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.
Diagram: RC Charging
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 |
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"
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
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.
Diagram: RC Discharging
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 \(\tau = L / R_{Th}\).
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.
Diagram: RL Circuit Charging
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 |
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
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.
Diagram: Exponential Properties
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: \(x(t) = x(\infty) + [x(0) - x(\infty)]e^{-t/\tau}\)
6.8 Worked Example: Complete Transient Analysis
This example walks through all five steps of a complete transient analysis — including a circuit that has both initial stored energy and a step input, which represents the most general case.
Problem: The circuit below has been in the position shown (switch open, 48 V source connected through R1 = 12 kΩ and R2 = 6 kΩ in series, with capacitor C = 5 μF across R2) for a long time. At \(t = 0\), the switch closes, connecting an additional parallel resistor R3 = 6 kΩ in parallel with R2 and C. Find \(V_C(t)\) for \(t \geq 0\).
Step 1 — Initial Condition from Pre-Switching Steady State
Before \(t = 0\) (long time, switch open), the capacitor acts as an open circuit (no DC current through it). The circuit is simply R1 and R2 in series from the 48 V source, with the capacitor across R2:
By the continuity principle: \(V_C(0^+) = V_C(0^-) = 16\) V. This is the initial condition.
Step 2 — Final Condition After Switching
After \(t = 0\) (long time, switch closed), the capacitor is again open circuit. Now R2 and R3 are in parallel, and that combination is in series with R1:
The final capacitor voltage has dropped from 16 V to 9.6 V — the switch disturbed the steady state.
Step 3 — Time Constant After Switching
Kill the 48 V source (replace with short circuit). Looking into the capacitor terminals, R1 appears in parallel with the series path through the closed switch. The Thévenin resistance seen by C is:
Step 4 — Complete Response
Applying the universal step-response formula with \(V_C(0) = 16\) V, \(V_C(\infty) = 9.6\) V, \(\tau = 12\) ms:
The voltage starts at 16 V and decays exponentially toward 9.6 V, with the transient term \(6.4\,e^{-83.3t}\) dying away with time constant 12 ms.
Step 5 — Verification
At \(t = 0^+\): \(V_C = 9.6 + 6.4\,e^0 = 9.6 + 6.4 = 16\) V ✓ (matches initial condition)
At \(t \to \infty\): \(V_C = 9.6 + 6.4 \times 0 = 9.6\) V ✓ (matches final condition)
At \(t = \tau = 12\) ms: \(V_C = 9.6 + 6.4\,e^{-1} = 9.6 + 6.4 \times 0.368 = 9.6 + 2.35 = 11.95\) V
This is 63.2% of the way from the initial value (16 V) down to the final value (9.6 V):
\(16 - 0.632 \times (16 - 9.6) = 16 - 4.04 = 11.96\) V ✓
6.9 Chapter Summary
Complete formula reference for first-order transients:
| Circuit | Condition | State Variable | Formula | Time Constant |
|---|---|---|---|---|
| RC | Charging (0 → \(V_S\)) | \(V_C(t)\) | \(V_S\!\left(1-e^{-t/\tau}\right)\) | \(\tau = RC\) |
| RC | Discharging (\(V_0\) → 0) | \(V_C(t)\) | \(V_0\,e^{-t/\tau}\) | \(\tau = RC\) |
| RL | Energizing (0 → \(V_S/R\)) | \(I_L(t)\) | \(\dfrac{V_S}{R}\!\left(1-e^{-t/\tau}\right)\) | \(\tau = L/R\) |
| RL | De-energizing (\(I_0\) → 0) | \(I_L(t)\) | \(I_0\,e^{-t/\tau}\) | \(\tau = L/R\) |
Universal step-response formula — works 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
Key milestones on the exponential curve:
| Time elapsed | % complete (from \(x(0)\) to \(x(\infty)\)) |
|---|---|
| \(\tau\) | 63.2% |
| \(2\tau\) | 86.5% |
| \(3\tau\) | 95.0% |
| \(5\tau\) | 99.3% — treated as "done" |
6.10 Common Mistakes and Misconceptions
Even after mastering the procedure, students frequently make a handful of predictable errors. Here are the most common ones — with the correct reasoning.
Mistake 1: Assuming Capacitor Voltage or Inductor Current Can Jump
The error: Setting the initial condition to the new steady-state value because "the switch just changed everything."
The truth: The state variable of the storage element (\(V_C\) or \(I_L\)) is continuous. It cannot jump at \(t = 0\). Energy cannot change instantaneously. The initial condition \(x(0^+)\) always equals the pre-switching value \(x(0^-)\).
Continuity Rule — Never Violate This
\(V_C(0^+) = V_C(0^-)\) always.
\(I_L(0^+) = I_L(0^-)\) always.
Everything else in the circuit can jump — resistor currents, capacitor current, inductor voltage — but not these two.
Example of the error: A capacitor charged to 5 V is connected to a 10 V source at \(t = 0\). A student writes \(V_C(0) = 10\) V. Wrong. The capacitor starts at 5 V and exponentially approaches 10 V.
Mistake 2: Confusing Natural Response and Step Response
Natural response is the circuit's behavior when there is no external source — only the stored energy decaying away:
Step response (or forced response) is the circuit's behavior driven by a suddenly applied DC source, starting from zero initial conditions:
Complete response is what you almost always want — both initial energy and a source driving the circuit:
The mistake is applying the natural-response formula (\(x(0)\,e^{-t/\tau}\)) when a source is present, which gives the wrong final value.
Mistake 3: Using the Wrong Thévenin Resistance for τ
To find \(\tau\), you must find the Thévenin resistance seen by the energy storage element, not the Thévenin resistance looking into the terminals of the whole circuit.
Procedure: 1. Remove the capacitor (or inductor) from the circuit. 2. Kill all independent sources (short voltage sources, open current sources). 3. Find the equivalent resistance looking into the terminals where the element was connected.
Keep Dependent Sources Active
When killing sources to find \(R_{Th}\), only independent sources are killed. Dependent sources (those controlled by a voltage or current elsewhere in the circuit) must remain. If dependent sources are present, use the test-voltage or \(V_{oc}/I_{sc}\) method.
Mistake 4: Sign Errors in the Exponential Term
The universal formula \(x(\infty) + [x(0) - x(\infty)]\,e^{-t/\tau}\) always has a negative exponent (\(-t/\tau\)), never positive. A positive exponent would produce a signal growing without bound, which violates energy conservation in a passive circuit.
If you find yourself writing \(e^{+t/\tau}\), stop and check your algebra — the sign error is almost certainly in the step where you match initial conditions.
Mistake 5: Applying Steady-State Assumptions Mid-Transient
In DC steady state: capacitor → open circuit, inductor → short circuit. These substitutions are only valid at \(t = 0^-\) and \(t \to \infty\). During the transient, the capacitor has an impedance \(1/(j\omega C)\) and the inductor has impedance \(j\omega L\) — but since we're in the time domain, we represent them by their differential equations, not by open/short substitutions.
Never draw a circuit diagram with the capacitor as an open and then write KCL for a mid-transient time. The open-circuit substitution only applies at the two steady-state endpoints used to find \(x(0)\) and \(x(\infty)\).