Chapter 6 — Transient Analysis of RC and RL Circuits
Chapter Overview (click to expand)
When a switch opens or closes in a circuit containing a capacitor or inductor, the circuit transitions between two steady states through an exponential transient governed by a characteristic time constant. This chapter develops the complete analytical framework for first-order RC and RL circuits, covering both the natural response (energy decay) and the step response (energy charging). **Key Takeaways** 1. The time constant tau (τ = RC or τ = L/R) determines how quickly a first-order circuit responds — after five time constants, the circuit is essentially at its new steady state. 2. The natural response is a decaying exponential representing the release of stored energy with no external forcing. 3. The complete response combines the natural response with the forced (particular) response to describe circuit behavior for any initial conditions and source values.6.1 Transient vs. Steady-State Response
Every circuit exists in one of two states: steady state or transient.
In steady state, all voltages and currents are constant (for DC) or periodic (for AC). Nothing is changing — the circuit has settled to its final behavior. This is the comfortable world of Chapters 1–4.
A transient occurs whenever the circuit is disturbed — a switch opens or closes, a source turns on or off. During the transient, the circuit transitions from one steady state to another. The transient is temporary; eventually the circuit reaches the new steady state.
Why does the transition take time at all? Because capacitors and inductors store energy and cannot release or absorb it instantaneously: - A capacitor's voltage cannot change instantaneously (stored charge must flow through finite resistance) - An inductor's current cannot change instantaneously (stored magnetic energy must be released through finite resistance)
This chapter focuses on first-order circuits — circuits with exactly one energy storage element (one capacitor or one inductor). These circuits always produce exponential transients.
6.2 The RC Circuit: Charging
Consider a resistor \(R\) in series with a capacitor \(C\), connected to a voltage source \(V_S\) via a switch closed at \(t = 0\). The capacitor starts uncharged (\(V_C(0) = 0\)).
Applying KVL around the loop:
Since \(i = C\,dV_C/dt\):
Solving this first-order differential equation with initial condition \(V_C(0) = 0\):
The capacitor voltage starts at 0 and rises exponentially toward \(V_S\). The current starts at \(V_S/R\) (as if the capacitor were a short circuit) and decays to zero as the capacitor charges.
Charging behavior at key moments:
| Time | \(V_C\) | \(i\) |
|---|---|---|
| \(t = 0\) | 0 | \(V_S/R\) |
| \(t = \tau\) | \(0.632\,V_S\) | \(0.368\,V_S/R\) |
| \(t = 2\tau\) | \(0.865\,V_S\) | \(0.135\,V_S/R\) |
| \(t = 5\tau\) | \(0.993\,V_S\) | \(0.007\,V_S/R\) |
| \(t \to \infty\) | \(V_S\) | 0 |
Diagram: RC Charging Circuit
6.3 The Time Constant
The product \(RC\) appears in the exponent and has units of seconds. It is called the time constant \(\tau\) (tau):
where \(R\) is in ohms (Ω), \(C\) is in farads (F), and \(\tau\) is in seconds (s).
The time constant is the single most important parameter describing a first-order transient:
- At \(t = \tau\): the circuit is 63.2% of the way from its initial to final value
- At \(t = 5\tau\): the circuit is 99.3% complete — engineers consider this "done"
Definition: Time Constant (τ) The time constant \(\tau\) is the characteristic time that determines how quickly a first-order circuit responds to a change. For an RC circuit \(\tau = RC\); for an RL circuit \(\tau = L/R\). After one time constant, the state variable has traveled 63.2% of the way from its initial to its final value. After five time constants, the response is 99.3% complete.
Physical interpretation: \(\tau\) is the time it would take the capacitor to charge to \(V_S\) if the initial rate of change were maintained. It's the "natural pace" of the circuit.
Example: An RC circuit with \(R = 10\) kΩ and \(C = 47\) μF has \(\tau = 10\,000 \times 47 \times 10^{-6} = 0.47\) s. After \(5\tau = 2.35\) s, the capacitor is fully charged.
The time constant applies equally to the discharging case.
Diagram: Time Constant Explorer
6.4 The RC Circuit: Discharging
Now consider a capacitor initially charged to \(V_0\), connected to a resistor at \(t = 0\) (with the source removed). The capacitor discharges through the resistor.
Applying KVL: \(V_C = -iR\) (current flows opposite to charging direction), which gives:
The capacitor voltage decays exponentially from \(V_0\) toward 0. The time constant \(\tau = RC\) governs the discharge rate just as it governs charging.
Finding τ in a Complex Circuit
When finding the time constant, reduce the circuit seen by the capacitor (or inductor) to its Thévenin equivalent. Then \(\tau = R_{Th} \cdot C\) or \(\tau = L / R_{Th}\).
Pro Tip: Find τ Using Thévenin Resistance To find the time constant in any circuit, temporarily remove the capacitor (or inductor), kill all independent sources, and calculate the resistance looking into the two terminals where the component was connected. That is \(R_{Th}\), and \(\tau = R_{Th} \cdot C\) (or \(L/R_{Th}\) for RL). This Thévenin approach works for any circuit complexity and is far easier than solving the differential equation from scratch.
Diagram: RC Discharging Circuit
6.5 The RL Circuit
An RL circuit pairs a resistor \(R\) with an inductor \(L\). It behaves as the dual of the RC circuit, with current replacing voltage and inductance replacing capacitance.
Energizing (switch closed at \(t = 0\), inductor initially uncharged):
where \(\tau = \frac{L}{R}\)
De-energizing (switch opened, inductor initially carrying current \(I_0\)):
The RL time constant:
where \(L\) is in henrys (H), \(R\) is in ohms (Ω), and \(\tau\) is in seconds (s).
Inductor Voltage Spike
When you abruptly open a switch carrying inductor current, the inductor tries to maintain its current instantaneously, producing a very large voltage spike: \(v = L\,di/dt\) with \(di/dt \to \infty\). This can damage switches and semiconductors. In practice, a freewheeling diode or snubber circuit is placed across the inductor to provide a controlled discharge path.
Safety Warning: Inductor Voltage Spikes Can Destroy Components Never abruptly disconnect an inductor carrying current — the resulting voltage spike can reach hundreds or thousands of volts, destroying transistors, MOSFETs, and control ICs in microseconds. Always include a freewheeling diode (flyback diode) across any inductive load — motors, solenoids, relay coils — when driven by a semiconductor switch. The diode provides a safe discharge path and clamps the spike. This is not optional in a real circuit design.
Common Mistake: RL Time Constant Behaves Opposite to RC In an RC circuit, \(\tau = RC\) — increasing resistance slows the response. In an RL circuit, \(\tau = L/R\) — increasing resistance speeds up the response because higher R dissipates the inductor's stored energy faster. Students often apply RC intuition to RL circuits and get the direction of change backwards. Keep the formula visible: τ and R are in the same direction for RC, and opposite directions for RL.
RC and RL duality:
| RC Circuit | RL Circuit |
|---|---|
| \(V_C\) is continuous | \(I_L\) is continuous |
| \(\tau = RC\) | \(\tau = L/R\) |
| \(V_C\) charges toward \(V_S\) | \(I_L\) energizes toward \(V_S/R\) |
| At steady state: open circuit | At steady state: short circuit |
Diagram: RL Circuit Charging
6.6 Initial and Final Conditions
To solve any first-order transient, you need two pieces of information: the initial condition and the final condition.
Initial condition \(x(0^+)\):
The value of the state variable immediately after switching. Due to continuity: - Capacitor voltage immediately after switching = capacitor voltage immediately before: \(V_C(0^+) = V_C(0^-)\) - Inductor current immediately after switching = inductor current immediately before: \(I_L(0^+) = I_L(0^-)\)
To find the initial condition, analyze the circuit at \(t = 0^-\) (DC steady state before switching): - Capacitor → open circuit (in DC steady state, \(i_C = 0\)) - Inductor → short circuit (in DC steady state, \(v_L = 0\))
Final condition \(x(\infty)\):
The DC steady-state value as \(t \to \infty\). Analyze the circuit at DC steady state with the new switch configuration: - Capacitor → open circuit - Inductor → short circuit
Common Mistake: Continuity Applies Only to V_C and I_L — Not to Everything Continuity at a switching instant means capacitor voltage and inductor current cannot jump. Other quantities — resistor voltages, resistor currents, inductor voltage, capacitor current — can jump instantaneously. A classic error is writing \(i_R(0^+) = i_R(0^-)\) for a resistor in series with a capacitor; in fact, \(i_R(0^+)\) equals \(V_S / R\) (the initial surge), which can be far from its pre-switch value.
Diagram: Initial and Final Conditions
6.7 Natural, Forced, and Complete Response
The complete response of a first-order circuit has two components:
Natural response \(x_n(t)\): The circuit's response due to its initial stored energy, with no external forcing. It decays to zero:
Forced response \(x_f(t)\): The response driven by the external source(s), which the circuit eventually settles to. For DC sources, this is a constant equal to the final value \(x(\infty)\).
Complete response = natural + forced:
This is the universal step-response formula — it solves any first-order DC transient. Once you know \(x(0)\), \(x(\infty)\), and \(\tau\), you're done.
Key Formula: Universal First-Order Step Response
This single formula solves any first-order DC transient — RC or RL, charging or discharging. Find three numbers: the initial value \(x(0)\), the final value \(x(\infty)\), and the time constant \(\tau = R_{Th}C\) or \(L/R_{Th}\). Plug in and you're done.
Procedure for first-order transient analysis:
- Find \(x(0)\): analyze circuit at DC steady state before switching (C → open, L → short)
- Find \(x(\infty)\): analyze circuit at DC steady state after switching (C → open, L → short)
- Find \(\tau\): find Thévenin resistance seen by the energy storage element; \(\tau = R_{Th}C\) or \(\tau = L/R_{Th}\)
- Write the solution: \(x(t) = x(\infty) + [x(0) - x(\infty)]e^{-t/\tau}\)
Diagram: RC and RL Circuit Applications
6.8 Worked Example: Complete Transient Analysis
Problem: A 24 V source drives a circuit with R1 = 6 kΩ in series with a parallel combination of R2 = 12 kΩ and C = 10 μF. The switch closes at \(t = 0\). Find \(V_C(t)\) for \(t \geq 0\).
Assume the capacitor is initially uncharged.
Step 1 — Initial condition: At \(t = 0^-\), capacitor is uncharged: \(V_C(0) = 0\) V.
Step 2 — Final condition:
At \(t \to \infty\), capacitor is open circuit. Current through R1 and R2 in series:
\(I = 24 / (6\,\text{k} + 12\,\text{k}) = 24/18\,\text{k} = 1.33\) mA
\(V_C(\infty) = I \times R2 = 1.33\,\text{m} \times 12\,\text{k} = 16\) V
Step 3 — Time constant:
Kill the source (short); Thévenin resistance seen by C = R1 ∥ R2 = (6k × 12k)/(6k + 12k) = 4 kΩ
\(\tau = R_{Th} C = 4\,000 \times 10 \times 10^{-6} = 40\) ms
Step 4 — Complete response:
At \(t = \tau = 40\) ms: \(V_C = 16(1 - e^{-1}) = 16 \times 0.632 = 10.1\) V ✓
6.9 Chapter Summary
Key results:
| Circuit | Time Constant | Energizing | De-energizing |
|---|---|---|---|
| RC | \(\tau = RC\) | \(V_C = V_S(1-e^{-t/\tau})\) | \(V_C = V_0 e^{-t/\tau}\) |
| RL | \(\tau = L/R\) | \(I_L = (V_S/R)(1-e^{-t/\tau})\) | \(I_L = I_0 e^{-t/\tau}\) |
Universal formula for any first-order DC transient:
Procedure: 1. Find \(x(0)\): DC steady state before switching (C = open, L = short) 2. Find \(x(\infty)\): DC steady state after switching (C = open, L = short) 3. Find \(\tau = R_{Th}C\) or \(\tau = L/R_{Th}\) (Thévenin resistance seen by storage element) 4. Apply the universal formula
Rules: - At \(t = \tau\): circuit is 63.2% from initial to final - After \(5\tau\): circuit is 99.3% complete — treated as fully settled