ELECTENG291 · Fundamentals of Electrical Engineering
First-Order Transients: RC & RL Circuits
Module 2 of University of Auckland ELECTENG 291 analyses circuits with a single (equivalent) energy-storage element — the RC and RL circuits — whose behaviour obeys a first-order ODE. The chapter covers the natural and step responses, the time constant (τ = RC or τ = L/R), and the general first-order solution y(t) = y(∞) + [y(0⁺) − y(∞)]e^(−t/τ), which is printed on the provided exam formula page. This three-step method is a reliable source of marks in the tests and the final exam.
What this chapter covers
- 01First-order circuit = one equivalent energy-storage element → first-order ODE a1 dy/dt + a0 y = x(t)
- 02Time constant: τ = RC for an RC circuit, τ = L/R for an RL circuit — where R is the Thévenin resistance seen by the C or L
- 03The general first-order solution y(t) = y(∞) + [y(0⁺) − y(∞)]e^(−t/τ) (on the exam formula page)
- 04The three-step method: find y(0⁺) via continuity, find y(∞) at the new steady state, find τ
- 05Natural response (source removed, decay to zero) vs step response (settle to a new final value)
- 06Settling: the response is within about 1% of its final value after roughly 5τ
- 07At DC endpoints, the capacitor is open and the inductor is short — used to get y(0⁺) and y(∞)
RC step response by the three-step method
- +1Time constant: τ = RC = (2×10³)(100×10⁻⁶) = 0.2 s.
- +1Initial value by continuity: the capacitor was uncharged, so v_C(0⁺) = v_C(0⁻) = 0 V.
- +1Final value: at the new DC steady state the capacitor is an open circuit, so no current flows and it charges to the full source voltage, v_C(∞) = 10 V.
- +1Assemble the general solution y(t) = y(∞) + [y(0⁺) − y(∞)]e^(−t/τ): v_C(t) = 10 + (0 − 10)e^(−t/0.2) = 10(1 − e^(−5t)) V, t > 0. It is essentially settled after 5τ = 1 s.
Key terms
- First-order circuit
- A circuit with one equivalent energy-storage element (one capacitor or one inductor, after reduction), whose dynamics obey a first-order differential equation. Only two kinds exist: RC and RL.
- Time constant (τ)
- The characteristic time of the exponential response: τ = RC for an RC circuit and τ = L/R for an RL circuit, where R is the Thévenin resistance seen by the storage element. Larger τ means slower response.
- General first-order solution
- y(t) = y(∞) + [y(0⁺) − y(∞)]e^(−t/τ): the final value plus the (initial − final) difference decaying exponentially. This form is provided on the exam formula page.
- Natural response
- The response of a source-free first-order circuit as its stored energy decays to zero — a pure decaying exponential set by τ and the initial condition.
- Step response
- The response when a DC source is suddenly applied (or changed): the variable moves exponentially from its initial value y(0⁺) to a new steady value y(∞) with time constant τ.
- Three-step method
- The standard recipe for any first-order transient: (1) find y(0⁺) using continuity of v_C or i_L, (2) find y(∞) at the post-switch DC steady state, (3) find τ from R and C (or L), then substitute into the general solution.
First-Order Transients: RC & RL Circuits FAQ
What is the time constant and how do I find it?
The time constant τ sets how fast a first-order circuit responds: τ = RC for an RC circuit and τ = L/R for an RL circuit. The R is the Thévenin resistance the capacitor or inductor 'sees' looking back into the rest of the circuit — so if there is more than one resistor, first reduce the network to its Thévenin equivalent at the storage element's terminals. After about 5τ the response is within roughly 1% of its final value.
How does the three-step method work?
For any first-order transient: (1) find the initial value y(0⁺) using continuity of v_C or i_L; (2) find the final value y(∞) by analysing the post-switch DC steady state, with the capacitor as an open circuit and the inductor as a short circuit; (3) find τ from the Thévenin R and the C or L. Then plug into y(t) = y(∞) + [y(0⁺) − y(∞)]e^(−t/τ). The exam formula page provides that solution form, so the marks are in the three inputs.
What is the difference between the natural and step response?
The natural response is what a source-free circuit does as its stored energy drains away — a decay to zero. The step response is what happens when a DC source is suddenly applied or changed — an exponential move to a new steady value. Both share the same time constant τ; the general solution y(∞) + [y(0⁺) − y(∞)]e^(−t/τ) covers both, with y(∞) = 0 for the pure natural response.
Can Sia help me with RC and RL transient problems in ELECTENG 291?
Yes, as a study aid. Sia can run the three-step method, find τ from the Thévenin resistance, get the initial and final values, and write and sketch y(t). It explains and drills you on fresh numbers; it does not do graded assessment for you, and University of Auckland academic-integrity rules apply — confirm what is permitted on Canvas.
Exam move
Turn the three-step method into muscle memory, because it solves every first-order problem the same way: y(0⁺) from continuity, y(∞) from the post-switch DC steady state (capacitor open, inductor short), and τ from the Thévenin resistance seen by the storage element times C (or L over R). The general solution is on the exam formula page, so drill the three inputs and the bracket order (initial − final), which is the classic sign slip. Practise recognising RC vs RL and computing τ quickly, and be able to sketch the exponential with the 5τ settling landmark. Because τ needs the Thévenin resistance at the element's terminals, keep your Module 1 reduction skills sharp. Confirm assessment details on Canvas.
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