1 ​ d y = g ( t ) d t

Integrating both sides directly yields the implicit or explicit solution for y ( t ) y(t) y ( t ) .

A general first-order linear differential equation takes the standard form: d y d t + P ( t ) y = Q ( t ) \frac{dy}{dt} + P(t)y = Q(t) d t d y ​ + P ( t ) y = Q ( t )

To solve this non-homogeneous linear equation, we multiply the entire equation by an Integrating Factor I ( t ) I(t) I ( t ) : I ( t ) = e ∫ P ( t ) d t I(t) = e^{\int P(t) dt} I ( t ) = e ∫ P ( t ) d t

Multiplying through yields d d t [ I ( t ) y ( t ) ] = I ( t ) Q ( t ) \frac{d}{dt}[I(t) y(t)] = I(t) Q(t) d t d ​ [ I ( t ) y ( t )] = I ( t ) Q ( t ) , which can be integrated directly: y ( t ) = 1 I ( t ) [ ∫ I ( t ) Q ( t ) d t + C ] y(t) = \frac{1}{I(t)} \left[ \int I(t) Q(t) dt + C \right] y ( t ) = I ( t ) 1 ​ [ ∫ I ( t ) Q ( t ) d t + C ]

2. First-Order RC and RL Transient Analysis

First-order circuit transients occur when a DC voltage or current source is suddenly connected to or disconnected from an R C RC RC or R L RL R L network.

2.1 RC Circuit Charging and Discharging

For a series R C RC RC circuit connected to a DC source V C C V_{CC} V CC ​ at t = 0 t = 0 t = 0 : R i ( t ) + v C ( t ) = V C C ⟹ R C d v C ( t ) d t + v C ( t ) = V C C R i(t) + v_C(t) = V_{CC} \implies R C \frac{dv_C(t)}{dt} + v_C(t) = V_{CC} R i ( t ) + v C ​ ( t ) = V CC ​ ⟹ RC d t d v C ​ ( t ) ​ + v C ​ ( t ) = V CC ​

This is a first-order linear non-homogeneous differential equation. Solving with initial condition v C ( 0 ) = V 0 v_C(0) = V_0 v C ​ ( 0 ) = V 0 ​ : v C ( t ) = V C C + ( V 0 − V C C ) e − t / τ v_C(t) = V_{CC} + (V_0 - V_{CC}) e^{-t/\tau} v C ​ ( t ) = V CC ​ + ( V 0 ​ − V CC ​ ) e − t / τ

Where the time constant τ \tau τ for an R C RC RC circuit is defined as: τ = R e q C e q \tau = R_{eq} C_{eq} τ = R e q ​ C e q ​

Operational Phase Capacitor Voltage v C ( t ) v_C(t) v C ​ ( t ) Capacitor Current i C ( t ) i_C(t) i C ​ ( t )

Charging ( V 0 = 0 V_0 = 0 V 0 ​ = 0 ) v C ( t ) = V C C ( 1 − e − t / τ ) v_C(t) = V_{CC} \left(1 - e^{-t/\tau}\right) v C ​ ( t ) = V CC ​ ( 1 − e − t / τ ) i C ( t ) = V C C R e − t / τ i_C(t) = \frac{V_{CC}}{R} e^{-t/\tau} i C ​ ( t ) = R V CC ​ ​ e − t / τ

Discharging ( V C C = 0 V_{CC} = 0 V CC ​ = 0 ) v C ( t ) = V 0 e − t / τ v_C(t) = V_0 e^{-t/\tau} v C ​ ( t ) = V 0 ​ e − t / τ i C ( t ) = − V 0 R e − t / τ i_C(t) = -\frac{V_0}{R} e^{-t/\tau} i C ​ ( t ) = − R V 0 ​ ​ e − t / τ

2.2 RL Circuit Current Growth and Decay

For an R L RL R L circuit energized by a DC voltage source V V V : L d i ( t ) d t + R i ( t ) = V L \frac{di(t)}{dt} + R i(t) = V L d t d i ( t ) ​ + R i ( t ) = V

The solution for current i L ( t ) i_L(t) i L ​ ( t ) given initial current I 0 I_0 I 0 ​ is: ∋ L ( t ) = V R + ( I 0 − V R ) e − t / τ \ni_L(t) = \frac{V}{R} + \left(I_0 - \frac{V}{R}\right) e^{-t/\tau} ∋ L ​ ( t ) = R V ​ + ( I 0 ​ − R V ​ ) e − t / τ

Where the time constant τ \tau τ for an R L RL R L circuit is: τ = L e q R e q \tau = \frac{L_{eq}}{R_{eq}} τ = R e q ​ L e q ​ ​

2.3 Physical Meaning of the Time Constant τ \tau τ

At t = 1 τ t = 1\tau t = 1 τ : The transient quantity reaches 63.2 % 63.2\% 63.2% of its final change during charging, or decays to 36.8 % 36.8\% 36.8% of its initial value during discharging.

At t = 5 τ t = 5\tau t = 5 τ : The transient response reaches 99.3 % 99.3\% 99.3% of its final value, which is universally considered in engineering practice as the achievement of steady-state .

3. Second-Order Homogeneous ODEs & RLC Transients

A series R L C RLC R L C circuit connected in a closed loop without an external source yields Kirchhoff's Voltage Law equation: L d i ( t ) d t + R i ( t ) + 1 C ∫ 0 t i ( τ ) d τ + v C ( 0 ) = 0 L \frac{d i(t)}{dt} + R i(t) + \frac{1}{C} \int_0^t i(\tau) d\tau + v_C(0) = 0 L d t d i ( t ) ​ + R i ( t ) + C 1 ​ ∫ 0 t ​ i ( τ ) d τ + v C ​ ( 0 ) = 0

Differentiating with respect to t t t gives the second-order homogeneous linear ODE: d 2 i ( t ) d t 2 + R L d i ( t ) d t + 1 L C i ( t ) = 0 \frac{d^2 i(t)}{dt^2} + \frac{R}{L} \frac{di(t)}{dt} + \frac{1}{LC} i(t) = 0 d t 2 d 2 i ( t ) ​ + L R ​ d t d i ( t ) ​ + L C 1 ​ i ( t ) = 0

3.1 Characteristic Equation and Damping Regimes

Substituting a trial solution i ( t ) = A e r t i(t) = A e^{rt} i ( t ) = A e r t yields the characteristic equation : r 2 + 2 α r + ω 0 2 = 0 r^2 + 2\alpha r + \omega_0^2 = 0 r 2 + 2 α r + ω 0 2 ​ = 0

Where:

Neper Frequency (Damping Factor) α \alpha α :

• Series RLC: α = R 2 L \alpha = \frac{R}{2L} α = 2 L R ​
• Parallel RLC: α = 1 2 R C \alpha = \frac{1}{2RC} α = 2 RC 1 ​

Undamped Natural Angular Frequency ω 0 \omega_0 ω 0 ​ :

ω 0 = 1 L C \omega_0 = \frac{1}{\sqrt{LC}} ω 0 ​ = L C ​ 1 ​

The characteristic roots r 1 , r 2 r_1, r_2 r 1 ​ , r 2 ​ are given by: r 1 , 2 = − α ± α 2 − ω 0 2 r_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} r 1 , 2 ​ = − α ± α 2 − ω 0 2 ​ ​

Depending on the relative magnitudes of α \alpha α and ω 0 \omega_0 ω 0 ​ , three distinct transient solution cases occur: Damping Regimes in RLC Circuits

• alpha > omega_0 alpha = omega_0 alpha < omega_0
• +---------------+ +---------------+ +---------------+
• | Overdamped | | Critically | | Underdamped |
• | Real & Distinct| | Real & Equal | | Complex Conj |
• | Roots | | Roots | | Roots |
• +---------------+ +---------------+ +---------------+
• Case 1: Overdamped ( α > ω 0 \alpha > \omega_0 α > ω 0 ​ )

Roots r 1 r_1 r 1 ​ and r 2 r_2 r 2 ​ are real, negative, and distinct.

The current response is non-oscillatory and returns slowly to zero: ∋ ( t ) = A 1 e r 1 t + A 2 e r 2 t \ni(t) = A_1 e^{r_1 t} + A_2 e^{r_2 t} ∋ ( t ) = A 1 ​ e r 1 ​ t + A 2 ​ e r 2 ​ t

Case 2: Critically Damped ( α = ω 0 \alpha = \omega_0 α = ω 0 ​ )

Roots r 1 = r 2 = − α r_1 = r_2 = -\alpha r 1 ​ = r 2 ​ = − α are real and equal.

This represents the boundary state yielding the fastest non-oscillatory return to equilibrium: ∋ ( t ) = ( A 1 + A 2 t ) e − α t \ni(t) = (A_1 + A_2 t) e^{-\alpha t} ∋ ( t ) = ( A 1 ​ + A 2 ​ t ) e − α t

Case 3: Underdamped ( α < ω 0 \alpha < \omega_0 α < ω 0 ​ )

Roots are complex conjugates: r 1 , 2 = − α ± j ω d r_{1,2} = -\alpha \pm j \omega_d r 1 , 2 ​ = − α ± j ω d ​ , where ω d \omega_d ω d ​ is the 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 ​

The response exhibits decaying sinusoidal oscillations: ∋ ( t ) = e − α t ( A 1 cos ⁡ ( ω d t ) + A 2 sin ⁡ ( ω d t ) ) \ni(t) = e^{-\alpha t} \left( A_1 \cos(\omega_d t) + A_2 \sin(\omega_d t) \right) ∋ ( t ) = e − α t ( A 1 ​ cos ( ω d ​ t ) + A 2 ​ sin ( ω d ​ t ) )

4. Second-Order Non-Homogeneous ODEs & Continuity Rules

When a driving force f ( t ) f(t) f ( t ) is applied, the ODE becomes non-homogeneous: d 2 y d t 2 + a 1 d y d t + a 0 y = f ( t ) \frac{d^2 y}{dt^2} + a_1 \frac{dy}{dt} + a_0 y = f(t) d t 2 d 2 y ​ + a 1 ​ d t d y ​ + a 0 ​ y = f ( t )

The complete general solution is the sum of the natural (complementary) solution y h ( t ) y_h(t) y h ​ ( t ) and the forced (particular) solution y p ( t ) y_p(t) y p ​ ( t ) : y ( t ) = y h ( t ) + y p ( t ) y(t) = y_h(t) + y_p(t) y ( t ) = y h ​ ( t ) + y p ​ ( t )

4.1 Initial Condition Continuity Rules

To evaluate arbitrary constants A 1 A_1 A 1 ​ and A 2 A_2 A 2 ​ , boundary conditions at t = 0 + t = 0^+ t = 0 + must be established using physical continuity laws:

Capacitor Voltage Continuity : Voltage across an ideal capacitor cannot change instantaneously without infinite current: v C ( 0 + ) = v C ( 0 − ) v_C(0^+) = v_C(0^-) v C ​ ( 0 + ) = v C ​ ( 0 − )

Inductor Current Continuity : Current through an ideal inductor cannot change instantaneously without infinite voltage: i L ( 0 + ) = i L ( 0 − ) i_L(0^+) = i_L(0^-) i L ​ ( 0 + ) = i L ​ ( 0 − )

• 5. PRC ECE Board Exam Worked Examples
• Worked Example 1: RC Discharging Circuit

Problem: A 10 μ F 10\,\mu\text{F} 10 μ F capacitor is initially charged to 50 V 50\,\text{V} 50 V . At t = 0 t = 0 t = 0 , it is connected across a 100 k Ω 100\,\text{k}\Omega 100 k Ω resistor. Calculate the capacitor voltage at t = 2.0 seconds t = 2.0\,\text{seconds} t = 2.0 seconds .

Solution:

Compute the time constant τ \tau τ : τ = R C = ( 100 × 10 3 Ω ) ( 10 × 10 − 6 F ) = 1.0 second \tau = R C = (100 \times 10^3\,\Omega)(10 \times 10^{-6}\,\text{F}) = 1.0\,\text{second} τ = RC = ( 100 × 1 0 3 Ω ) ( 10 × 1 0 − 6 F ) = 1.0 second

Express v C ( t ) v_C(t) v C ​ ( t ) using the discharging formula: v C ( t ) = V 0 e − t / τ = 50 e − t / 1.0 v_C(t) = V_0 e^{-t/\tau} = 50 e^{-t/1.0} v C ​ ( t ) = V 0 ​ e − t / τ = 50 e − t /1.0

Substitute t = 2.0 s t = 2.0\,\text{s} t = 2.0 s : v C ( 2 ) = 50 e − 2 / 1.0 = 50 × 0.135335 = 6.767 V v_C(2) = 50 e^{-2/1.0} = 50 \times 0.135335 = 6.767\,\text{V} v C ​ ( 2 ) = 50 e − 2/1.0 = 50 × 0.135335 = 6.767 V

Worked Example 2: Series RLC Damping Determination

Problem: A series R L C RLC R L C circuit has R = 20 Ω R = 20\,\Omega R = 20 Ω , L = 0.1 H L = 0.1\,\text{H} L = 0.1 H , and C = 10 μ F C = 10\,\mu\text{F} C = 10 μ F . Determine the damping regime and compute the damped natural frequency ω d \omega_d ω d ​ .

Solution:

Compute Neper frequency α \alpha α : α = R 2 L = 20 2 ( 0.1 ) = 100 rad/s \alpha = \frac{R}{2L} = \frac{20}{2(0.1)} = 100\,\text{rad/s} α = 2 L R ​ = 2 ( 0.1 ) 20 ​ = 100 rad/s

Compute undamped natural frequency ω 0 \omega_0 ω 0 ​ : ω 0 = 1 L C = 1 0.1 × 10 − 5 = 1 10 − 6 = 1000 rad/s \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\,\text{rad/s} ω 0 ​ = L C ​ 1 ​ = 0.1 × 1 0 − 5 ​ 1 ​ = 1 0 − 6 ​ 1 ​ = 1000 rad/s

Compare α \alpha α and ω 0 \omega_0 ω 0 ​ :

Since α = 100 < ω 0 = 1000 \alpha = 100 < \omega_0 = 1000 α = 100 < ω 0 ​ = 1000 , the system is underdamped .

Calculate damped natural frequency ω d \omega_d ω d ​ : ω d = ω 0 2 − α 2 = 1000 2 − 100 2 = 1 , 000 , 000 − 10 , 000 = 990 , 000 ≈ 994.98 rad/s \omega_d = \sqrt{\omega_0^2 - \alpha^2} = \sqrt{1000^2 - 100^2} = \sqrt{1,000,000 - 10,000} = \sqrt{990,000} \approx 994.98\,\text{rad/s} ω d ​ = ω 0 2 ​ − α 2 ​ = 100 0 2 − 10 0 2 ​ = 1 , 000 , 000 − 10 , 000 ​ = 990 , 000 ​ ≈ 994.98 rad/s