3.1 Differential Equations & Circuit Transients

Key Takeaways

  • First-order differential equations model RL and RC transient circuits, with time constants \(\tau = RC\) and \(\tau = L/R\) governing exponential charging/discharging rates.
  • Second-order linear ODEs model RLC circuits, exhibiting three distinct transient responses based on the damping factor \(\alpha\) and natural frequency \(\omega_0\): overdamped, critically damped, and underdamped.
  • The complete transient response of a non-homogeneous ODE consists of the natural response \(y_h(t)\) plus the particular (forced) response \(y_p(t)\).
  • Physical continuity of capacitor voltage \(v_C(0^-) = v_C(0^+)\) and inductor current \(i_L(0^-) = i_L(0^+)\) 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 $y(t)$ to its derivatives with respect to a single independent variable, typically time $t$. 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:

dydt=g(t)h(y)    1h(y)dy=g(t)dt\frac{dy}{dt} = g(t) h(y) \implies \frac{1}{h(y)} dy = g(t) dt

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

A general first-order linear differential equation takes the standard form:

dydt+P(t)y=Q(t)\frac{dy}{dt} + 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)=eP(t)dtI(t) = e^{\int P(t) dt}

Multiplying through yields $\frac{d}{dt}[I(t) y(t)] = I(t) Q(t)$, which can be integrated directly:

y(t)=1I(t)[I(t)Q(t)dt+C]y(t) = \frac{1}{I(t)} \left[ \int I(t) Q(t) dt + C \right]


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 $RC$ or $RL$ network.

2.1 RC Circuit Charging and Discharging

For a series $RC$ circuit connected to a DC source $V_{CC}$ at $t = 0$:

Ri(t)+vC(t)=VCC    RCdvC(t)dt+vC(t)=VCCR i(t) + v_C(t) = V_{CC} \implies R C \frac{dv_C(t)}{dt} + v_C(t) = V_{CC}

This is a first-order linear non-homogeneous differential equation. Solving with initial condition $v_C(0) = V_0$:

vC(t)=VCC+(V0VCC)et/τv_C(t) = V_{CC} + (V_0 - V_{CC}) e^{-t/\tau}

Where the time constant $\tau$ for an $RC$ circuit is defined as:

τ=ReqCeq\tau = R_{eq} C_{eq}

Operational PhaseCapacitor Voltage $v_C(t)$Capacitor Current $i_C(t)$
Charging ($V_0 = 0$)$v_C(t) = V_{CC} \left(1 - e^{-t/\tau}\right)$$i_C(t) = \frac{V_{CC}}{R} e^{-t/\tau}$
Discharging ($V_{CC} = 0$)$v_C(t) = V_0 e^{-t/\tau}$$i_C(t) = -\frac{V_0}{R} e^{-t/\tau}$

2.2 RL Circuit Current Growth and Decay

For an $RL$ circuit energized by a DC voltage source $V$:

Ldi(t)dt+Ri(t)=VL \frac{di(t)}{dt} + R i(t) = V

The solution for current $i_L(t)$ given initial current $I_0$ is:

iL(t)=VR+(I0VR)et/τi_L(t) = \frac{V}{R} + \left(I_0 - \frac{V}{R}\right) e^{-t/\tau}

Where the time constant $\tau$ for an $RL$ circuit is:

τ=LeqReq\tau = \frac{L_{eq}}{R_{eq}}

2.3 Physical Meaning of the Time Constant $\tau$

  • At $t = 1\tau$: The transient quantity reaches $63.2%$ of its final change during charging, or decays to $36.8%$ of its initial value during discharging.
  • At $t = 5\tau$: The transient response reaches $99.3%$ 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 $RLC$ circuit connected in a closed loop without an external source yields Kirchhoff's Voltage Law equation:

Ldi(t)dt+Ri(t)+1C0ti(τ)dτ+vC(0)=0L \frac{d i(t)}{dt} + R i(t) + \frac{1}{C} \int_0^t i(\tau) d\tau + v_C(0) = 0

Differentiating with respect to $t$ gives the second-order homogeneous linear ODE:

d2i(t)dt2+RLdi(t)dt+1LCi(t)=0\frac{d^2 i(t)}{dt^2} + \frac{R}{L} \frac{di(t)}{dt} + \frac{1}{LC} i(t) = 0

3.1 Characteristic Equation and Damping Regimes

Substituting a trial solution $i(t) = A e^{rt}$ yields the characteristic equation:

r2+2αr+ω02=0r^2 + 2\alpha r + \omega_0^2 = 0

Where:

  • Neper Frequency (Damping Factor) $\alpha$:
    • Series RLC: $\alpha = \frac{R}{2L}$
    • Parallel RLC: $\alpha = \frac{1}{2RC}$
  • Undamped Natural Angular Frequency $\omega_0$:
    • $\omega_0 = \frac{1}{\sqrt{LC}}$

The characteristic roots $r_1, r_2$ are given by:

r1,2=α±α2ω02r_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}

Depending on the relative magnitudes of $\alpha$ and $\omega_0$, 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 ($\alpha > \omega_0$)

Roots $r_1$ and $r_2$ are real, negative, and distinct. The current response is non-oscillatory and returns slowly to zero:

i(t)=A1er1t+A2er2ti(t) = A_1 e^{r_1 t} + A_2 e^{r_2 t}

Case 2: Critically Damped ($\alpha = \omega_0$)

Roots $r_1 = r_2 = -\alpha$ are real and equal. This represents the boundary state yielding the fastest non-oscillatory return to equilibrium:

i(t)=(A1+A2t)eαti(t) = (A_1 + A_2 t) e^{-\alpha t}

Case 3: Underdamped ($\alpha < \omega_0$)

Roots are complex conjugates: $r_{1,2} = -\alpha \pm j \omega_d$, where $\omega_d$ is the damped natural frequency:

ωd=ω02α2\omega_d = \sqrt{\omega_0^2 - \alpha^2}

The response exhibits decaying sinusoidal oscillations:

i(t)=eαt(A1cos(ωdt)+A2sin(ωdt))i(t) = e^{-\alpha t} \left( A_1 \cos(\omega_d t) + A_2 \sin(\omega_d t) \right)


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

When a driving force $f(t)$ is applied, the ODE becomes non-homogeneous:

d2ydt2+a1dydt+a0y=f(t)\frac{d^2 y}{dt^2} + a_1 \frac{dy}{dt} + a_0 y = f(t)

The complete general solution is the sum of the natural (complementary) solution $y_h(t)$ and the forced (particular) solution $y_p(t)$:

y(t)=yh(t)+yp(t)y(t) = y_h(t) + y_p(t)

4.1 Initial Condition Continuity Rules

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

  1. Capacitor Voltage Continuity: Voltage across an ideal capacitor cannot change instantaneously without infinite current: vC(0+)=vC(0)v_C(0^+) = v_C(0^-)
  2. Inductor Current Continuity: Current through an ideal inductor cannot change instantaneously without infinite voltage: iL(0+)=iL(0)i_L(0^+) = i_L(0^-)

5. PRC ECE Board Exam Worked Examples

Worked Example 1: RC Discharging Circuit

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

Solution:

  1. Compute the time constant $\tau$: τ=RC=(100×103Ω)(10×106F)=1.0second\tau = R C = (100 \times 10^3\,\Omega)(10 \times 10^{-6}\,\text{F}) = 1.0\,\text{second}
  2. Express $v_C(t)$ using the discharging formula: vC(t)=V0et/τ=50et/1.0v_C(t) = V_0 e^{-t/\tau} = 50 e^{-t/1.0}
  3. Substitute $t = 2.0,\text{s}$: vC(2)=50e2/1.0=50×0.135335=6.767Vv_C(2) = 50 e^{-2/1.0} = 50 \times 0.135335 = 6.767\,\text{V}

Worked Example 2: Series RLC Damping Determination

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

Solution:

  1. Compute Neper frequency $\alpha$: α=R2L=202(0.1)=100rad/s\alpha = \frac{R}{2L} = \frac{20}{2(0.1)} = 100\,\text{rad/s}
  2. Compute undamped natural frequency $\omega_0$: ω0=1LC=10.1×105=1106=1000rad/s\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\,\text{rad/s}
  3. Compare $\alpha$ and $\omega_0$: Since $\alpha = 100 < \omega_0 = 1000$, the system is underdamped.
  4. Calculate damped natural frequency $\omega_d$: ωd=ω02α2=100021002=1,000,00010,000=990,000994.98rad/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}
Test Your Knowledge

For a series RC circuit with R = 50 kΩ and C = 20 µF connected to a 100 V DC source at t = 0, what is the capacitor voltage after t = 2.0 seconds?

A
B
C
D
Test Your Knowledge

A series RLC circuit has R = 40 Ω, L = 0.2 H, and C = 5 µF. Which type of transient response does this circuit exhibit when energized?

A
B
C
D
Test Your Knowledge

An inductor of L = 0.5 H carries a steady-state DC current of 4 A. At t = 0, a switch opens instantly inserting a series resistor. What is the current i_L(0⁺) through the inductor immediately after switch opening?

A
B
C
D