RC circuits — charging, discharging, time constant τ = RC: Worked examples

Before anything, one symbol reminder so nothing is used unearned:

$E$ (script E) = the battery's emf, its steady push in volts.
$R$ = resistance in ohms ( $Ω$ ); it limits current via $V_{R} = I R$ (see Ohm's Law and Resistance).
$C$ = capacitance in farads (F); the cap holds charge $Q$ with $V_{C} = Q / C$ (see Capacitors and Capacitance).
$τ = R C$ = the time constant, measured in seconds.
$e \approx 2.718$ = the exponential base. $e^{- 1} \approx 0.368$ , and $e^{- t / τ}$ answers "what fraction is left after time $t$ ?"

The scenario matrix

Every RC problem you can be asked lives in one of these cells. The right column names the example that covers it.

#	Case class	What makes it distinct	Covered by
A	Charging — find $V$ or $Q$ at given $t$	plug $t$ into $1 - e^{- t / τ}$	Ex 1
B	Charging — find $t$ for a target $V$	invert with $ln$	Ex 2
C	Discharging — find $V$ at given $t$	plug into $e^{- t / τ}$	Ex 3
D	Discharging — find $t$ for a target $V$	invert with $ln$	Ex 4
E	Current at $t = 0$ vs later	current is same-shape decay for BOTH modes	Ex 5
F	Limiting / degenerate inputs	$t \to 0$ , $t \to \infty$ , $R \to 0$ , $C \to 0$	Ex 6
G	Half-life style ("time to halve")	special target $V_{0} /2$ → $τ ln 2$	Ex 7
H	Real-world word problem	translate a story (camera flash) into $R, C$	Ex 8
I	Exam twist: two resistors / find unknown $C$	rearrange for a hidden quantity	Ex 9

The two master laws we keep reusing:

Figure — RC circuits — charging, discharging, time constant τ = RC

The figure above is your map: the rising teal curve is charging, the falling orange curve is discharging, and they cross the $t = τ$ line at 63.2% and 36.8% — the two numbers this whole page orbits.

The worked examples

Cell A — charging, find voltage at a time

Worked example Example 1 — how full after 3 seconds?

$E = 9$ V, $R = 200 k Ω$ , $C = 10 μ$ F, capacitor starts empty. Find $V_{C}$ at $t = 3$ s.

Forecast: $τ$ will be a couple of seconds, so at 3 s we're past one time constant — guess somewhere around 70–80% of 9 V. Write your number down.

$τ = R C = (2 \times 1 0^{5}) (1 \times 1 0^{- 5}) = 2$ s. Why this step? Everything is measured in multiples of $τ$ ; without it we can't say whether 3 s is "a lot" or "a little."
$t / τ = 3/2 = 1.5$ . Why this step? The formula only ever sees the ratio $t / τ$ , never $t$ alone.
$V_{C} = 9 (1 - e^{- 1.5}) = 9 (1 - 0.2231) = 9 (0.7769) = 6.99$ V. Why this step? Case A is a direct plug-in of the charging law.

Verify: $1.5 τ$ should sit between the $1 τ$ value ( $0.632 \times 9 = 5.69$ V) and the $2 τ$ value ( $0.865 \times 9 = 7.78$ V). Our $6.99$ V is neatly between them. ✓ Units: volts × (dimensionless) = volts. ✓

Cell B — charging, find the time for a target voltage

Cell C — discharging, find voltage at a time

Cell D — discharging, find the time for a target voltage

Cell E — current at $t = 0$ versus later (both modes)

Worked example Example 5 — the current story

Charging circuit: $E = 6$ V, $R = 2 k Ω$ , $C = 500 μ$ F. Find the current at $t = 0$ , at $t = τ$ , and at $t = 2 τ$ .

Forecast: Current is largest at the very start (empty cap = a plain wire) and decays with the same $τ$ as everything else. Guess it starts at a few milliamps and roughly thirds each $τ$ .

$I_{0} = \frac{E}{R} = \frac{6}{2000} = 3 \times 1 0^{- 3}$ A $= 3$ mA. Why this step? At $t = 0$ the cap holds zero volts, so the entire $E$ sits across $R$ ; Ohm's law gives the peak current.
$τ = R C = (2000) (5 \times 1 0^{- 4}) = 1$ s (used only to label the times).
$I (τ) = I_{0} e^{- 1} = 3 (0.3679) = 1.10$ mA. Why this step? Current follows $I_{0} e^{- t / τ}$ — the same decay shape whether charging or discharging.
$I (2 τ) = I_{0} e^{- 2} = 3 (0.1353) = 0.406$ mA.

Verify: the ratio $I (τ) / I_{0} = 0.368$ and $I (2 τ) / I_{0} = 0.135$ — exactly the standard $e^{- 1}$ and $e^{- 2}$ fractions. ✓ Note the current decays even while charge rises: no contradiction, because rising $V_{C}$ eats the driving voltage $E - V_{C}$ .

Cell F — degenerate and limiting inputs

Worked example Example 6 — the edge cases (no calculator needed)

Take the charging law $V_{C} = E (1 - e^{- t / τ})$ and the decay $V = V_{0} e^{- t / τ}$ . Reason out each extreme.

Forecast: guess each limit before reading the step.

$t \to 0$ (charging): $e^{0} = 1$ , so $V_{C} = E (1 - 1) = 0$ . Why? At the instant the switch closes, the cap is still empty — it can't jump.
$t \to \infty$ (charging): $e^{- \infty} = 0$ , so $V_{C} = E (1 - 0) = E$ . Why? Eventually the cap fully opposes the battery; current stops; $V_{C}$ equals the source.
$R \to 0$ : $τ = R C \to 0$ , so $e^{- t / τ} \to 0$ instantly for any $t > 0$ . Charging is immediate. Why? No resistor means nothing throttles the current — but physically $I_{0} = E / R \to \infty$ , a warning that ideal wires can't limit current.
$C \to 0$ : $τ \to 0$ too, so again instant — but there's essentially no charge to store ( $Q_{m a x} = E C \to 0$ ).
$R \to \infty$ or $C \to \infty$ : $τ \to \infty$ , so $e^{- t / τ} \to 1$ for all finite $t$ : charging never perceptibly happens.

Verify (numeric spot-check): take $E = 10$ , $τ = 1$ . At $t = 0.001$ s, $V_{C} = 10 (1 - e^{- 0.001}) = 10 (0.0009995) = 0.009995$ V — essentially 0, confirming step 1. At $t = 20$ s, $V_{C} = 10 (1 - e^{- 20}) = 9.99999998$ V — essentially $E$ , confirming step 2. ✓

Cell G — the "time to halve" special target

Worked example Example 7 — the half-life of a discharge

A capacitor discharges. Show that the time to fall to half its voltage is $t_{1/2} = τ ln 2$ , and compute it for $τ = 4$ s.

Forecast: $ln 2 \approx 0.693$ , so half-life should be a bit under one $τ$ — around 2.8 s here.

Set the target: $\frac{V _{0}}{2} = V_{0} e^{- t / τ}$ . Why this step? "Halve" means the ratio $V / V_{0} = 1/2$ ; the actual voltages cancel.
Cancel $V_{0}$ : $\frac{1}{2} = e^{- t / τ}$ . Why this step? This is the reason half-life is independent of the starting voltage — the same trick as radioactive decay in Exponential Decay and Differential Equations and Newton's Law of Cooling.
$- t / τ = ln (1/2) = - ln 2 \Rightarrow t_{1/2} = τ ln 2$ .
For $τ = 4$ s: $t_{1/2} = 4 (0.6931) = 2.77$ s.

Verify: at $t = 2.77$ s, $e^{- 2.77/4} = e^{- 0.6931} = 0.5000$ — exactly half. ✓ Because $ln 2 < 1$ , the half-life is shorter than $τ$ , which makes sense: reaching 50% comes before falling to 36.8%.

Cell H — real-world word problem

Worked example Example 8 — a camera flash capacitor

A camera flash uses a $1000 μ$ F capacitor charged to $300$ V. It dumps its charge through the flash tube, whose effective resistance is $10 Ω$ , and the flash "ends" when voltage drops to $37%$ of peak. How long does the flash last, and what is the peak current?

Forecast: $37%$ is exactly the $e^{- 1}$ point, so the flash lasts about one $τ$ . And $τ = R C$ with a small $R$ but big $C$ — guess a small fraction of a second. Peak current: $300/10$ is big, tens of amps.

Peak current: $I_{0} = \frac{V _{0}}{R} = \frac{300}{10} = 30$ A. Why this step? At the instant of firing, the full $300$ V sits across the tube (Ohm's law), giving a huge, bright burst.
$τ = R C = (10) (1 \times 1 0^{- 3}) = 0.010$ s $= 10$ ms.
Voltage at $37%$ means $V = 0.37 V_{0} = e^{- t / τ} V_{0}$ , i.e. $t = τ$ (since $e^{- 1} = 0.368$ ). Why this step? The problem handed us the special $e^{- 1}$ fraction, so no logarithm is needed — recognizing it saves work.
So the flash lasts $t \approx τ = 10$ ms.

Verify: $τ = 10$ ms is a believable flash duration (real flashes are 1–10 ms). ✓ Energy sanity via Energy Stored in a Capacitor: $\frac{1}{2} C V_{0}^{2} = \frac{1}{2} (1 0^{- 3}) (300)^{2} = 45$ J released — a genuinely bright pop. ✓

Cell I — exam twist: find the hidden capacitance

Worked example Example 9 — reverse-engineer

$C$ from a measurement An engineer charges an unknown capacitor through $R = 50 k Ω$ from a $10$ V source and measures $V_{C} = 8$ V at $t = 4$ s. Find $C$ .

Forecast: 8 V out of 10 V is past one $τ$ (which reaches 6.32 V), so $τ$ is under 4 s → $C$ under $\sim 80 μ$ F.

Charging law with the measurement: $8 = 10 (1 - e^{- 4/ τ})$ . Why this step? The unknown is buried in $τ$ ; one data point pins it down.
Isolate the exponential: $e^{- 4/ τ} = 1 - 0.8 = 0.2$ .
Take the log: $- \frac{4}{τ} = ln 0.2 = - 1.609 \Rightarrow τ = \frac{4}{1.609} = 2.486$ s. Why $ln$ ? Same inverse-of-exponential tool as Cells B and D — the workhorse of every "solve for time or $τ$ " problem.
Now unpack $τ = R C$ : $C = \frac{τ}{R} = \frac{2.486}{5 \times 1 0 ^{4}} = 4.97 \times 1 0^{- 5}$ F $\approx 49.7 μ$ F. Why this step? $R$ is known, so dividing recovers $C$ .

Verify: with $C = 49.7 μ$ F, $τ = R C = (5 \times 1 0^{4}) (4.97 \times 1 0^{- 5}) = 2.486$ s, and $V_{C} (4) = 10 (1 - e^{- 4/2.486}) = 10 (1 - 0.2) = 8.00$ V — matches the measurement. ✓

Recall checkpoint

Recall Which cell is each of these?

A cap discharging — you're told $V$ and asked for $t$ . ::: Cell D — invert with $ln$ . You're told $V_{C}$ at one time and asked for $C$ . ::: Cell I — solve for $τ$ then divide by $R$ . "How long until it halves?" ::: Cell G — answer is $τ ln 2$ , independent of $V_{0}$ . Current the instant the switch closes. ::: Cell E — $I_{0} = E / R$ , the peak. What happens as $R \to \infty$ ? ::: Cell F — $τ \to \infty$ , charging never perceptibly completes.

Connections

Parent topic — the derivations these examples apply.
Ohm's Law and Resistance — every "initial current" step uses $I_{0} = V / R$ .
Capacitors and Capacitance — $V_{C} = Q / C$ , and why the cap starts as a wire and ends as an open switch.
Kirchhoff's Voltage Law — the loop equation behind both laws.
Exponential Decay and Differential Equations · Newton's Law of Cooling — same half-life algebra as Ex 7.
Energy Stored in a Capacitor — the 45 J sanity check in Ex 8.
LR Circuits — swap $τ = R C$ for $τ = L / R$ and every example transfers.

Worked examples — RC circuits — charging, discharging, time constant τ = RC