lecture_EE1C1_lec7_first_order_transient

📌 Overview

First-order transient circuits involve RC or RL combinations responding to disturbances like switches. Analysis uses time-domain differential equations or algorithmic methods to find voltages/currents during transients. Key: time constant τ = RC or L/R, step responses via unit step u(t). Covers recap of C/L, DE solving, examples (charging/discharging), complex circuits, and exam-style problems. Ensures ability to solve post-lecture exercises on transients.

🎯 Learning Objectives

Understand capacitance/inductance recap and energy storage/release.
Analyze first-order RC/RL circuits under transients (e.g., step inputs).
Solve differential equations for transients: homogeneous + particular solutions.
Compute time constants τ and interpret their role in response speed.
Apply Method 1 (DE) and Method 2 (algorithmic: initial/final values, Thévenin) to find i(t)/v(t).
Handle step responses, pulses, and continuity conditions (v_C, i_L continuous).
Work exam examples: find v_C(0+), v_C(t) for t>0.

💡 Key Concepts & Definitions

Transient Circuits

First-order: One reactive element (C or L) + resistors/sources. Steady-state (DC) vs. transient (time-varying after disturbance).
Disturbance: Switch, step input. Abrupt changes can’t instantly affect C voltage or L current (continuity: v_C(0+) = v_C(0-), i_L(0+) = i_L(0-)).
Time Constant (τ): Measures response speed. After τ, response drops by e^{-1} (~63%). Smaller τ = faster transient.
- RC: τ = R_{Th} C (Thevenin resistance at C terminals).
- RL: τ = L / R_{Th} (Thevenin resistance at L terminals).
Step Response: Response to unit step u(t) = {0 (t<0), 1 (t≥0)}. Pulses: A [u(t - t_0) - u(t - t_0 - T)].

➗ Formulas

Differential Equation (DE) General Form

The general form of the first-order linear differential equation is

\frac{d x ( t )}{d t} + a x (t) = f (t)

Homogeneous solution:

\frac{d x _{h} ( t )}{d t} + a x_{h} (t) = 0 \Rightarrow x_{h} (t) = K e^{- t / τ}, τ = \frac{1}{a}

Particular solution (for constant $f (t) = A$ ):

x_{p} (t) = \frac{A}{a}

General solution:

x (t) = x_{\infty} + (x (0^{+}) - x_{\infty}) e^{- t / τ}

where $x_{\infty}$ is the steady-state value ( $t \to \infty$ ).

Algorithmic Method Steps

Assume x(t) = K_1 + K_2 e^{-t/τ}.
Find x(0-): Steady-state before switch.
x(0+) = x(0-) (continuity).
Find x(∞): Steady-state after (t>5τ, C→open, L→short).
Compute τ via Thévenin at reactive terminals (deactivate sources: V=0, I=short).
Solve: K_1 = x(∞), K_2 = x(0+) - x(∞).

When Allowed: Linear circuits, independent sources, no dependent sources or nonlinearities. Rare: If τ varies (e.g., switching changes R), recompute per interval. Initial conditions from prior steady-state.

RC Charging (Source $V_{S}$ on at t=0, initial $v_{C} = 0$ )

v_{C} (t) = V_{S} (1 - e^{- t / (RC)})

RC Discharging (initial $v_{C} = V_{0}$ , source off)

v_{C} (t) = V_{0} e^{- t / (RC)}

✍️ Notes

Recap: Capacitors & Inductors

Capacitor: q(t) = C v(t), i = C dv/dt. Energy: (1/2) C v^2. v continuous.
Inductor: v = L di/dt, Φ = L i. Energy: (1/2) L i^2. i continuous.
Interconnections: Series/parallel C_eq, L_eq formulas (from prior lectures).

Transient Analysis Basics

Circuits store/release energy slowly. Time-domain: Solve DE from KCL/KVL on reactive variable (v_C or i_L).

Camera flash example: Charging C via R, discharging: dv_C/dt + v_C/(RC) = 0 → v_C(t) = V_0 e^{-t/τ}, τ=RC.

Method 1: Differential Equations

Write DE for reactive var (e.g., KCL on C node: C dv/dt = currents).
Solve: Particular (steady) + homogeneous (transient). Use initial condition.

Example 1: RC Charging (Worked Out)
Circuit: V_S — R — C (ground), switch closes at t=0, v_C(0-)=0.
Step 1: KCL at C node: $C \frac{d v _{C}}{d t} = \frac{V _{S} - v _{C}}{R}$ .
This gives the DE:

\frac{d v _{C}}{d t} + \frac{v _{C}}{RC} = \frac{V _{S}}{RC}

Step 2: Particular solution: $v_{p} = V_{S}$ (steady-state).
Homogeneous solution: $v_{h} = K e^{- t / τ}$ , with $τ = RC$ .
General solution: $v_{C} (t) = V_{S} + K e^{- t / (RC)}$ .
Step 3: Apply initial condition: $v_{C} (0^{+}) = 0 = V_{S} + K ⟹ K = - V_{S}$ .
Thus, the solution is

v_{C} (t) = V_{S} (1 - e^{- t / (RC)})

Plot: Rises from 0 to $V_{S}$ , reaches 63% at $t = τ$ .
(Allowed for constant sources; if initial $\neq = 0$ , adjust $K$ accordingly.)

\usepackage{circuitikz}
\begin{circuitikz}
\draw (0,0) to[V, v=V_S] (0,2) -- (2,2) to[R, l=R] (2,0) to[C, l=C] (0,0);
\draw (2,2) -- (2,3) node[anchor=south] {Switch closes at t=0};
\end{circuitikz}

Method 2: Algorithmic (Initial-Final Values)

Faster for circuits. Use Thévenin for τ.

Example 2: RC Discharging (Worked Out)
Circuit: DC source $V_{S} = 12$ V, $R_{1} = 6$ k $Ω$ in series with parallel combination of $R_{2} = 3$ k $Ω$ and $C = 100 μ$ F. At $t = 0$ , the switch disconnects $V_{S}$ and $R_{1}$ , so $C$ discharges through $R_{2}$ only. Find the current $i (t)$ through $R_{2}$ for $t > 0$ .
Step 1: Assume the form $i (t) = K_{1} + K_{2} e^{- t / τ}$ .
Step 2: For $t < 0$ , steady-state (C as open): $v_{C} (0^{-}) = V_{S} \frac{R _{2}}{R _{1} + R _{2}} = 12 \times \frac{3}{9} = 4$ V.
Thus, $i (0^{-}) = \frac{v _{C} ( 0 ^{-} )}{R _{2}} = \frac{4}{3000} = 1.333$ mA.
By continuity of $v_{C}$ , $v_{C} (0^{+}) = 4$ V, so $i (0^{+}) = 1.333$ mA.
Step 3: For $t \to \infty$ , steady-state (C open, no source): $i (\infty) = 0$ .
Step 4: Time constant $τ = R_{T h} C$ , where $R_{T h} = R_{2} = 3$ k $Ω$ (seen by C, sources deactivated). Thus, $τ = 3000 \times 1 0^{- 4} = 0.3$ s.
Step 5: $K_{1} = i (\infty) = 0$ , $K_{2} = i (0^{+}) - i (\infty) = 1.333$ mA.
Therefore,

i (t) = \frac{4}{3} e^{- t /0.3} mA (t > 0)

Plot: Decays exponentially from 1.333 mA to 0.
Rare Case: If the capacitor is partially charged at switch time, use the corresponding $v_{C} (0^{+})$ .

Example 3: Thévenin for τ (Worked Out, Exam-Style)
Circuit: Complex RC with 36 V source, 2 k $Ω$ , 4 k $Ω$ , 2 mA current source, C=0.5 mF at node A to ground. At t=0, current source is shorted (deactivated). Find branch current i(t) through 2 k $Ω$ for t>0 (refer to slides for exact config).
Step 1: Initial condition (t=0+): From DC analysis and continuity, i(0+) = 16 mA.
Step 2: Steady-state (t→∞, C open): i(∞) = 9 mA.
Step 3: Time constant τ = R_{Th} C = 0.15 s (R_{Th} computed via Thévenin at C terminals: deactivate sources, find eq. resistance).
Step 4:

i (t) = 9 + (16 - 9) e^{- t /0.15} = 9 + 7 e^{- t /0.15} mA (t > 0)

Plot: Transitions exponentially from 16 mA to 9 mA.
To compute R_{Th}: Deactivate independent sources (V shorted to 0 V, I opened), find resistance seen by C.

Exam Example (Worked Out)
Circuit: 20 V — 5 k $Ω$ — node — 10 k $Ω$ to ground, C=2 $μ$ F parallel to 10 k $Ω$ . Assume at t=0, switch causes discharge (e.g., shorts source, refer to slides). Find v_C(0+), v_C(t) for t>0.
a) t<0 steady-state (C open):

v_{C} (0^{-}) = 20 \times \frac{10}{5 + 10} = 20 \times \frac{2}{3} \approx 13.33 V

By continuity, v_C(0+) = 13.33 V.
b) t>0: Assume v_C(∞) = 0 V (full discharge). R_{Th} = 5 k $Ω$ || 10 k $Ω$ = $\frac{5 \times 10}{15} \approx 3.33$ k $Ω$ .

τ = 3.33 \times 1 0^{3} \times 2 \times 1 0^{- 6} \approx 6.66 μ s

v_{C} (t) = 0 + (13.33 - 0) e^{- t / τ} = 13.33 e^{- t /6.66 \times 1 0^{- 6}} V (t > 0)

Explanation: Applies general first-order method. For complex switches, segment by time intervals, enforcing continuity.

graph TD
    A[20V] --> B[5kΩ]
    B --> C[node]
    C --> D[10kΩ to gnd]
    C --> E[C=2μF to gnd]
    F[Switch at t=0] -->|closes| G[Additional path]

(Visual: Equivalent circuit for τ: Deactivate sources, R_Th seen by C.)

Step Response & Pulses

u(t): Models switch-on DC source V_0 u(t).
Pulse: V_0 [u(t) - u(t-T)]. Response: Superposition of two steps.
Rare: If T << τ, approx impulse; if T>>τ, two separate transients.

🔗 Resources

Presentation:
SGH6 exercises on transients.

❓ Post lecture

Why τ determines speed? Simulate with large/small C.
Difference RC vs RL: Duals (v↔i, R↔1/R? No, τ=RC vs L/R).
When DE vs algorithmic? Algorithm faster for circuits, DE for understanding.

📖 Homework

Complete SGH6: Practice RC/RL charging/discharging, complex with Thévenin.
Seminars Tue/Fri: Solve exam-like transients.
Register for end-term exam.
Next: Second-order circuits (RCL).

JKoch notes

Explorer

lecture_EE1C1_lec7_first_order_transient_2025-10-20

Metadata

📌 Overview

🎯 Learning Objectives

💡 Key Concepts & Definitions

Transient Circuits

➗ Formulas

Differential Equation (DE) General Form

Algorithmic Method Steps

RC Charging (Source $V_{S}$ on at t=0, initial $v_{C} = 0$ )

RC Discharging (initial $v_{C} = V_{0}$ , source off)

✍️ Notes

Recap: Capacitors & Inductors

Transient Analysis Basics

Method 1: Differential Equations

Method 2: Algorithmic (Initial-Final Values)

Step Response & Pulses

🔗 Resources

❓ Post lecture

📖 Homework

Graph View

Table of Contents

Backlinks

JKoch notes

Explorer

lecture_EE1C1_lec7_first_order_transient_2025-10-20

Metadata

📌 Overview

🎯 Learning Objectives

💡 Key Concepts & Definitions

Transient Circuits

➗ Formulas

Differential Equation (DE) General Form

Algorithmic Method Steps

RC Charging (Source VS​ on at t=0, initial vC​=0)

RC Discharging (initial vC​=V0​, source off)

✍️ Notes

Recap: Capacitors & Inductors

Transient Analysis Basics

Method 1: Differential Equations

Method 2: Algorithmic (Initial-Final Values)

Step Response & Pulses

🔗 Resources

❓ Post lecture

📖 Homework

Graph View

Table of Contents

Backlinks

RC Charging (Source $V_{S}$ on at t=0, initial $v_{C} = 0$ )

RC Discharging (initial $v_{C} = V_{0}$ , source off)