Understand RC charging - discharging time constants: Visual walkthrough

Before line one, three plain words:

Voltage — how hard electricity is being "pushed" at a point, measured in volts ( $V$ ). Think of it as water pressure.
Current — how fast electric charge flows past a point, measured in amps ( $A$ ). Think of flow rate of water.
Charge — the actual "stuff" (electrons) that piles up, measured in coulombs ( $C$ ). Think of litres of water collected.

Step 1 — Meet the three parts on one loop

Figure — Understand RC charging - discharging time constants

WHAT. We draw the circuit: a battery of voltage $V_{s}$ , a resistor $R$ , and a capacitor $C$ , all wired in a single closed ring (a series loop).

WHY this picture first. Every quantity we will ever write lives on this ring. If you cannot point to where a symbol sits, the algebra later is just noise. So we anchor them now:

$V_{s}$ (amber battery) — the fixed push the battery supplies. It never changes.
$R$ (the zig-zag) — a "narrow pipe" that fights the flow of charge.
$C$ (the two parallel plates) — a "bucket" that collects charge and builds up its own back-pressure $V_{C}$ .

PICTURE. Notice the single cyan arrow marked $I$ : because it's ONE loop, the same current flows through every part. That single-arrow fact is what lets us chain the components together in Step 4.

We rely on Kirchhoff's Voltage Law and the capacitor law from Capacitors and stored energy. Both are rebuilt below.

Step 2 — The capacitor's one rule: charge makes pressure

WHAT. We zoom into the capacitor alone and state its defining law:

$charge collected (C) Q = bucket size (F) C \times back-pressure (V) V_{C}$

WHY this tool. We need a link between how much charge has arrived and the voltage it creates, because current (charge arriving) and voltage (what we want to plot) are two different things. This equation is the bridge. It says pressure is proportional to how full the bucket is.

PICTURE. In the figure the bucket fills with cyan charge; the amber gauge on the side (the voltage) rises in lock-step. Small bucket $C$ → the same charge makes the gauge shoot up high. Big bucket $C$ → the same charge barely moves the gauge.

Step 3 — Current is charge arriving per second

WHAT. Current is the rate at which charge flows:

$I = \frac{d Q}{d t}$

Read $\frac{d Q}{d t}$ as "the tiny change in charge $d Q$ during a tiny slice of time $d t$ " — a flow rate.

WHY the derivative. We could ask "how much charge is in the bucket?" (that's $Q$ ) or "how fast is charge going in right now?" (that's the slope of $Q$ versus time). Current is the second question, and the tool that answers "how fast is something changing right now" is the derivative. That is why $d / d t$ appears and no other tool does.

Now combine with Step 2. Since $Q = C V_{C}$ and $C$ is constant:

$I = \frac{d Q}{d t} = C \frac{d V _{C}}{d t}$

Term by term: $\frac{d V _{C}}{d t}$ is how fast the pressure is rising, and multiplying by the bucket size $C$ turns "pressure rising" into "charge arriving."

PICTURE. The figure shows the $Q$ -versus- $t$ curve; the current $I$ is the steepness of that curve (the little slope triangle). Steep curve = lots of current. Flat curve = no current.

Step 4 — Walk the loop: Kirchhoff's Voltage Law

WHAT. Kirchhoff's Voltage Law says: add up every voltage change as you walk once around the loop and you must return to where you started — the pushes and drops cancel to the source:

$battery pushes up V_{s} = resistor drops V_{R} + capacitor holds back V_{C}$

WHY. This is our master equation — it's the only thing tying all three characters together at once. Without it we have three separate facts and no way to combine them.

PICTURE. The figure walks the loop as a "voltage staircase": the battery lifts you up by $V_{s}$ ; the resistor drops you by $V_{R}$ ; the capacitor drops you by $V_{C}$ ; you land back at the start. The two drops together must equal the one lift.

Substitute Ohm's law for the resistor, $V_{R} = I R$ (voltage drop = current × resistance), and then the current from Step 3, $I = C d V_{C} / d t$ :

$V_{s} = drop across R R C \frac{d V _{C}}{d t} + drop across C V_{C}$

Step 5 — Read the equation as a rule of motion (before any calculus)

WHAT. Rearrange to isolate the rate:

$\frac{d V _{C}}{d t} = \frac{V _{s} - V _{C} gap to target}{τ R C}$

WHY this rewrite. The left side is "how fast $V_{C}$ climbs." The right side tells us exactly what sets that speed at every instant — no solving needed yet. This is the physical heart of the whole topic.

Read it aloud: the speed of charging is proportional to the remaining gap between where you are ( $V_{C}$ ) and where you're heading ( $V_{s}$ ).

PICTURE. The figure shows the target line $V_{s}$ and the current level $V_{C}$ . The amber double-arrow is the gap $V_{s} - V_{C}$ . When the gap is big (start), the climb is steep. As $V_{C}$ rises, the gap shrinks, so the climb flattens. This one sentence is the shape of the curve.

At $t = 0$ : $V_{C} = 0$ , gap $= V_{s}$ → fastest climb.
Later: $V_{C}$ near $V_{s}$ , gap $\to 0$ → climb $\to 0$ , curve goes flat.
Never: gap exactly zero in finite time → the curve approaches $V_{s}$ but only reaches it at infinity.

Step 6 — Do the integral (add up the tiny steps)

WHAT. To go from the rate rule to the actual curve we must sum up every tiny step — that's integration. First separate so each side holds one variable:

$\frac{d V _{C}}{V _{s} - V _{C}} = \frac{d t}{R C}$

WHY separate. Integrating needs each side to depend on only one thing. Left: only $V_{C}$ . Right: only $t$ . Now each side can be summed independently.

Integrate both sides from the start ( $t = 0$ , $V_{C} = 0$ ):

$- ln (V_{s} - V_{C})_{0}^{V_{C}} = \frac{t}{R C}$

$ln (\frac{V _{s} - V _{C}}{V _{s}}) = - \frac{t}{R C}$

Why the $ln$ appears: the left integrand $1/ (V_{s} - V_{C})$ is exactly the derivative-pattern that produces a natural logarithm. We didn't choose $ln$ ; the algebra demanded it. To undo a $ln$ , we raise $e$ to both sides — that's how the exponential re-enters:

$\frac{V _{s} - V _{C}}{V _{s}} = e^{- t / R C}$

Solve for $V_{C}$ and name $τ = R C$ :

PICTURE. The figure overlays the finished cyan curve on the "shrinking gap" arrows of Step 5, so you see the steep-then-flat shape is literally the gap arrows getting shorter.

Step 7 — What one $τ$ means, on the curve

WHAT. Plug $t = τ$ into the boxed result:

$V_{C} (τ) = V_{s} (1 - e^{- 1}) = V_{s} (1 - 0.368) = 0.632 V_{s}$

WHY it matters. $τ$ is the natural clock-tick of the circuit. It's the time to close $63.2%$ of the gap — and it stays $63.2%$ no matter where you start, because each tick the gap shrinks by the same fraction.

PICTURE. The figure marks $t = τ, 2 τ, 3 τ, 4 τ, 5 τ$ . At each tick the curve climbs $63.2%$ of whatever gap remains, giving $63.2, 86.5, 95.0, 98.2, 99.3%$ . The amber dashed line at $5 τ$ shows why engineers call it "done."

Recall The five-tick ladder

Each tick closes 63.2% of the remaining gap ::: so $2 τ \to 86.5%$ , $3 τ \to 95%$ , $5 τ \to 99.3%$ .

Step 8 — The mirror case: discharging

WHAT. Remove the battery ( $V_{s} = 0$ ). A capacitor charged to $V_{0}$ now drives current backward through $R$ . KVL becomes $0 = I R + V_{C}$ , so $R C \frac{d V _{C}}{d t} + V_{C} = 0$ . Same integration with target $0$ :

$V_{C} (t) = V_{0} e^{- t / τ}$

WHY it's the mirror. Same rate rule — "speed ∝ gap" — but now the target is zero, so the gap is $V_{C}$ itself. The curve falls fast when full, slow when nearly empty.

PICTURE. Charging (rising, ceiling $V_{s}$ ) and discharging (falling toward $0$ ) drawn together. At $1 τ$ the falling curve is at $36.8%$ — the exact complement of the $63.2%$ rising one.

Step 9 — Edge and degenerate cases

Every scenario the reader could hit, so nothing surprises them:

$R \to 0$ (wire, no resistor). $τ = R C \to 0$ : the curve becomes a vertical jump — voltage does change instantly because nothing limits current. This is the idealised (and physically dangerous) case.
$C \to 0$ (no capacitor). $τ \to 0$ again: no bucket to fill, $V_{C}$ instantly equals $V_{s}$ .
$R \to \infty$ (open circuit). $τ \to \infty$ : the curve is flat — no current ever flows, capacitor never charges.
$t < 0$ / before switch-on. Formula only valid for $t \geq 0$ ; before that $V_{C}$ holds its initial value.
Very large $t$ . $e^{- t / τ} \to 0$ , so charging $\to V_{s}$ and discharging $\to 0$ , but never exactly in finite time — the flat tail is asymptotic.

PICTURE. Four mini-curves: the vertical jump ( $τ \to 0$ ), the flat line ( $τ \to \infty$ ), and the normal charge/discharge bends between them, so you can see $τ$ as a dial stretching the same shape.

The one-picture summary

This final figure stacks the whole story: the loop (Step 1) feeds the master equation (Step 4), which becomes the rate rule (Step 5), which integrates into the boxed curve (Step 6), with the $τ$ -ladder (Step 7) and the discharge mirror (Step 8) marked on it.

Recall Feynman: the whole walkthrough in plain words

You've got a battery, a narrow pipe (the resistor), and a bucket (the capacitor) in one loop. The battery pushes water toward the bucket, but the narrow pipe only lets so much through at once. When the bucket is empty, the whole battery pressure pushes water in fast. As the bucket fills, its own water level pushes back, so the pressure difference shrinks and the flow slows down. The speed of filling always matches how much room is left — and any time "speed matches what's left," you get that famous flattening exponential curve. One "tick of the clock" ( $τ = R \times C$ ) is how long it takes to fill $63%$ of whatever gap remains. After five ticks it looks full ( $99%$ ), though technically it sneaks up on the top forever. Pull the battery out and it all runs in reverse: the bucket empties fast when full, slow when nearly dry — the same curve, flipped upside down. Make the pipe wider or the bucket smaller and the whole thing just happens quicker; the shape stays exactly the same.

Connections

← Parent: RC time constants
Capacitors and stored energy
Kirchhoff's Voltage Law
First-order differential equations
RC low-pass and high-pass filters
555 timer circuits
Digital signal rise time and integrity

Visual walkthrough — Understand RC charging - discharging time constants