RL circuit — growth and decay of current: Worked examples

Before we start, one reminder of the two symbols we lean on constantly:

$τ = L / R$ — the time constant, the natural "clock tick" of the circuit (seconds).
$I_{m a x} = ε / R$ — the steady current, the value growth climbs toward and decay falls from.

The scenario matrix

Think of every RL problem as landing in one cell of this grid. The columns are which quantity is unknown; the rows are what phase the circuit is in.

Phase → / Unknown ↓	Find current at a time	Find the time	Find a component ( $L$ , $R$ , $ε$ )	Voltage / rate / energy
Growth (rising)	Ex 1	Ex 2	Ex 3	Ex 6 (voltage), Ex 8 (energy)
Decay (falling)	Ex 4	Ex 5	Ex 7	—
Degenerate / limiting	Ex 9 ( $t = 0$ , $t \to \infty$ , $L \to 0$ )	—	—	—
Word problem / exam twist	Ex 10 (two-stage: charge then dump)

Every cell that can hold a distinct idea gets its own example below. As you read each one, the label tells you which cell it fills so you can see the coverage build.

Figure — RL circuit — growth and decay of current

Growth-phase examples

Worked example Example 1 — Growth: current at a given time

(cell: Growth / find current) $ε = 20 V$ , $R = 5 Ω$ , $L = 10 H$ . Find $I$ at $t = 4$ s.

Forecast: First guess whether $t = 4$ s is "early" or "late". Since $τ = L / R = 2$ s, then $4$ s $= 2 τ$ . Two time constants means we're past the 63% mark but not yet saturated — expect somewhere around 85% of max.

Step 1 — Compute $τ = L / R = 10/5 = 2$ s. Why this step? Every exponential needs its clock. Without $τ$ the " $- t / τ$ " in the exponent is meaningless.

Step 2 — Compute $I_{m a x} = ε / R = 20/5 = 4$ A. Why this step? Growth is a fraction of the ceiling. We need the ceiling first.

Step 3 — Plug into growth: $I = 4 (1 - e^{- 4/2}) = 4 (1 - e^{- 2})$ . Why this step? This is the actual growth law; $t / τ = 4/2 = 2$ .

Step 4 — Evaluate: $e^{- 2} = 0.1353$ , so $I = 4 (1 - 0.1353) = 4 (0.8647) = 3.459$ A.

Verify: $3.459/4 = 86.5%$ of max — right in the "around 85%" range we forecast. Units: (V/Ω) = A ✓.

Worked example Example 2 — Growth: solve for the

time (cell: Growth / find time) Same circuit ( $τ = 2$ s, $I_{m a x} = 4$ A). When does the current first reach $3$ A?

Forecast: $3$ A is $75%$ of max. That's between $τ$ (63%) and $2 τ$ (86.5%), so expect $t$ between $2$ s and $4$ s — probably close to $2.8$ s.

Step 1 — Write $3 = 4 (1 - e^{- t /2})$ , then divide: $1 - e^{- t /2} = 0.75$ . Why this step? We isolate the exponential so the unknown $t$ sits alone inside it.

Step 2 — Rearrange: $e^{- t /2} = 0.25$ . Why this step? Now the exponential equals a bare number; the only tool that "undoes" $e^{(\cdot)}$ is its inverse, the natural log $ln$ . That is why $ln$ enters — it is the exact question "to what power must $e$ be raised to give $0.25$ ?"

Step 3 — Take $ln$ : $- t /2 = ln 0.25 = - 1.386$ . Why this step? $ln$ pulls $t$ down from the exponent so we can solve linearly.

Step 4 — Solve: $t = 2 \times 1.386 = 2.77$ s.

Verify: Plug back: $4 (1 - e^{- 2.77/2}) = 4 (1 - e^{- 1.386}) = 4 (1 - 0.25) = 3$ A ✓. And $2.77$ s sits between $2$ and $4$ s as forecast.

Decay-phase examples

Component & energy examples

Worked example Example 6 — Growth: voltage across the inductor

(cell: Growth / voltage) $ε = 24$ V, $τ = 2$ s. Find the voltage across the inductor $V_{L}$ at $t = 0$ , $t = 2$ s, and $t = 6$ s.

Forecast: During growth the inductor's voltage is a decaying exponential $V_{L} = ε e^{- t / τ}$ (opposite behaviour to the current!). At $t = 0$ it's the full $24$ V; at one $τ$ it's $37%$ ( $\approx 9$ V); at $3 τ$ it's $\approx 5%$ ( $\approx 1.2$ V).

Step 1 — At $t = 0$ : $V_{L} = 24 e^{0} = 24$ V. Why this step? At the instant of switching, current can't jump, so all the EMF appears across the coil — this is why the inductor behaves like an open circuit at $t = 0$ .

Step 2 — At $t = 2$ s: $V_{L} = 24 e^{- 1} = 24 \times 0.3679 = 8.83$ V. Why this step? As current rises, the resistor takes a bigger share and the coil's share falls.

Step 3 — At $t = 6$ s: $V_{L} = 24 e^{- 3} = 24 \times 0.0498 = 1.20$ V.

Verify: At any time $V_{L} + I R = ε$ . At $t = 2$ s: $I = \frac{ε}{R} (1 - e^{- 1})$ , so $I R = ε (1 - e^{- 1}) = 24 \times 0.632 = 15.17$ V; then $V_{L} + I R = 8.83 + 15.17 = 24$ V $= ε$ ✓ (KVL from Kirchhoff's voltage law).

Worked example Example 8 — Energy stored at steady state

(cell: Growth / energy) $ε = 12$ V, $R = 3 Ω$ , $L = 2$ H. Find the magnetic energy stored once the current is fully grown, and the energy at $t = τ$ .

Forecast: Full current $I_{m a x} = 4$ A, so full energy $\frac{1}{2} L I^{2} = \frac{1}{2} (2) (16) = 16$ J. At $t = τ$ current is $63.2%$ of max, and energy $\propto I^{2}$ , so energy $\approx 0.63 2^{2} = 40%$ of $16 \approx 6.4$ J.

Step 1 — $I_{m a x} = ε / R = 12/3 = 4$ A. Why this step? Energy depends on current, so find the final current first.

Step 2 — $U_{\infty} = \frac{1}{2} L I_{m a x}^{2} = \frac{1}{2} (2) (4^{2}) = 16$ J. Why this step? This is the Energy stored in a magnetic field formula — energy hides in the field the coil builds.

Step 3 — At $t = τ$ : $I = 4 (1 - e^{- 1}) = 2.528$ A, so $U = \frac{1}{2} (2) (2.52 8^{2}) = 6.39$ J. Why this step? Same energy formula, current-at- $τ$ plugged in.

Verify: $6.39/16 = 39.9% \approx 0.63 2^{2} = 0.399$ ✓. Units: H·A² = (V·s/A)·A² = V·A·s = W·s = J ✓.

Degenerate & limiting cases

Every good student must know what happens at the edges: the very first instant, the far future, and pathological components. This one example sweeps all of them so you never freeze on a "trick" limit.

Worked example Example 9 — The three limits

(cell: degenerate / limiting) For a growth circuit $ε = 10$ V, $R = 2 Ω$ , $L = 4$ H, evaluate the current at (a) $t = 0$ , (b) $t \to \infty$ , and (c) discuss the degenerate limit $L \to 0$ .

Forecast: (a) current can't jump, so $I = 0$ . (b) coil becomes a plain wire, so $I = ε / R = 5$ A. (c) with no inductance there's nothing to slow the current — it should jump instantly to $5$ A.

Step 1 — At $t = 0$ : $I = 5 (1 - e^{0}) = 5 (1 - 1) = 0$ A. Why this step? $e^{0} = 1$ , killing the bracket. This confirms the "inductor = open circuit at $t = 0$ " rule.

Step 2 — At $t \to \infty$ : $e^{- t / τ} \to 0$ , so $I \to 5 (1 - 0) = 5$ A. Why this step? The shrink factor vanishes; the coil offers no opposition once $d I / d t = 0$ , so it acts like bare wire and current is set purely by $R$ .

Step 3 — $L \to 0$ : then $τ = L / R \to 0$ , so $e^{- t / τ} \to 0$ for any $t > 0$ . The current is $5$ A immediately — the exponential collapses to a step. Why this step? Zero inductance means zero "electrical inertia" — with $τ \to 0$ the smooth rise happens infinitely fast, recovering the plain resistor circuit (Inductance and self-induction is what created the delay).

Verify: $I (0) = 0$ ✓, $I (\infty) = ε / R = 5$ ✓, and $lim_{τ \to 0^{+}} 5 (1 - e^{- 1/ τ}) = 5$ ✓ (checked at a tiny $τ$ below).

Word problem / exam twist

Worked example Example 10 — Two-stage: charge up, then dump

(cell: word problem / exam twist) An electromagnet coil ( $L = 0.8$ H, $R = 4 Ω$ ) is connected to a $16$ V supply. It is left connected for $0.2$ s, then the supply is instantly replaced by a short so the current decays. Find the current at the moment of switching, and the current $0.1$ s after the switch to short.

Forecast: $τ = 0.2$ s, so $0.2$ s of charging = one $τ$ → about $63%$ of $I_{m a x} = 4$ A $\approx 2.5$ A. Then $0.1$ s of decay is half a $τ$ , factor $e^{- 0.5} \approx 0.61$ , so final $\approx 1.5$ A.

Step 1 — $τ = L / R = 0.8/4 = 0.2$ s; $I_{m a x} = ε / R = 16/4 = 4$ A. Why this step? Both phases share the same $τ$ (same $L$ , same $R$ ), so we compute it once.

Step 2 — Growth for $0.2$ s: $I_{1} = 4 (1 - e^{- 0.2/0.2}) = 4 (1 - e^{- 1}) = 4 \times 0.6321 = 2.528$ A. Why this step? This ending current becomes the starting current $I_{0}$ of the decay phase — the two stages are stitched at this value.

Step 3 — Decay for $0.1$ s starting from $I_{0} = 2.528$ A: $I_{2} = 2.528 e^{- 0.1/0.2} = 2.528 e^{- 0.5} = 2.528 \times 0.6065 = 1.533$ A. Why this step? Once shorted, Lenz's law keeps current flowing but it now decays from wherever it happened to be.

Verify: Stage-1 current $2.528$ A $= 63.2%$ of $4$ A ✓. Stage-2 factor $e^{- 0.5} = 0.6065$ ; $2.528 \times 0.6065 = 1.533$ A ✓, matching the $\approx 1.5$ A forecast.

Active recall

Recall Which cell is each of these?

"Given $ε, R, L$ , find $I$ at $t$ ." ::: Growth / find current (Ex 1).
"Current falls from $I_{0}$ to a value; find the time." ::: Decay / find time (Ex 5).
"Reaches X% of max in time $t$ ; find $L$ ." ::: Growth / find component (Ex 3).
"Voltage across the coil at $t$ ." ::: Growth / voltage — use $V_{L} = ε e^{- t / τ}$ (Ex 6).
"Charge for a while, then short and decay." ::: Two-stage exam twist — stitch end of growth as $I_{0}$ of decay (Ex 10).

Connections

Parent topic (Hinglish) — the derivations these examples exercise.
Kirchhoff's voltage law — the KVL check used in Example 6.
Inductance and self-induction — why $L \to 0$ kills the delay (Example 9).
Lenz's law — why current keeps flowing on shorting (Example 10).
Energy stored in a magnetic field — the $\frac{1}{2} L I^{2}$ used in Example 8.
RC circuit — charging and discharging — same log-based time solving, dual $τ = R C$ .