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3.1 Differential Equations & Circuit Transients

Key Takeaways

  • First-order differential equations model RL and RC transient circuits, with time constants τ=RC\tau = RC and τ=L/R\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 α\alpha and natural frequency ω0\omega_0: overdamped, critically damped, and underdamped.

  • The complete transient response of a non-homogeneous ODE consists of the natural response yh(t)y_h(t) plus the particular (forced) response yp(t)y_p(t).

  • Physical continuity of capacitor voltage vC(0−)=vC(0+)v_C(0^-) = v_C(0^+) and inductor current iL(0−)=iL(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) to its derivatives with respect to a single independent variable, typically time tt. 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)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):

I(t)=e∫P(t)dtI(t) = e^{\int P(t) dt}

Multiplying through yields ddt[I(t)y(t)]=I(t)Q(t)\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 RCRC or RLRL network.

2.1 RC Circuit Charging and Discharging

For a series RCRC circuit connected to a DC source VCCV_{CC} at t=0t = 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 vC(0)=V0v_C(0) = V_0:

vC(t)=VCC+(V0−VCC)e−t/τv_C(t) = V_{CC} + (V_0 - V_{CC}) e^{-t/\tau}

Where the time constant τ\tau for an RCRC circuit is defined as:

τ=ReqCeq\tau = R_{eq} C_{eq}
Operational PhaseCapacitor Voltage vC(t)v_C(t)Capacitor Current iC(t)i_C(t)
Charging (V0=0V_0 = 0)vC(t)=VCC(1−e−t/τ)v_C(t) = V_{CC} \left(1 - e^{-t/\tau}\right)iC(t)=VCCRe−t/τi_C(t) = \frac{V_{CC}}{R} e^{-t/\tau}
Discharging (VCC=0V_{CC} = 0)vC(t)=V0e−t/τv_C(t) = V_0 e^{-t/\tau}iC(t)=−V0Re−t/τi_C(t) = -\frac{V_0}{R} e^{-t/\tau}

2.2 RL Circuit Current Growth and Decay

For an RLRL circuit energized by a DC voltage source VV:

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

The solution for current iL(t)i_L(t) given initial current I0I_0 is:

∋L(t)=VR+(I0−VR)e−t/τ\ni_L(t) = \frac{V}{R} + \left(I_0 - \frac{V}{R}\right) e^{-t/\tau}

Where the time constant τ\tau for an RLRL circuit is:

τ=LeqReq\tau = \frac{L_{eq}}{R_{eq}}

2.3 Physical Meaning of the Time Constant τ\tau

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

Ldi(t)dt+Ri(t)+1C∫0ti(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 tt 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)=Aerti(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) α\alpha:
    • Series RLC: α=R2L\alpha = \frac{R}{2L}
    • Parallel RLC: α=12RC\alpha = \frac{1}{2RC}
  • Undamped Natural Angular Frequency ω0\omega_0:
    • ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

The characteristic roots r1,r2r_1, r_2 are given by:

r1,2=−α±α2−ω02r_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}

Depending on the relative magnitudes of α\alpha and ω0\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 (α>ω0\alpha > \omega_0)

Roots r1r_1 and r2r_2 are real, negative, and distinct. The current response is non-oscillatory and returns slowly to zero:

∋(t)=A1er1t+A2er2t\ni(t) = A_1 e^{r_1 t} + A_2 e^{r_2 t}

Case 2: Critically Damped (α=ω0\alpha = \omega_0)

Roots r1=r2=−αr_1 = r_2 = -\alpha are real and equal. This represents the boundary state yielding the fastest non-oscillatory return to equilibrium:

∋(t)=(A1+A2t)e−αt\ni(t) = (A_1 + A_2 t) e^{-\alpha t}

Case 3: Underdamped (α<ω0\alpha < \omega_0)

Roots are complex conjugates: r1,2=−α±jωdr_{1,2} = -\alpha \pm j \omega_d, where ωd\omega_d is the damped natural frequency:

ωd=ω02−α2\omega_d = \sqrt{\omega_0^2 - \alpha^2}

The response exhibits decaying sinusoidal oscillations:

∋(t)=e−αt(A1cos⁡(ωdt)+A2sin⁡(ωdt))\ni(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)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 yh(t)y_h(t) and the forced (particular) solution yp(t)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 A1A_1 and A2A_2, boundary conditions at t=0+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 μF10\,\mu\text{F} capacitor is initially charged to 50 V50\,\text{V}. At t=0t = 0, it is connected across a 100 kΩ100\,\text{k}\Omega resistor. Calculate the capacitor voltage at t=2.0 secondst = 2.0\,\text{seconds}.

Solution:

  1. Compute the time constant τ\tau: τ=RC=(100×103 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}
  2. Express vC(t)v_C(t) using the discharging formula: vC(t)=V0e−t/τ=50e−t/1.0v_C(t) = V_0 e^{-t/\tau} = 50 e^{-t/1.0}
  3. Substitute t=2.0 st = 2.0\,\text{s}: vC(2)=50e−2/1.0=50×0.135335=6.767 Vv_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 RLCRLC circuit has R=20 ΩR = 20\,\Omega, L=0.1 HL = 0.1\,\text{H}, and C=10 μFC = 10\,\mu\text{F}. Determine the damping regime and compute the damped natural frequency ωd\omega_d.

Solution:

  1. Compute Neper frequency α\alpha: α=R2L=202(0.1)=100 rad/s\alpha = \frac{R}{2L} = \frac{20}{2(0.1)} = 100\,\text{rad/s}
  2. Compute undamped natural frequency ω0\omega_0: ω0=1LC=10.1×10−5=110−6=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}
  3. Compare α\alpha and ω0\omega_0: Since α=100<ω0=1000\alpha = 100 < \omega_0 = 1000, the system is underdamped.
  4. Calculate damped natural frequency ωd\omega_d: ωd=ω02−α2=10002−1002=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}
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

36.8 V

B

63.2 V

C

86.5 V

D

99.3 V

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

Overdamped

B

Critically damped

C

Underdamped

D

Undamped resonant

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

0 A

B

2 A

C

4 A

D

Infinite current

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