Worked examples — Understand RC charging - discharging time constants

Before anything, let me restate the two tools we will use again and again, in plain words.

Everything below is these two lines, rearranged.

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The scenario matrix

Every RC question is one cell of this grid. The last column names the worked example that covers it.

#	Case class	What is unknown	Worked example
A	Charging — find $V_C$ at a time	voltage	Ex 1
B	Charging — find the time to reach a voltage	time	Ex 2
C	Discharging — find $V_C$ at a time	voltage	Ex 3
D	Discharging — find the time to fall to a level	time	Ex 4
E	Degenerate: $t=0$ and limit $t\to \infty$	boundary values	Ex 5
F	Non-zero start — cap not empty when charging begins	general formula	Ex 6
G	Charging current (not voltage) at a moment	current	Ex 7
H	Real-world word problem (555 timer style)	design $R$ or $C$	Ex 8
I	Exam twist — two of $R,C,\tau$ hidden, solve backwards	a component value	Ex 9

We use a common circuit unless stated otherwise: $R=10\text{ k}\Omega$, $C=100\mu \text{F}$, so $\tau =RC=(10\times 10^3)(100\times 10^{-6})=1\text{ s}.$

The figure above is our map: the rising blue curve is charging, the falling pink curve is discharging, and each example is a labelled point or slope on one of them. Keep glancing back.

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Example 1 — Cell A: charging, find the voltage

1. Convert time into time constants. $\frac{t}{\tau }=\frac{2}{1}=2$. Why this step? The formula only cares about the ratio $t/\tau$, never the raw seconds — this is why the same-shaped curve fits every RC circuit.
2. Evaluate the exponential. $e^{-2}=0.135$. Why this step? $e^{-t/\tau }$ is the "fraction still missing"; here $13.5\%$ of the climb is left.
3. Plug in. $V_C=5(1-0.135)=5(0.865)=4.32\text{ V}$. Why this step? $1-e^{-t/\tau }$ is the fraction already achieved.

Verify: matches the $2\tau =86.5\%$ row of the parent's table, and $4.32<5$ as any charging voltage must be. ✔

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Example 2 — Cell B: charging, find the time

1. Write the charging equation with the target. $3=5(1-e^{-t})$. Why this step? We know the voltage, want the time — so time must come out of the equation, not go in.
2. Isolate the exponential. $1-e^{-t}=0.6\Rightarrow e^{-t}=0.4$. Why this step? Peel away everything wrapped around $e^{-t}$ so it stands alone.
3. Take the natural log. $-t=\ln (0.4)=-0.916\Rightarrow t=0.916\text{ s}$. Why this step? $\ln$ is the exact question "what power of $e$ gives this?" — it is the only tool that pulls $t$ down out of the exponent. Nothing else undoes $e^{-t}$.

Verify: $t=0.916\text{ s}$ is just below $1\text{ s}$, matching the forecast (we asked for $60\%$, slightly below the $63.2\%$ reached at $1\tau$). Plug back: $5(1-e^{-0.916})=5(1-0.400)=3.00\text{ V}$. ✔

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Example 3 — Cell C: discharging, find the voltage

1. Compute $\tau$. $\tau =(2000)(470\times 10^{-6})=0.94\text{ s}$. Why this step? The falling curve's speed is set entirely by $\tau$; nothing moves until we have it.
2. Convert time. $\frac{t}{\tau }=\frac{1}{0.94}=1.064$.
3. Use the discharge equation. $V_C=12e^{-1.064}=12(0.345)=4.14\text{ V}$. Why this step? Discharging falls, so we use $e^{-t/\tau }$ alone — no $(1-\dots )$. See the parent mistake box on this exact slip.

Verify: $4.14\text{ V}$ is close to $0.37\times 12=4.44\text{ V}$ (slightly lower because $1.064\tau >1\tau$). Consistent with the forecast. ✔

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Example 4 — Cell D: discharging, find the time

1. Set up the equation. $3=12e^{-t/0.94}$.
2. Isolate the exponential. $e^{-t/0.94}=\frac{3}{12}=0.25$. Why this step? The fraction remaining is exactly $V_C/V_0$ for discharging.
3. Take the log. $-\frac{t}{0.94}=\ln (0.25)=-1.386\Rightarrow t=0.94\times 1.386=1.303\text{ s}$. Why this step? Again $\ln$ is the only lever that frees $t$ from the exponent.

Verify: $1.303\text{ s}$ sits between $0.94$s and $1.88$s — forecast confirmed. Plug back: $12e^{-1.303/0.94}=12e^{-1.386}=12(0.25)=3.00\text{ V}$. ✔

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Example 5 — Cell E: the boundaries $t=0$ and $t\to \infty$

These are the "degenerate" cases people forget — but they are the sanity checks that catch every algebra error.

1. At $t=0$: $e^{-0/\tau }=e^0=1$, so $V_C=5(1-1)=0\text{ V}$. Why this step? $e^0=1$ because any nonzero number to the power $0$ is $1$. Physically the cap starts empty — voltage cannot jump (would need infinite current), so it must be $0$.
2. As $t\to \infty$: $e^{-t/\tau }$ shrinks toward $0$, so $V_C=5(1-0)=5\text{ V}$. Why this step? Once fully charged, no current flows, no drop across $R$, so the cap sits at the full source voltage forever.
3. Discharge boundaries ($V_0=12$): at $t=0$, $V_C=12e^0=12\text{ V}$; as $t\to \infty$, $V_C=12\times 0=0\text{ V}$.

Verify: charging goes $0\to V_s$; discharging goes $V_0\to 0$. Both curves in the figure start and end exactly there. ✔

Recall Why boundaries never lie

A charging formula that gives anything other than $0$ at $t=0$ or anything other than $V_s$ at $t\to \infty$ is wrong — check these two points before trusting any RC result. What should charging $V_C$ equal at $t=0$? ::: $0$ (cap starts empty, voltage can't jump) What should charging $V_C$ approach as $t\to \infty$? ::: $V_s$ (fully charged, no current, no $R$ drop)

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Example 6 — Cell F: charging from a non-zero start

Real circuits often begin with the cap already partly charged. The parent's $V_C=V_s(1-e^{-t/\tau })$ assumed an empty start; here is the general rule.

1. Identify the three ingredients. $V_{\text{start}}=2$, $V_{\text{final}}=5$, $t/\tau =1$. Why this step? The general formula needs exactly these three; naming them prevents mixing.
2. Shrink the gap. gap $\times e^{-1}=3\times 0.368=1.104\text{ V}$ still to go.
3. Add to the target-from-below. $V_C=5-1.104=3.896\approx 3.90\text{ V}$. Why this step? We approach $5$ from below, so subtract the leftover gap from $V_{\text{final}}$.

Verify: $2+0.632\times 3=2+1.896=3.90\text{ V}$ — identical, forecast confirmed. And $2<3.90<5$, correctly between start and target. ✔

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Example 7 — Cell G: charging current, not voltage

1. Recall the current law. $I(t)=\frac{V_s}{R}{e}^{-t/\tau }$. Why this step? Current is $Cd{V}_C/dt$; differentiating the rising voltage gives a falling exponential — as $V_C$ climbs, the drop across $R$ (which is $V_s-V_C$) shrinks, so less current flows.
2. At $t=0$: $I=\frac{5}{10000}{e}^0=0.5\text{ mA}$. Why this step? At the first instant $V_C=0$, so the full $5\text{ V}$ sits across $R$ — maximum current, $V_s/R$.
3. At $t=1\text{ s}$: $I=0.5\text{ mA}\times e^{-1}=0.5\times 0.368=0.184\text{ mA}$.

Verify: current fell to $36.8\%$ of its start after $1\tau$ — the same $37\%$ discharge number, because current always decays exponentially whether charging or discharging. ✔ Units: $\text{V}/\Omega =\text{A}$, and $5/10000=5\times 10^{-4}\text{ A}=0.5\text{ mA}$. ✔

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Example 8 — Cell H: real-world word problem

1. Set up the crossing equation. $6=9(1-e^{-t/\tau })$ at $t=0.5$s.
2. Solve for $t/\tau$. $e^{-t/\tau }=1-\frac{6}{9}=\frac{1}{3}\Rightarrow \frac{t}{\tau }=\ln 3=1.0986$. Why this step? The $\frac{2}{3}$ threshold always corresponds to $\ln 3\approx 1.1$ time constants — this is why the 555's timing formula famously carries a factor of $1.1$.
3. Get $\tau$, then $R$. $\tau =\frac{t}{1.0986}=\frac{0.5}{1.0986}=0.455\text{ s}$. Then $R=\frac{\tau }{C}=\frac{0.455}{10\times 10^{-6}}=45,500\Omega \approx 45.5\text{ k}\Omega$. Why this step? $\tau =RC$ rearranges to $R=\tau /C$; we solve backwards because the unknown is the component, not the behaviour.

Verify: with $R=45.5\text{ k}\Omega$, $\tau =0.455$s, at $t=0.5$s: $9(1-e^{-0.5/0.455})=9(1-e^{-1.099})=9(1-0.333)=6.00\text{ V}$. ✔ Right on the trip point.

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Example 9 — Cell I: exam twist, solve backwards for a component

1. Recognise the fraction. $\frac{3.68}{10}=0.368=e^{-1}$, so $\frac{t}{\tau }=1$. Why this step? Spotting $0.368=e^{-1}$ skips the log entirely — the fingerprint of one $\tau$.
2. Read off $\tau$. Since $t/\tau =1$ and $t=2\text{ ms}$, we get $\tau =2\text{ ms}=0.002\text{ s}$.
3. Solve for $R$. $R=\frac{\tau }{C}=\frac{0.002}{1\times 10^{-6}}=2000\Omega =2\text{ k}\Omega$. Why this step? Backwards use of $\tau =RC$ again — the exam hid two of the three variables and gave you a voltage instead.

Verify: with $R=2\text{ k}\Omega$, $\tau =(2000)(10^{-6})=2\text{ ms}$; discharge at $t=2\text{ ms}$: $10e^{-1}=3.68\text{ V}$. ✔

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Recall Self-test before you leave

Which fraction is the fingerprint of exactly one time constant, charging vs discharging? ::: Charging $63.2\%$ reached; discharging $36.8\%$ remaining (both are $e^{-1}$ facts). If only $V_{\text{start}}$, $V_{\text{final}}$, $\tau$ are known, which single formula covers charge AND discharge? ::: $V_C(t)=V_{\text{final}}+(V_{\text{start}}-V_{\text{final}})e^{-t/\tau }$. The 555's $\frac{2}{3}{V}_s$ threshold corresponds to how many time constants? ::: $\ln 3\approx 1.10\tau$.

Connections

• Understand RC charging - discharging time constants — the parent equations these examples exercise.
• First-order differential equations — where the general formula in Ex 6 comes from.
• 555 timer circuits — Ex 8's $\ln 3\approx 1.1$ threshold in action.
• RC low-pass and high-pass filters — the same $\tau$ seen in the frequency domain.
• Digital signal rise time and integrity — Ex 7's current decay sets how fast edges settle.
• Capacitors and stored energy · Kirchhoff's Voltage Law — the physics underneath every step.