Visual walkthrough — Understand RL transient behavior

We assume only two things you can picture: a battery pushes and a resistor pushes back in proportion to how much current flows. Everything else we build.

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Step 1 — Draw the loop and name the pieces

WHAT. We lay out one loop: a battery, a resistor, a coil (inductor), and a switch — all strung in a single line (in series, meaning the same current $i$ flows through every part).

WHY. Before any equation, we need a picture where each symbol has a home. If you can't point at $V$, $R$, $L$ on a drawing, the algebra later is just noise.

PICTURE.

Reading the figure:

• $V$ — the battery voltage (in volts). Think of it as a pump that always tries to push the same amount.
• $R$ — the resistor (in ohms $\Omega$). It burns energy; the more current, the bigger its push-back.
• $L$ — the inductor (a coil of wire, in henries $\text{H}$). This is our troublemaker — it resists changes in current.
• $i(t)$ — the current, measured in amps. The $(t)$ means "it changes with time" — that's the whole story.
• The switch is open before $t=0$ and snaps closed at $t=0$.

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Step 2 — The one law that makes an inductor special

WHAT. We write down what an inductor does: its voltage depends not on the current, but on how fast the current is changing.

$v_L=L\frac{di}{dt}$

Term by term:

• $v_L$ — voltage across the coil.
• $\frac{di}{dt}$ — the rate of change of current: "how many amps per second is $i$ climbing right now?" A steep current curve → big number; a flat curve → zero.
• $L$ — the coil's stubbornness. Bigger $L$ means the same change in current produces a bigger opposing voltage.

WHY this tool — why $\frac{di}{dt}$ and not just $i$? A resistor cares about how much current flows ($v_R=iR$). But a coil stores energy in a magnetic field, and that field only changes when the current changes. So the natural question is not "how much current?" but "how fast is it changing?" — and the mathematical object that answers "how fast is something changing?" is the derivative $\frac{di}{dt}$. That's exactly why the derivative enters here and nowhere in a pure-resistor circuit. See First-Order Differential Equations and Inductor Fundamentals.

PICTURE.

The figure shows two current curves. Where the curve is steep, $v_L$ is large; where it flattens, $v_L$ shrinks toward zero. A flat current (constant $i$) gives $\frac{di}{dt}=0$, so $v_L=0$ — the coil then behaves like a plain wire.

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Step 3 — Walk around the loop (Kirchhoff's Voltage Law)

WHAT. We add up every voltage as we walk once around the loop and set the total equal to the battery.

$\underset{\text{pump}}{\underbrace{V}}=\underset{\text{resistor drop}}{\underbrace{iR}}+\underset{\text{coil drop}}{\underbrace{L\frac{di}{dt}}}$

Term by term:

• $V$ — the push supplied.
• $iR$ — Ohm's law: the resistor's voll drop, proportional to current.
• $L\frac{di}{dt}$ — the coil's drop, from Step 2.

WHY. Energy can't appear or vanish going around a closed loop, so what the battery gives, the components must use up. That balance-sheet is Kirchhoff's Voltage Law. It converts a picture into one equation linking $i$ and its rate of change.

PICTURE.

The arrows show the "climb" at the battery and the two "drops" across $R$ and $L$; they must cancel back to zero when you return to the start.

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Step 4 — Separate current from time

WHAT. We shuffle the equation so everything with $i$ sits on the left and everything with $t$ sits on the right.

$L\frac{di}{dt}=V-iR⟹\frac{di}{V-iR}=\frac{dt}{L}$

• Left: $V-iR$ is "how much push is left over to change the current" — the battery minus what the resistor already ate.
• Right: pure time, scaled by $L$.

WHY this tool — separation of variables. The equation is a first-order differential equation: it mixes $i$ and $\frac{di}{dt}$. We can't just solve algebraically. The trick is to get $i$-stuff and $t$-stuff on opposite sides so each side can be integrated on its own — that's the standard machinery in First-Order Differential Equations.

PICTURE.

The figure shows the "leftover push" $V-iR$ (blue gap) shrinking as $i$ grows — when it hits zero, the current stops changing.

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Step 5 — Integrate both sides (add up all the tiny steps)

WHAT. Integration means "sum every infinitesimal slice from the start to now." We integrate from $t=0,i=0$ up to time $t$, current $i$:

$-\frac{1}{R}\ln (V-iR){|}_0^i=\frac{t}{L}$

• $\int \frac{di}{V-iR}=-\frac{1}{R}\ln (V-iR)$ — because the inside $V-iR$ changes at rate $-R$, and the integral of $\frac{1}{u}$ is $\ln u$. The $\ln$ appears precisely because we're integrating "one over something."
• The right side integrates trivially to $\frac{t}{L}$.

WHY the logarithm. We didn't choose $\ln$ for style — it's forced. Whenever a quantity's growth is proportional to itself (as here), summing $\frac{1}{u}du$ produces a natural log, and inverting it later produces the exponential. That's the mathematical seed of the whole exponential curve.

PICTURE.

The shaded strip is one tiny slice $\frac{di}{V-iR}$; the integral stacks all such strips into the area that equals $t/L$.

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Step 6 — Unwrap the log into an exponential

WHAT. Plug in the limits and solve for $i$.

$\ln \frac{V-iR}{V}=-\frac{R}{L}t⟹V-iR=V{e}^{-Rt/L}$

• The left combines both endpoints: $\ln (V-iR)-\ln V=\ln \frac{V-iR}{V}$.
• To undo $\ln$, we raise $e$ to both sides — $e^{\ln x}=x$. This is why the exponential $e^{-Rt/L}$ shows up: it's the inverse of the log we were forced into.

WHY. We want $i(t)$ as an explicit picture, not tangled inside a logarithm. Exponentiating frees it.

PICTURE.

The figure shows the "leftover push" $V-iR$ decaying as the pure exponential $V{e}^{-Rt/L}$ — starting at $V$, sliding toward $0$.

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Step 7 — Read off the current, and meet $\tau$

WHAT. Solve the last line for $i$:

$\boxed{i(t)=\frac{V}{R}(1-e^{-t/\tau })},\tau =\frac{L}{R}$

• $\frac{V}{R}$ — the final current $I_{\text{final}}$: once the coil stops fighting ($\frac{di}{dt}=0$), it's a wire and Ohm's law rules. This is the ceiling the curve rises toward. See Steady-State Analysis.
• $(1-e^{-t/\tau })$ — the shape: starts at $0$ (at $t=0$, $e^0=1$, bracket $=0$), climbs toward $1$.
• $\tau =\frac{L}{R}$ — the Time Constant: the natural clock of the circuit. We grouped $\frac{R}{L}t=\frac{t}{\tau }$ so time is measured in units of $\tau$.

WHY $\tau$ in the denominator of the exponent. Big $L$ → more inertia → slower rise (larger $\tau$). Big $R$ → burns stored energy faster → quicker settle (smaller $\tau$). This is why $R$ sits below $L$.

PICTURE.

The rising curve is marked at $1\tau$ (63.2% of the way up) and $5\tau$ (essentially done).

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Step 8 — The other case: discharge (source removed)

WHAT. Now open the battery out and short the loop while current $I_0$ still flows. Set $V=0$ in Step 3:

$0=iR+L\frac{di}{dt}⟹i(t)=I_0{e}^{-t/\tau }$

• No pump ($V=0$), so the coil's stored energy drains through $R$.
• Same $\tau =L/R$, same $e^{-t/\tau }$ — but now a pure decay from $I_0$ down to $0$.

WHY include this. A derivation must cover every case. Energizing and de-energizing are the two halves of "transient behaviour"; the second falls straight out of the same loop equation with $V=0$.

PICTURE.

Decay curve from $I_0$, dropping to 36.8% at $1\tau$.

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The one-picture summary

Everything on one board: the loop (top-left), the KVL equation, the rise curve toward $V/R$, its mirror inductor-voltage, and the decay curve — all breathing to the same clock $\tau =L/R$.

Recall Feynman retelling — the whole walkthrough in plain words

We drew a loop: pump, resistor, coil. The coil's special rule is that its voltage listens to how fast current changes, not how much there is — that's why a derivative shows up. We walked the loop and set pump = drops (that's KVL), giving one equation tying current to its own rate of change. We split current-stuff from time-stuff, summed up tiny slices (integrating), which coughed up a logarithm — and undoing a logarithm gives an exponential. Cleaning up, the current climbs like $\frac{V}{R}(1-e^{-t/\tau })$: it starts flat at zero (coil won't let it jump), curves up, and levels off at $V/R$ once the coil gives up and becomes a wire. Kill the pump and the same machinery gives a pure downhill $I_0{e}^{-t/\tau }$. The single number running the clock is $\tau =L/R$: heavier coil = slower, bigger resistor = quicker. Sixty-three percent by one $\tau$, done by five.

Recall Quick self-test

At $t=0^{+}$, what is the current in the energizing circuit? ::: Zero — the coil forbids an instant jump, so $i(0^{+})=i(0^{-})=0$. Why does a logarithm appear during integration? ::: Because we integrate $\frac{di}{V-iR}$, and $\int \frac{1}{u}du=\ln u$. Why does the exponential appear right after? ::: Undoing the log (exponentiating both sides) turns $\ln$ into $e$. What is the current's ceiling and why? ::: $V/R$ — at steady state $\frac{di}{dt}=0$, so the coil is a wire and only $R$ limits current. If $L=0$, what happens to the curve? ::: $\tau =0$, so current jumps instantly to $V/R$ — a plain resistor circuit.

Connections

• Understand RL transient behavior — the parent topic this walkthrough serves.
• Inductor Fundamentals — source of $v_L=Ldi/dt$.
• Kirchhoff's Voltage Law — the loop equation of Step 3.
• First-Order Differential Equations — the separate-and-integrate machinery.
• Time Constant — the clock $\tau =L/R$.
• Steady-State Analysis — the $V/R$ ceiling after $5\tau$.
• RC Transient Behavior — the same shape with $\tau =RC$.