Understand RL transient behavior: Worked examples — Hardware deep dive

Before anything, let's re-anchor the three symbols we will use constantly, in plain words:

The scenario matrix

Every RL problem is one (or a blend) of these case classes. Each row is a distinct "shape" of question; the examples below are labelled with the row they cover.

#	Case class	What makes it different	Covered by
A	Energizing (source switched on, $i$ starts at 0)	Current rises, use $1 - e^{- t / τ}$	Ex 1
B	De-energizing (source removed, $i$ starts at $I_{0}$ )	Current decays, use $e^{- t / τ}$	Ex 2
C	Non-zero initial current + new source (general case)	Neither pure rise nor pure decay	Ex 3
D	Voltage question ( $v_{L}$ , $v_{R}$ ) instead of current	Must differentiate / use KVL	Ex 4
E	Inverse problem (given $i$ , find $t$ or $L$ )	Solve for the unknown parameter, needs $ln$	Ex 5
F	Limiting / degenerate inputs ( $t = 0^{+}$ , $t \to \infty$ , $R \to 0$ )	Check the edges where formulas "break"	Ex 6
G	Real-world word problem (energy, relay, spark)	Translate words → symbols	Ex 7
H	Exam twist (two-stage: charge then discharge)	Output of stage 1 becomes $I_{0}$ of stage 2	Ex 8

We will hit all eight. Keep this figure of the two master curves nearby — every example lives somewhere on one of these two shapes:

Figure — Understand RL transient behavior

Example 1 — Case A: Energizing (current rise)

Worked example Statement

A battery $V = 12 V$ drives a series $R = 4 Ω$ , $L = 8 mH$ . The switch closes at $t = 0$ with zero current. Find (a) $τ$ , (b) the final current, (c) $i$ at $t = 4 ms$ .

Forecast: Will $i$ at $4 ms$ be above or below half of the final current? (Guess before reading — $4 ms$ is exactly $2 τ$ , so think about the 86.5% row.)

Step 1 — Compute $τ = L / R = 0.008/4 = 0.002 s = 2 ms$ . Why this step? The time constant is the clock; every later answer is measured in multiples of it. See Time Constant.

Step 2 — Final current $I_{final} = V / R = 12/4 = 3 A$ . Why this step? After a long time $d i / d t \to 0$ , so $v_{L} = 0$ and the inductor is just a wire — pure Ohm's law. This is Steady-State Analysis.

Step 3 — At $t = 4 ms = 2 τ$ : $i = 3 (1 - e^{- 2}) = 3 (1 - 0.13534) = 3 (0.86466) = 2.594 A$ . Why this step? We plug $t / τ = 2$ into the rise formula (Case A shape). The exponent is dimensionless — that's why we always work in units of $τ$ .

Verify: $2.594 A$ is $86.5%$ of $3 A$ , matching the $2 τ$ row of the parent's table. Above half — forecast confirmed. Units: $(V) / (Ω) = A$ ✓.

Example 2 — Case B: De-energizing (current decay)

Worked example Statement

An inductor is already carrying $I_{0} = 5 A$ . At $t = 0$ the source is shorted out, leaving just $R = 10 Ω$ and $L = 20 mH$ in a loop. Find $i$ at $t = 6 ms$ .

Forecast: After $6 ms = 3 τ$ , roughly what fraction of $5 A$ is left — half? A tenth? Twentieth?

Step 1 — $τ = L / R = 0.020/10 = 0.002 s = 2 ms$ . Why this step? Same clock, even though there is no source now — $τ$ depends only on $L$ and $R$ , not on $V$ .

Step 2 — Identify the shape: no source, so KVL gives $0 = i R + L d i / d t$ , whose solution is pure decay $i = I_{0} e^{- t / τ}$ . Why this step? With $V = 0$ the "target" the current heads toward is $0$ ; the exponential just eats the initial value away.

Step 3 — $t = 6 ms = 3 τ$ : $i = 5 e^{- 3} = 5 (0.049787) = 0.2489 A$ . Why this step? Substitute $t / τ = 3$ .

Verify: $0.2489/5 = 4.98%$ , matching the $3 τ$ "remaining $5.0%$ " row. Sanity: decaying answer is below the start $5 A$ ✓.

Example 3 — Case C: Non-zero start heading to a new final value

Worked example Statement

A coil already carries $I_{0} = 1 A$ when, at $t = 0$ , a source is switched in so the circuit would eventually settle at $I_{final} = 4 A$ . With $τ = 3 ms$ , find $i$ at $t = 3 ms$ .

Forecast: The current must climb from $1 A$ toward $4 A$ . After one $τ$ it covers $63.2%$ of the gap. What is the gap, and where does that put us?

Step 1 — Write the general first-order solution: $i (t) = I_{final} + (I_{0} - I_{final}) e^{- t / τ}$ . Why this step? Every first-order transient is "final value plus a decaying memory of how far you started off." This one formula contains Case A ( $I_{0} = 0$ ) and Case B ( $I_{final} = 0$ ) as special cases. See First-Order Differential Equations.

Step 2 — The gap is $I_{0} - I_{final} = 1 - 4 = - 3 A$ (negative because we start below the target). Why this step? The sign tells us the exponential term is negative, i.e. it adds current over time — the curve rises.

Step 3 — At $t = τ$ : $i = 4 + (- 3) e^{- 1} = 4 - 3 (0.36788) = 4 - 1.10364 = 2.896 A$ . Why this step? Substitute $t / τ = 1$ .

Verify: Gap covered $= (2.896 - 1) / (4 - 1) = 1.896/3 = 63.2%$ ✓ — exactly the $1 τ$ rule, just applied to the gap rather than to the final value. Answer sits between start ( $1$ ) and target ( $4$ ) ✓.

Example 4 — Case D: Voltage across the inductor and resistor

Worked example Statement

Same circuit as Example 1 ( $V = 12 V$ , $R = 4 Ω$ , $L = 8 mH$ , $τ = 2 ms$ , energizing from 0). Find $v_{L}$ and $v_{R}$ at (a) $t = 0^{+}$ , (b) $t = 2 ms$ .

Forecast: At the very instant of switch-on, who takes the full $12 V$ — the resistor or the inductor?

Step 1 — Get $v_{L} (t)$ . Differentiate the rise current, or use the shortcut $v_{L} = V e^{- t / τ}$ . Why this step? $v_{L} = L \frac{d i}{d t}$ ; differentiating $\frac{V}{R} (1 - e^{- t / τ})$ gives $\frac{V}{R} \cdot \frac{1}{τ} e^{- t / τ} \cdot L$ , and since $τ = L / R$ the $L$ and $R$ cancel to leave $V e^{- t / τ}$ . This is why we needed calculus here rather than just plugging numbers — voltage is about the rate of change of current, per Inductor Fundamentals.

Step 2 — At $t = 0^{+}$ : $v_{L} = 12 e^{0} = 12 V$ , and $v_{R} = i R = 0$ (current still zero). Why this step? At the switch instant current is continuous from before ( $= 0$ ), so the resistor drops nothing and the inductor must swallow the entire source. Forecast answer: the inductor takes all $12 V$ .

Step 3 — At $t = 2 ms = 1 τ$ : $v_{L} = 12 e^{- 1} = 4.4146 V$ ; $v_{R} = 12 - 4.4146 = 7.5854 V$ . Why this step? KVL says $v_{R} + v_{L} = V$ at every instant, so once we have $v_{L}$ the resistor voltage is just the leftover. See Kirchhoff's Voltage Law.

Verify: Cross-check $v_{R}$ from current: at $1 τ$ , $i = 3 (1 - e^{- 1}) = 1.8964 A$ , so $v_{R} = i R = 1.8964 \times 4 = 7.5854 V$ ✓ — matches. And $v_{R} + v_{L} = 7.5854 + 4.4146 = 12 V$ ✓ (KVL).

Example 5 — Case E: Inverse problem (solve for the unknown)

Worked example Statement

In an energizing RL circuit with $I_{final} = 2 A$ and $τ = 5 ms$ , at what time $t$ does the current first reach $1.5 A$ ?

Forecast: $1.5 A$ is $75%$ of the final value — between the $1 τ$ (63%) and $2 τ$ (86%) marks. So expect $t$ between $5$ and $10 ms$ .

Step 1 — Start from $i = I_{final} (1 - e^{- t / τ})$ and isolate the exponential: $e^{- t / τ} = 1 - \frac{i}{I _{final}} = 1 - \frac{1.5}{2} = 0.25$ . Why this step? We know $i$ and want $t$ , so we must undo the exponential — which means the logarithm is unavoidable. That's why $ln$ appears in every "find the time" problem.

Step 2 — Take natural log: $- \frac{t}{τ} = ln (0.25) = - 1.38629$ , so $t = 1.38629 τ$ . Why this step? $ln$ is the exact inverse of $e^{(\cdot)}$ ; it converts "what exponent gives 0.25?" into a number.

Step 3 — $t = 1.38629 \times 5 ms = 6.931 ms$ . Why this step? Convert back to real time by multiplying by $τ$ .

Verify: Plug back: $i = 2 (1 - e^{- 6.931/5}) = 2 (1 - e^{- 1.38629}) = 2 (1 - 0.25) = 1.5 A$ ✓. And $6.93 ms$ lies between $5$ and $10 ms$ — forecast confirmed.

Example 6 — Case F: Limiting and degenerate inputs

Worked example Statement

For the general rise $i (t) = \frac{V}{R} (1 - e^{- t / τ})$ , evaluate the three "edge" behaviours: (a) $t = 0^{+}$ , (b) $t \to \infty$ , (c) the degenerate case $R \to 0$ (ideal wire + coil, no resistor).

Forecast: With no resistor, what stops the current — does it settle, or keep climbing forever?

Step 1 — At $t = 0^{+}$ : $e^{0} = 1$ , so $i = \frac{V}{R} (1 - 1) = 0$ . Why this step? Confirms current is continuous through the switch — it starts exactly where it was ( $0$ ), no jump. This is the edge that kills the "current jumps to $V / R$ " mistake.

Step 2 — As $t \to \infty$ : $e^{- t / τ} \to 0$ , so $i \to \frac{V}{R} = I_{final}$ . Why this step? The exponential memory of the start fades completely; the circuit forgets the transient and reaches steady state.

Step 3 — Degenerate $R \to 0$ : here $τ = L / R \to \infty$ (infinitely slow clock) and $I_{final} = V / R \to \infty$ . Expanding for small $t$ , $i (t) \approx \frac{V}{R} \cdot \frac{t}{τ} = \frac{V}{R} \cdot \frac{R t}{L} = \frac{V}{L} t$ . Why this step? Notice the $R$ cancels! With no resistance, current rises linearly at rate $V / L$ and never levels off — matching physics ( $v_{L} = V$ constant forces constant $d i / d t$ ). Forecast answer: it ramps forever.

Verify: Take $V = 12$ , $L = 8 mH$ . At $t = 0.5 ms$ the ramp gives $i = \frac{12}{0.008} (0.0005) = 0.75 A$ . The full formula with a tiny $R = 0.001 Ω$ gives $i = 12000 (1 - e^{- 0.0005/8}) \approx 0.75 A$ ✓ — the linear ramp is the $R \to 0$ limit.

Example 7 — Case G: Real-world word problem (relay coil)

Worked example Statement

A relay coil ( $L = 50 mH$ , winding resistance $R = 25 Ω$ ) is driven by a $5 V$ supply. The relay pulls in when its current reaches $150 mA$ . How long after switch-on does it click? Also, how much energy is stored in the coil at that instant?

Forecast: Final current is $5/25 = 200 mA$ ; we need $150 mA = 75%$ of it. From Example 5's pattern, expect about $1.39 τ$ .

Step 1 — $τ = L / R = 0.050/25 = 0.002 s = 2 ms$ ; $I_{final} = V / R = 5/25 = 0.2 A$ . Why this step? Translate the words "supply $5 V$ , coil $R, L$ " into circuit numbers first.

Step 2 — Find $t$ for $i = 0.15 A$ : $e^{- t / τ} = 1 - \frac{0.15}{0.2} = 0.25 \Rightarrow t = τ ln 4 = 2 ms \times 1.38629 = 2.773 ms$ . Why this step? Same inverse-log method as Case E; the relay "clicks" the moment current hits threshold.

Step 3 — Energy stored: $E = \frac{1}{2} L i^{2} = \frac{1}{2} (0.050) (0.15)^{2} = \frac{1}{2} (0.050) (0.0225) = 5.625 \times 1 0^{- 4} J = 0.5625 mJ$ . Why this step? The inductor's magnetic energy is $\frac{1}{2} L I^{2}$ (from Inductor Fundamentals); the question asked for stored energy at the click instant, so we use the current at that time.

Verify: Time check — plug back: $i = 0.2 (1 - e^{- 2.773/2}) = 0.2 (1 - 0.25) = 0.15 A$ ✓. Energy units: $H \cdot A^{2} = (V\cdots/A) \cdot A^{2} = V\cdotA\cdots = W\cdots = J$ ✓.

Example 8 — Case H: Exam twist (charge then discharge)

Worked example Statement

Two-stage problem. Stage 1: $V = 20 V$ , $R = 10 Ω$ , $L = 5 mH$ , switch closed for a long time so the coil reaches steady current. Stage 2: at $t = 0$ the source is disconnected and the coil discharges through the same $R$ . Find the current $3 τ$ after disconnection.

Forecast: "A long time" in Stage 1 means the coil is fully charged to $V / R$ . After that, Stage 2 decays it. So the Stage-1 final current becomes the Stage-2 starting current — that's the whole trick.

Step 1 — Stage 1 steady current: $I_{0} = V / R = 20/10 = 2 A$ . Why this step? "A long time" ( $≫ 5 τ$ ) means the transient is over and $d i / d t = 0$ ; inductor is a wire, so pure Ohm's law. This value is the hand-off to Stage 2.

Step 2 — $τ = L / R = 0.005/10 = 0.0005 s = 0.5 ms$ (same $R, L$ in both stages). Why this step? The clock is unchanged because $R$ and $L$ are unchanged; only the source condition flipped.

Step 3 — Stage 2 decay at $t = 3 τ$ : $i = I_{0} e^{- 3} = 2 (0.049787) = 0.09957 A \approx 99.6 mA$ . Why this step? Case B pure decay, now with $I_{0} = 2 A$ inherited from Stage 1.

Verify: $0.09957/2 = 4.98%$ — the $3 τ$ "5% remaining" rule again ✓. Two-stage sanity: the answer must be below the Stage-1 steady $2 A$ , and it is ✓.

Recall Quick self-test

A coil at $2 A$ discharges through $R$ with $τ = 1 ms$ . Current at $2 ms$ ? ::: $2 e^{- 2} = 0.271 A$ . Rising circuit reaches what fraction of final at $t = τ ln 4$ ? ::: $75%$ (since $e^{- l n 4} = 0.25$ , and $1 - 0.25 = 0.75$ ). With no resistor, current does what over time? ::: Rises linearly at $V / L$ , never settles.

Connections

Understand RL transient behavior — the parent note with the core derivation.
Time Constant — the $τ = L / R$ clock every example uses.
Steady-State Analysis — the "long time" endpoint (Examples 1, 8).
First-Order Differential Equations — the general-solution shape in Example 3.
Inductor Fundamentals — the $v_{L} = L d i / d t$ and $\frac{1}{2} L I^{2}$ used in Examples 4, 7.
Kirchhoff's Voltage Law — the loop identity checked in Example 4.
RC Transient Behavior — the mirror-image case (swap current↔voltage, $τ = R C$ ).