Visual walkthrough — RL circuit — growth and decay of current

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Step 1 — Draw the circuit and name every arrow

WHAT. We have a loop: a battery, a switch, a resistor, and an inductor, all connected end-to-end in a single ring. Nothing more.

WHY start here. Every equation later is just "the total voltage around this loop adds to zero." You cannot write that sentence until you can see each thing that pushes or resists in the loop.

PICTURE. Look at the figure. Follow the loop clockwise starting at the battery.

Let me name what each label means, in plain words:

• $\epsilon$ (the battery) — a fixed push, measured in volts. Think of it as a pump that always tries to shove the same amount of "electrical pressure" into the loop. (Volt = push per unit charge.)
• $R$ (the resistor) — a friction element. When current $I$ flows through it, it eats up voltage $IR$. Bigger current → bigger loss.
• $L$ (the inductor) — a coil of wire. It is the stubborn element. We build its behaviour carefully in Step 3.
• $I$ — the current: how fast charge flows past a point, measured in amperes (amps). This is the single number whose story we are telling.

The whole page answers one question: starting from $I=0$ the instant we close the switch, how does $I$ grow with time?

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Step 2 — The rate of change $\frac{dI}{dt}$, drawn as a slope

WHAT. We introduce the symbol $\frac{dI}{dt}$. In plain words it is the slope of the current-versus-time graph — how many amps the current is gaining per second, right now.

WHY we need this tool and not another. The whole drama of an inductor is about change: it reacts to how fast the current is changing, not to how big it is. To talk about "how fast something changes at an instant," the only tool that fits is the derivative — the slope of the curve at a single point. That is exactly what $\frac{dI}{dt}$ is.

PICTURE. On the graph of $I$ against $t$, pick any instant. Zoom in until the curve looks like a straight line. The steepness of that little line is $\frac{dI}{dt}$.

Reading the figure:

• Where the curve is steep (early on, red tangent), $\frac{dI}{dt}$ is large — current is gaining amps fast.
• Where the curve is flat (late, mint tangent), $\frac{dI}{dt}\approx 0$ — current barely changes.

Hold this thought: a stubborn inductor is going to force the slope to start big and shrink to zero. That shrinking slope is the bending of the curve.

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Step 3 — The inductor's rule: it fights the slope

WHAT. The inductor produces a voltage of its own, called the back-EMF, equal to $L\frac{dI}{dt}$. It points against whatever change is happening.

WHY this form. By Faraday's law of electromagnetic induction, a changing current makes a changing magnetic field, which induces a voltage. By Lenz's law, that induced voltage always opposes the change that made it. The size is proportional to how fast the current changes ($\frac{dI}{dt}$) and to how good the coil is at this ($L$). Multiply them: opposition $=L\frac{dI}{dt}$.

PICTURE. Two panels. Current trying to grow → inductor pushes backward (a drop). The faster the attempted change, the bigger the backward push.

Term-by-term:

$V_L=\underset{\text{how stubborn}}{\underbrace{L}}\underset{\text{how fast current changes}}{\underbrace{\frac{dI}{dt}}}$

• $L$ measures the coil's stubbornness. Double $L$ → double the back-push for the same rate of change.
• $\frac{dI}{dt}$ is the rate from Step 2. If current is not changing, $\frac{dI}{dt}=0$, so the inductor produces nothing — it becomes an ordinary wire.

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Step 4 — Add up the loop: Kirchhoff's voltage law

WHAT. Walk once around the ring and add every voltage. It must total zero — you cannot come back to where you started at a different "height." This is Kirchhoff's voltage law.

WHY. Voltage is like height on a hill. Go around any closed path and your net rise must be zero. So (battery push) minus (resistor drop) minus (inductor back-push) $=0$.

PICTURE. The loop with a "+" gain at the battery and two "$-$" drops, one at $R$, one at $L$.

$\underset{\text{battery raises}}{\underbrace{\epsilon }}-\underset{\text{resistor eats}}{\underbrace{IR}}-\underset{\text{inductor fights}}{\underbrace{L\frac{dI}{dt}}}=0$

• $\epsilon$ — the constant push, positive because we cross the battery from $-$ to $+$.
• $IR$ — the resistor drop; subtracted because current flows through it, losing voltage.
• $L\frac{dI}{dt}$ — the inductor's opposition while current grows ($\frac{dI}{dt}>0$), so it acts like a drop too.

Now rearrange to get the slope alone:

$\frac{dI}{dt}=\frac{\epsilon -IR}{L}$

Read this as a rule for the slope: at any current $I$, the slope of the curve equals $(\epsilon -IR)/L$. This single line already tells the whole story of the shape — Step 5 reads it off.

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Step 5 — Read the shape straight off the slope rule

WHAT. Before any integration, just think about $\frac{dI}{dt}=\frac{\epsilon -IR}{L}$ at the two ends.

WHY. A good physicist reads the qualitative shape from the equation first, then confirms with algebra. This is where the bending comes from.

PICTURE. The curve with slope arrows drawn at start, middle and end.

• At the start ($t=0$, so $I=0$): $\frac{dI}{dt}=\frac{\epsilon -0}{L}=\frac{\epsilon }{L}$ — the largest slope. The curve leaves the origin steeply.
• As $I$ grows, the term $IR$ gets bigger, so $\epsilon -IR$ shrinks, so the slope shrinks — the curve bends over.
• At the finish, when $IR=\epsilon$, the top and slope hit zero: $\frac{dI}{dt}=0$. The current stops growing. This special value is

$I_{\max }=\frac{\epsilon }{R}.$

So even without solving, we know: starts steep, bends over, flattens at $\epsilon /R$. That is an exponential approach.

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Step 6 — Turn the slope rule into the curve (integration)

WHAT. We now solve $\frac{dI}{dt}=\frac{\epsilon -IR}{L}$ to get $I$ as a formula in $t$.

WHY integration. The derivative gave us the slope at every point. To rebuild the actual curve from its slopes, we run the derivative backwards — that operation is called integration (adding up all the little slope-steps). It is the natural undo of $\frac{d}{dt}$.

How — separate the variables. Put everything with $I$ on the left, everything with $t$ on the right, so each side can be added up on its own:

$\frac{dI}{\epsilon -IR}=\frac{dt}{L}$

Add up (integrate) from the switch-on ($I=0$ at $t=0$) to a general later moment ($I$ at $t$):

${\int }_0^I\frac{dI}{\epsilon -IR}={\int }_0^t\frac{dt}{L}$

The right side is easy: $\frac{t}{L}$. The left side, worked in the parent note, gives $-\frac{1}{R}\ln \left(\frac{\epsilon -IR}{\epsilon }\right)$. Setting them equal:

$\ln \left(\frac{\epsilon -IR}{\epsilon }\right)=-\frac{Rt}{L}$

• $\ln$ is the natural logarithm — it answers "$e$ to what power gives this?" We use it because integrating $1/(\text{stuff})$ always produces a log.
• The right side carries the combination $\frac{R}{L}$, which has units of $1/\text{seconds}$ — its reciprocal $\tau =L/R$ is a time. This is the timescale we name next.

PICTURE. The left integral shown as the shaded area under the $\frac{1}{\epsilon -IR}$ curve — "adding up the slopes."

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Step 7 — Undo the log: the exponential appears

WHAT. Remove the logarithm by raising $e$ to both sides. The exponential $e^x$ is exactly the function that undoes $\ln$.

WHY the exponential. $\ln$ and $e^{(\cdot )}$ are perfect opposites: $e^{\ln x}=x$. Applying $e$ frees $I$ from inside the log.

$\frac{\epsilon -IR}{\epsilon }=e^{-Rt/L}$

Solve for $I$ and write $\tau =L/R$:

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

Term-by-term, right where each lives:

$I(t)=\underset{\text{ceiling }{I}_{\max }}{\underbrace{\frac{\epsilon }{R}}}(1-\underset{\begin{array}{c}\text{starts at }1,\\ \text{dies to }0\end{array}}{\underbrace{e^{-t/\tau }}})$

• $\frac{\epsilon }{R}$ — the flat ceiling from Step 5.
• $e^{-t/\tau }$ — a factor that is $1$ at $t=0$ and decays toward $0$. So $(1-e^{-t/\tau })$ climbs from $0$ to $1$.
• $\tau =L/R$ — the time constant: the clock speed of the whole process.

PICTURE. The exponential term $e^{-t/\tau }$ falling, and its mirror $1-e^{-t/\tau }$ rising, plotted together with $t$ measured in units of $\tau$.

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Step 8 — Why $\tau =L/R$ sets the pace (edge behaviours)

WHAT. Look at the two extreme circuits — very stubborn ($L$ huge) and very lossy ($R$ huge) — plus the degenerate case $R\to 0$.

WHY. A formula you can only plug into is half-understood. Pushing $L$ and $R$ to extremes shows why $\tau =L/R$ and not $LR$ or $R/L$.

PICTURE. Three growth curves on one axis: big $\tau$ (slow, lavender), medium (coral), small $\tau$ (fast, mint).

• Big $L$ → big $\tau$ → the coil is very stubborn → current creeps up slowly (lavender curve stretched right).
• Big $R$ → small $\tau$ → strong friction settles things quickly (mint curve snaps up).
• Units check (the parent's mistake-buster): $[L]/[R]=\text{H}/\Omega =(\text{V⋅s/A})/(\text{V/A})=\text{s}$ ✓. $LR$ would give the wrong units, so $\tau =L/R$ is forced.
• Degenerate case $R\to 0$: then $\tau =L/R\to \infty$ and $I_{\max }=\epsilon /R\to \infty$. The formula says the current rises forever with slope $\epsilon /L$ (a straight line, no ceiling). Physically: with no resistor to cap it, an ideal battery drives current up without bound — the curve never flattens.

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The one-picture summary

Everything above, compressed: the slope rule at the left (steep→flat), the integration in the middle, the finished exponential curve on the right, with the three landmark dots (0, 63.2%, ~100%) marked.

Recall Feynman retelling of the whole walkthrough

Close the switch. The battery wants to push current, but the coil hates sudden change, so at the very first instant the current is still zero — the curve starts at the floor. The battery's full push has nowhere to go except into the coil, so the current climbs at its steepest right at the start. As current builds, the resistor starts eating more and more of the battery's push ($IR$ grows), leaving less push to keep speeding the current up. So the climb gets lazier and lazier. Eventually the resistor eats the entire battery push; now there is nothing left to change the current, the coil falls silent, and the current sits flat at $\epsilon /R$. Mathematically that "start steep, bend over, flatten" is an exponential approach, and one number — $\tau =L/R$ — is the stopwatch: after one $\tau$ you are 63.2% of the way up, after about five $\tau$ you are basically there. Bigger coil, slower climb; bigger resistor, faster finish. (For the mirror-image story of a capacitor, see RC circuit — charging and discharging; for where the coil's energy went, see Energy stored in a magnetic field.)

Recall Quick self-check on the pictures

• In Step 5, why does the slope shrink as current grows? ::: Because slope $=(\epsilon -IR)/L$; as $I$ rises, $IR$ rises, so $\epsilon -IR$ shrinks.
• What operation rebuilt the curve from its slope in Step 6? ::: Integration (the undo of the derivative).
• Why does raising $e$ to both sides appear in Step 7? ::: $e^{(\cdot )}$ undoes $\ln$, freeing $I$ from inside the logarithm.
• What happens to the curve if $R\to 0$? ::: $\tau \to \infty$, no ceiling — current rises as a straight line forever with slope $\epsilon /L$.

Connections

• Kirchhoff's voltage law — the loop sum in Step 4.
• Inductance and self-induction — source of the $LdI/dt$ term (Step 3).
• Faraday's law of electromagnetic induction — why a changing current induces voltage.
• Lenz's law — why that induced voltage opposes the change.
• Energy stored in a magnetic field — where the built-up current's energy sits.
• RC circuit — charging and discharging — same-shaped curve, capacitor plays the mirror role.

Concept Map