First-Order Transient CircuitsIntroduction and contextA transient is the time-dependent interval during which a circuit moves from one steady operating condition to another. Switching, energizing, de-energizing, fault interruption, and signal changes can all produce transients. In a first-order circuit, one independent energy-storage element is present: a capacitor or an inductor. Consequently, the circuit response is described by one first-order differential equation and one exponential mode. First-order analysis is important because it connects circuit laws with practical quantities such as rise time, delay, relay operation, inrush current, filtering, snubber design, and protection. The central strategy is to determine the initial value, the final value, and the time constant. Learning objectivesAfter studying this chapter, you should be able to: - distinguish natural, forced, and complete responses;
- derive and apply the time constants of RC and RL circuits;
- use continuity of capacitor voltage and inductor current at ideal switching instants;
- analyze source-free, step, and switched circuits;
- calculate voltages and currents during charging, discharging, and energizing;
- use initial- and final-value reasoning to check results;
- recognize practical limitations of ideal first-order models.
1. Physical basis of first-order behaviorA capacitor stores electric-field energy according to wC = 1/2 C vC2 where wC is energy in joules, C is capacitance in farads, and vC is capacitor voltage in volts. Its current-voltage relation is iC(t) = C dvC(t)/dt An ideal capacitor voltage cannot change instantaneously unless an impulsive, unbounded current is applied. Thus, for ordinary finite switching currents, vC(0^+) = vC(0^-) An inductor stores magnetic-field energy: wL = 1/2 L iL2 where L is inductance in henries and iL is inductor current in amperes. Its voltage-current relation is vL(t) = L diL(t)/dt For finite voltages, inductor current is continuous: iL(0^+) = iL(0^-) These continuity rules apply to ideal lumped circuits without impulses. They do not mean that every other voltage or current is continuous; resistor variables may change abruptly after switching. 2. The universal first-order responseFor a stable first-order circuit, the response of any selected voltage or current can be written as x(t) = x(∞) + [x(0^+) − x(∞)]e^(−t/τ), t ≥ 0 Here x(t) is the desired variable, x(0^+) is its value immediately after switching, x(∞) is its final dc value, and τ is the time constant in seconds. This expression assumes the post-switching circuit is linear, time invariant, and stable, and that the independent sources are constant for t ≥ 0. The same result is often separated into x(t) = xn(t) + xf(t) where xn(t) is the natural (source-free or homogeneous) response and xf(t) is the forced (particular or steady-state) response. For a stable circuit, xn(t) = K e^(−t/τ) The complete response is the sum of both parts. The coefficient K is determined by the initial condition. At characteristic times, the remaining initial error is especially useful: | Time | Fraction of initial error remaining | Fraction of transition completed |
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| τ | 36.8% | 63.2% | | 2τ | 13.5% | 86.5% | | 3τ | 5.0% | 95.0% | | 4τ | 1.83% | 98.17% | | 5τ | 0.67% | 99.33% |
The time constant is not the time required for complete charging or discharging; an exponential approaches its limiting value asymptotically. Engineering practice commonly treats five time constants as practical settling. 3. RC circuits3.1 Source-free RC circuitConsider a charged capacitor connected across a resistor, with no independent source. Applying KCL gives C dvC/dt + vC/R = 0 or dvC/dt + vC/(RC) = 0 The solution is vC(t) = V0 e^(−t/(RC)) where V0 = vC(0^+) and τ = RC The resistor current, with a reference direction from the capacitor’s positive terminal through the resistor, is iR(t) = vC(t)/R = (V0/R)e^(−t/τ) If the current reference is chosen in the opposite direction, its sign is negative. The capacitor voltage and stored energy decay exponentially; energy decays as e^(−2t/τ), not as e^(−t/τ). 3.2 Series RC step responseSuppose a dc source Vs is connected through R to an initially uncharged capacitor. KVL gives Vs = Ri + vC Using i = C dvC/dt yields RC dvC/dt + vC = Vs The final capacitor voltage is Vs, and the initial voltage is zero. Therefore, vC(t) = Vs[1 − e^(−t/RC)] and i(t) = (Vs/R)e^(−t/RC) At t = 0^+, the capacitor initially behaves like a voltage source fixed at its prior voltage; an initially uncharged capacitor has zero voltage and therefore behaves like a short circuit for the purpose of finding the initial current. At dc steady state, the capacitor current is zero and the capacitor behaves as an open circuit. For a nonzero initial voltage V0, the same source is applied at t=0 and vC(t) = Vs + (V0 − Vs)e^(−t/RC) This is the general step response for the series topology. Worked example 1: charging capacitorA 12 V source is connected through a 4.7 kΩ resistor to a 22 μF capacitor initially at 0 V. Determine τ, vC(τ), and the initial current. The time constant is τ = RC = (4700)(22 × 10^(−6)) = 0.1034 s The initial current is i(0^+) = 12/4700 = 2.55 mA. At one time constant, vC(τ) = 12(1 − e^(−1)) ≈ 7.59 V. The result is physically consistent: the voltage has completed 63.2% of its 12 V transition. 3.3 Finding an RC time constant in a general networkThe formula τ=RC uses the resistance seen by the capacitor, not necessarily a single visible resistor. To find the Thevenin resistance Rth seen from the capacitor terminals: - remove the capacitor;
- deactivate independent sources: replace independent voltage sources by shorts and independent current sources by opens;
- calculate the resistance looking into the terminals.
Then τ = Rth C Dependent sources must remain active; in that case Rth is found using a test source or an appropriate circuit method. If the network contains multiple independent capacitors, the response is generally higher order and this single-time-constant method is not sufficient. 4. RL circuits4.1 Source-free RL circuitFor an inductor discharging through a resistor, KVL gives L diL/dt + RiL = 0 Therefore, iL(t) = I0 e^(−t/(L/R)) with τ = L/R The resistor voltage follows from the selected polarity and Ohm’s law. With current flowing through the resistor, its magnitude is R I0 e^(−t/τ). At an ideal dc steady state, an inductor behaves as a short circuit, although its current may be nonzero. 4.2 Series RL step responseA dc source Vs is connected in series with R and L. KVL gives Vs = Ri + L di/dt The final current is I_∞ = Vs/R, and the general response from initial current I0 is i(t) = I_∞ + (I0 − I_∞)e^(−t/τ), τ = L/R For an initially unenergized inductor, i(t) = (Vs/R)[1 − e^(−tR/L)] and the inductor voltage is vL(t) = Vs e^(−t/τ) for the usual polarity. Initially, an unenergized inductor behaves as an open circuit for current determination; at dc steady state it behaves as a short circuit. These statements describe limiting behavior and must not be used without considering the surrounding circuit. For a general network containing one inductor, τ = L/Rth where Rth is the resistance seen by the inductor after independent sources are deactivated. The reciprocal relationship between RC and RL time constants reflects the dimensions of the storage element: capacitance multiplies resistance, whereas inductance is divided by resistance. Worked example 2: RL energizationA 24 V source is applied to a series circuit with R=8 Ω and L=40 mH, with zero initial current. Then τ=L/R=0.040/8=0.005 s=5 ms, and I_∞=24/8=3 A. Thus, i(t)=3(1−e^(−t/5 ms)) A. At t=10 ms=2τ, i=3(1−e^(−2))≈2.59 A. The inductor voltage at that instant is 24e^(−2)≈3.25 V. 5. Switched circuits: a systematic methodA reliable procedure is: - Identify the variable: capacitor voltage or inductor current.
- Analyze the circuit for t<0: determine the stored initial energy. If the circuit has been in dc steady state for a long time, capacitors are open circuits and inductors are short circuits in that pre-switch network.
- Apply continuity: use vC(0^+)=vC(0^−) or iL(0^+)=iL(0^−) where the ideal finite-source assumptions apply.
- Analyze the post-switch dc circuit: determine x(∞).
- Find the time constant: use τ=RthC for RC or τ=L/Rth for RL.
- Write the complete response:
x(t)=x(∞)+[x(0^+)−x(∞)]e^(−t/τ). - Check limits and units: verify the response at t=0^+ and as t→∞.
A switch can change topology, source connections, and the resistance seen by the storage element. Consequently, the time constant before switching need not equal the time constant after switching. Worked example 3: nonzero initial capacitor voltageA capacitor initially has 10 V. After switching, it is connected to a 2 V ideal source through a resistance such that τ=4 ms. The capacitor voltage is vC(t)=2+(10−2)e^(−t/4 ms)=2+8e^(−t/4 ms) V. At one time constant, vC(τ)=2+8e^(−1)≈2+2.943=4.94 V. The voltage moves 63.2% of the way from 10 V toward 2 V: the change is approximately −5.06 V, leaving 4.94 V. This initial/final check prevents the common error of applying the zero-initial-voltage formula to a precharged capacitor. 6. Initial- and final-value reasoningThe initial value is determined from the pre-switch circuit and continuity. The final value is determined by the post-switch dc circuit, provided the circuit reaches a stable dc state. For a capacitor, set its dc current to zero; for an inductor, set its dc voltage to zero while retaining its current path. The response must satisfy x(0^+) = x(∞) + [x(0^+) − x(∞)] = x(0^+) and lim(t→∞) x(t) = x(∞) The initial slope can also be checked from the governing law. For an RC circuit, dvC/dt = iC/C and for an RL circuit, diL/dt = vL/L A capacitor voltage can have a finite initial slope but not a finite-current instantaneous jump. An inductor current can have a finite initial slope but not a finite-voltage instantaneous jump. These checks are valuable when signs or switch positions are complicated. 7. ApplicationsTiming and delayAn RC network can generate a threshold delay. For a charging capacitor reaching threshold VT from zero, tT = −RC ln(1 − VT/Vs), 0 < VT < Vs For a discharging capacitor from V0 toward zero, tT = −RC ln(VT/V0), 0 < VT < V0 These equations assume an ideal source, linear resistor, constant capacitance, and a threshold measurement that does not significantly load the capacitor. FilteringA first-order RC low-pass network attenuates rapid changes because the capacitor voltage cannot change quickly. Its sinusoidal transfer function, for an output across the capacitor, is H(jω) = 1/(1+jωRC) where ω is angular frequency in rad/s and j=√(−1). The cutoff angular frequency is ωc=1/(RC), or fc=1/(2πRC). This frequency-domain result complements, but does not replace, the step-response analysis. Relay and solenoid protectionAn inductive load can produce a large voltage when its current path is interrupted. A flyback diode provides an alternate current path, reducing the voltage across the switching device. The diode changes the effective discharge circuit and therefore changes the decay time; its forward voltage and the coil resistance must be included for accurate design. A resistor-capacitor snubber or a voltage-suppression device may be preferable where release speed, polarity, or switching frequency makes a diode unsuitable. Power-up and reset circuitsRC networks are used to delay reset release and suppress short disturbances. However, component tolerances, leakage, threshold variation, temperature, and input loading make a simple RC a poor precision timer. A comparator, monostable, or digital timer is generally needed when timing accuracy is important. 8. Illustrative applied case: relay-driver flyback protectionBackground. An embedded controller drives a 12 V relay coil. The coil has R=120 Ω and L=240 mH. A transistor switches the coil on and later turns it off. Problem. Without a suppression path, interrupting the inductor current can produce a potentially damaging voltage. The design must limit stress while allowing predictable current decay. Analysis. When energized from 12 V, the steady current is I_∞=12/120=0.10 A. The coil time constant during energization is τ=L/R=0.240/120=0.002 s=2 ms. When the transistor opens, the current cannot instantly become zero. A flyback diode provides a loop containing the coil resistance and the diode. If the diode drop is approximated as 0.7 V and the coil current is defined in the original direction, the decay approximately satisfies L di/dt + Ri + VD = 0. Thus the current approaches the final value −VD/R in the mathematical loop model, rather than zero asymptotically under the constant-drop approximation. A more physically useful expression over the interval while the diode conducts is i(t)=(I0+VD/R)e^(−tR/L)−VD/R, with I0=0.10 A, until the current reaches zero and the diode turns off. The initial decay slope is di/dt=−(RI0+VD)/L=−(12+0.7)/0.240≈−52.9 A/s. The diode greatly reduces transistor voltage but increases the relay release time compared with an unclamped high-voltage turn-off. Solution choice. Use the diode when transistor protection and low cost dominate. Use a zener or TVS clamp when faster release is required, selecting its clamp voltage within the transistor’s safe operating limits. Outcome and learning points. The storage element determines the unavoidable continuity condition; the suppression network determines the subsequent decay path. Protection design is therefore a transient-design problem, not merely a component-selection problem. 9. Assumptions, limitations, and professional practiceThe standard exponential formulas assume lumped, linear, time-invariant components; ideal switching; constant independent sources; and a stable post-switching network. Real capacitors have leakage, equivalent series resistance, voltage-dependent capacitance, and dielectric absorption. Real inductors have winding resistance, core losses, saturation, parasitic capacitance, and sometimes nonlinear inductance. Switches have finite resistance, bounce, turn-on time, and voltage/current limits. For high-speed switching, parasitic elements may create second- or higher-order behavior. For large-signal magnetic circuits, L is not constant. A circuit with several storage elements may require differential-equation, state-space, Laplace-transform, or numerical simulation methods. SPICE simulation is useful, but its results should be checked against continuity, energy, limiting values, and component ratings. Good engineering practice includes specifying tolerances, worst-case time constants, temperature range, leakage and loading, maximum capacitor voltage, inductor saturation current, switch stress, electromagnetic compatibility, and safe discharge paths. A stored capacitor charge or coil energy can remain hazardous after power removal; design must include appropriate isolation, bleeder resistors, discharge verification, and applicable safety standards. Chapter SummaryA first-order transient circuit contains one independent energy-storage element. Capacitor voltage and inductor current are continuous at an ideal finite-current or finite-voltage switching instant. The complete response is the sum of natural and forced responses and is most efficiently written as x(t)=x(∞)+[x(0^+)−x(∞)]e^(−t/τ). For an RC circuit, τ=RthC; for an RL circuit, τ=L/Rth. Source-free responses decay toward zero, while step responses move exponentially toward a final dc value. Initial-value, final-value, unit, and energy checks are essential. These methods support timing, filtering, relay protection, reset circuits, and practical switching design, while real components and tolerances limit ideal-model accuracy. Key TermsTransient: Time interval during which a circuit changes between operating states. Natural response: Response produced by stored energy with independent sources set to zero. Forced response: Response produced by external sources in the post-switching circuit. Complete response: Sum of natural and forced responses. Time constant: Time scale τ governing exponential change. Capacitance: Ratio of stored charge to voltage, measured in farads. Inductance: Property relating flux linkage to current, measured in henries. Continuity condition: Requirement that ideal capacitor voltage or inductor current cannot jump under finite excitation. Thevenin resistance: Resistance seen by the storage element after independent sources are deactivated. Source-free circuit: Circuit whose independent sources have been removed or set to zero. Step response: Response to a sudden change in a constant source. Steady state: Limiting condition after transient effects have decayed. Initial value: Circuit variable evaluated immediately after switching. Final value: Circuit variable approached as time tends to infinity. Flyback diode: Diode providing a current path for a switched-off inductive load. Cutoff frequency: Frequency at which a first-order low-pass magnitude is 1/√2 of its low-frequency value. Settling time: Time required to enter a specified tolerance band around the final value. Snubber: Suppression network used to control switching voltage or current transients. Review QuestionsShort Answer Questions- Why is capacitor voltage continuous at an ideal switching instant under finite current?
- State the time constants for a general first-order RC and RL circuit.
- Distinguish natural response from forced response.
- What are the dc steady-state models of a capacitor and an inductor?
- What fraction of the initial error remains after five time constants?
Long Answer Questions- Derive the source-free RC response from KCL and identify its time constant.
- Derive the series RL step response for a nonzero initial inductor current.
- Explain a systematic procedure for solving a switched first-order circuit.
- Discuss how a flyback diode changes inductive turn-off behavior.
- Explain the assumptions and limitations of the ideal first-order model.
Analytical / Application Questions- A 5 V source charges a 10 μF capacitor through 2 kΩ from zero initial voltage. Find the time constant, initial current, and capacitor voltage at 3τ.
- An inductor has 80 mH and discharges through 16 Ω from an initial current of 2 A. Find its time constant and current at 4 ms.
- A capacitor initially at 10 V is connected through a network with τ=4 ms to a 2 V final value. Determine its voltage at 4 ms and explain the physical direction of change.
Multiple Choice Questions with AnswersQuestion 1For a source-free RC circuit, which expression gives the capacitor voltage when its initial voltage is V₀? A. vC(t)=V₀e^(−t/RC) B. vC(t)=V₀(1−e^(−t/RC)) C. vC(t)=V₀e^(t/RC) D. vC(t)=RCV₀e^(−t) Correct Answer: A. vC(t)=V₀e^(−t/RC) Question 2A capacitor is connected to a resistor and has reached one time constant during charging from zero toward Vₛ. What voltage does it reach? A. 36.8% of Vₛ B. 50.0% of Vₛ C. 63.2% of Vₛ D. 86.5% of Vₛ Correct Answer: C. 63.2% of Vₛ Question 3A 4 kΩ resistor and a 25 μF capacitor form a first-order RC circuit. What is the time constant? A. 0.010 ms B. 0.100 s C. 100 s D. 625 s Correct Answer: B. 0.100 s Question 4For a series RL circuit with R = 10 Ω and L = 0.20 H, what is the time constant? A. 0.020 s B. 0.20 s C. 2.0 s D. 50 s Correct Answer: A. 0.020 s Question 5What continuity condition applies to an ideal inductor at an ordinary switching instant? A. Its voltage must be zero B. Its current cannot change instantaneously C. Its resistance must be infinite D. Its power must remain constant Correct Answer: B. Its current cannot change instantaneously Question 6A capacitor voltage changes from 10 V toward a final value of 2 V with τ = 4 ms. What is its voltage at t = τ? A. 2.94 V B. 4.94 V C. 5.06 V D. 7.06 V Correct Answer: B. 4.94 V Question 7In finding the time constant of an RC network, what is done to independent voltage sources? A. Replace them with open circuits B. Replace them with short circuits C. Double their voltages D. Leave them active while removing the resistor Correct Answer: B. Replace them with short circuits Question 8An initially unenergized series RL circuit is connected to a dc source. At t = 0⁺, the inductor most appropriately behaves as: A. An open circuit for current determination B. A short circuit with zero current C. A resistor equal to L/R D. An ideal voltage source equal to the supply Correct Answer: A. An open circuit for current determination Question 9After three time constants, approximately what percentage of the initial transient error remains? A. 1.8% B. 5.0% C. 36.8% D. 63.2% Correct Answer: B. 5.0% Question 10A 12 V source drives a 120 Ω relay coil. What is the ideal steady-state coil current? A. 0.010 A B. 0.10 A C. 1.44 A D. 10 A Correct Answer: B. 0.10 A
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