Physics · Electric circuits

RC time constant

One ideal capacitor with constant positive capacitance and finite positive effective resistance in a linear network that is time-invariant after switching. The time constant describes voltage deviation from the final value; multiple independent storage elements can produce multiple time constants.

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Formula

Calculate

τ = RC, q (t) = Q (1 - e^{- t / τ})

τ in seconds, R in ohms and C in farads; q and Q in coulombs.

Variables & units

Variables & units
Symbol	Meaning	Unit
$τ$	Time constant	s
$R$	Resistance	Ω
$C$	Capacitance	F
$Q$	Final charge during charging	C
$t$	Time since switching	s

LaTeX

\tau = R C, \quad q(t)=Q(1-e^{-t/\tau})

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Exponential rise from 1 V toward a horizontal 5 V asymptote. Markers at one and two time constants show 3.5285 and 4.4587 V; after five time constants, 5 V is still not reached exactly. — The capacitor starts at 1 V and approaches 5 V with τ = 103.4 ms. After τ, 63.2% of the initial 4 V deviation has decayed.

RC time constant calculator

Find the time constant τ = R · C, the capacitor voltage at a given time, or the time to reach a voltage when charging or discharging.

Calculation model

Find

Equivalent resistance R_eq

Resistance the capacitor charges through, e.g. 10 kΩ; with several resistors the equivalent one.

Capacitance C

Capacitance of the capacitor, e.g. 100 µF for an electrolytic capacitor.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for τ, substitute, evaluate

1. Rearrange for τ
$\tau=R_{eq}C$
2. Substitute values (in SI units)
- R_eq = 10 kΩ = 10 000 Ω
- C = 100 µF = 1 · 10⁻⁴ F
$\tau=10000\cdot 0.0001$
3. Result
τ = 1 s

Result: Time constant τ

τ = 1 s

With R_eq = 10 kΩ and C = 100 µF, τ = 1 s.

Quick tips

Rule of thumb: after 1 τ about 63 % of the way to the final voltage is done, after 5 τ over 99 %. In the model the capacitor never quite gets there.
Unit check: Ω times F gives seconds. 1 kΩ · 1 µF = 1 ms, and 10 kΩ · 100 µF = 1 s.

Assumptions and limits

One ideal linear capacitor with C > 0 and finite positive Req in the constant resistive network after switching. Req is resistance seen by the capacitor, not automatically one labelled resistor. Suppress independent ideal sources when finding Req; dependent sources remain active. Further independent storage elements are excluded.

Derivation & background

After switching, consider one ideal capacitor connected to a time-invariant resistive network. Replace the network at the capacitor by its Thévenin voltage and effective resistance; dependent sources remain active. The loop gives $R_{\mathrm{eq}}i+U_C=U_f$ , while the capacitor gives $i=C\frac{\mathrm dU_C}{\mathrm dt}$ . Set $x=U_C-U_f$ : then $\frac{\mathrm dx}{\mathrm dt}=-\frac{x}{R_{\mathrm{eq}}C}$ . With initial voltage $U_i$ , $U_C(t)=U_f+(U_i-U_f)e^{-\frac t\tau}$ , where $\tau=R_{\mathrm{eq}}C$ . After one time constant, the reciprocal of e of the initial deviation remains. The familiar 63.2 percent of final voltage specifically applies to charging from zero volts.

Exam blueprint

Validity range

Derivation steps

Resistance limits charging current while capacitor voltage reduces the remaining driving voltage difference.

1After switching, consider one ideal capacitor connected to a time-invariant resistive network.
2Replace the network at the capacitor by its Thévenin voltage and effective resistance; dependent sources remain active.
3The loop gives $R_{\mathrm{eq}}i+U_C=U_f$ , while the capacitor gives $i=C\frac{\mathrm dU_C}{\mathrm dt}$ .
4Set $x=U_C-U_f$ : then $\frac{\mathrm dx}{\mathrm dt}=-\frac{x}{R_{\mathrm{eq}}C}$ .
5With initial voltage $U_i$ , $U_C(t)=U_f+(U_i-U_f)e^{-\frac t\tau}$ , where $\tau=R_{\mathrm{eq}}C$ .
6After one time constant, the reciprocal of e of the initial deviation remains. The familiar 63.2 percent of final voltage specifically applies to charging from zero volts.

Rearrangements

Find the charge after a given time

\frac{q(t)}{Q_f}=1-e^{-\frac{t}{R_{\mathrm{eq}}C}}

Only for an initially uncharged linear capacitor after a constant voltage step with positive equivalent resistance; Qf=CUf must be nonzero. With precharge, the exponential instead describes remaining deviation from the final state.

Task variant

R=10 kΩ and C=100 μF. Find the time constant τ.

With $R=10\,\mathrm{k\Omega}$ and $C=100\,\mathrm{\mu F}$ , $\tau=RC=1.0\,\mathrm s$ .
Only for the explicitly assumed zero start toward constant nonzero final voltage does one second give $\frac q{Q_f}=1-e^{-1}\approx0.632$ .
In general, 36.8% of initial deviation remains.

Check: $\frac{1\,\mathrm s}{RC}=1$ , so the exponent is dimensionless.

Common mistakes

Failing to convert microfarads to farads or confusing charging and discharging.

1 μF = 10⁻⁶ F. Only charging an initially uncharged capacitor towards positive final charge starts q at zero and increases it. In general q approaches the final value from its actual initial value; when discharging towards zero its magnitude decreases.

Exam context

Charging curves, timing elements and simple RC low-pass circuits.

Formula cluster

Electrical time response

The time constant turns component values into an intuitive characteristic time.

C = Q/U

Capacitance of a Capacitor

U = R·I

Ohm's Law

y′ + a y = r(t)

First-order linear differential equation

Worked example

With $R=10\,\mathrm{k\Omega}$ and $C=100\,\mathrm{\mu F}$ , $\tau=RC=1\,\mathrm s$ . Charging from zero volts reaches about 63.2 percent of the final voltage after one second; generally about 36.8 percent of the initial deviation remains.

Applications

Timing and filtering in circuits, flash units, sensors and time-dependent voltage or charge measurements.

How do you calculate with RC time constant?

Here is how to work through a typical RC time constant (τ = RC; q(t)=Q(1−e^(−t/τ))) task step by step:

1
Task
R=10 kΩ and C=100 μF. Find the time constant τ.
Solution path
1. With $R=10\,\mathrm{k\Omega}$ and $C=100\,\mathrm{\mu F}$ , $\tau=RC=1.0\,\mathrm s$ .
2. Only for the explicitly assumed zero start toward constant nonzero final voltage does one second give $\frac q{Q_f}=1-e^{-1}\approx0.632$ .
3. In general, 36.8% of initial deviation remains.
Check: $\frac{1\,\mathrm s}{RC}=1$ , so the exponent is dimensionless.

Understand and apply the equation

After switching, capacitor voltage in an ideal RC model does not jump instantly to source voltage. Current first flows while the remaining voltage difference decreases. The time constant describes that approach’s timescale. Initial state, final state and switch position jointly determine the response, so another circuit can charge or discharge the same capacitor differently.

An exponential approach, not a completion instant

An ideal linear capacitor has plate charge proportional to voltage. Current changes charge and hence voltage. Resistance limits that current according to the remaining voltage difference. Prerequisites are the resistor relation, capacitance and an exponential function; the derivation explains why it is specifically the gap to the final value that decays exponentially.[1][2]

\begin{aligned}\tau&=R_{\mathrm{eq}}C\\{}u_C(t)&=U_f+(U_i-U_f)e^{-\frac{t}{\tau}}\end{aligned}

One ideal linear capacitor and a constant stable resistive equivalent after switching; t=0 at the switching instant.[1][2]

After τ, $e^{-1}$ of the initial deviation from the final value remains. Only charging from zero to Uf gives an achieved fraction $1-e^{-1}\approx0.632$ . Discharge toward zero leaves 36.8% of initial voltage. Capacitor voltage is continuous without an ideal impulse current. A source voltage step therefore does not automatically make capacitor voltage jump instantaneously.[1][2]

Voltage and energy decay at different rates

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Both curves describe a linear capacitor discharging toward zero:

\frac{u_C}{U_i}=e^{-\frac t\tau}

\frac{W_C}{W_i}=e^{-\frac{2t}\tau}

. This normalization requires nonzero initial voltage.[2][4]

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Combine the component relation and node balance

Always find the final value first: in steady DC an ideal capacitor carries no current. Next find the resistance seen at its terminals in that same switch position. Finally insert the voltage already present initially. This describes charging, discharge and transitions between two nonzero voltages with one model.[1]

\begin{aligned}i&=C\frac{\mathrm du_C}{\mathrm dt}=\frac{U_f-u_C}{R_{\mathrm{eq}}}\\{}\tau\frac{\mathrm du_C}{\mathrm dt}+u_C&=U_f\end{aligned}

Current is positive into the positive capacitor terminal; capacitance is constant.[1]

\begin{aligned}w&=u_C-U_f\\{}\frac{\mathrm dw}{w}&=-\frac{\mathrm dt}{\tau}\\{}w(t)&=(U_i-U_f)e^{-\frac{t}{\tau}}\end{aligned}

Solve the deviation from steady state; the special case Ui=Uf remains constant.[1]

The initial condition fixes the coefficient. If Ui exceeds Uf, voltage falls; if it is lower, voltage rises. The same differential equation explains both directions. $q=CU_C$ denotes signed charge on the chosen plate, not growing net charge of the complete ideal capacitor.[1]

Capacitance of a capacitor connects plate charge with voltage. For constant C its time derivative gives capacitor current in the balance above.[1]

Find the correct resistance after switching

Req is resistance seen by the capacitor in the applicable switch position, not necessarily one labelled resistor. In a purely resistive network with independent sources, ideal voltage sources are shorted and ideal current sources opened when finding resistance. Dependent sources remain active and require a test excitation. Charge and discharge positions may have different Req.[1]

A 4 kΩ/4 kΩ divider across an ideal 12 V source has midpoint Uf=6 V. A 22 μF capacitor connected from that midpoint to reference sees 4||4=2 kΩ when the independent source is suppressed, giving τ=44 ms. The sum 8 kΩ is not the resistance seen by this capacitor. This equivalent applies to this single storage element and topology.[1][3]

Series and parallel resistor combinations help determine Req after the switch position and source treatment have been established. A resistance sum without this equivalent network does not reliably give the RC time constant.[3][1]

Initial charge, threshold time and switching

Charge from 1 V to 5 V rather than from zero

Given: $R=2.2\,\mathrm{k\Omega}$ , $C=47\,\mathrm{\mu F}$ , $U_i=1.0\,\mathrm V$ , $U_f=5.0\,\mathrm V$ , $t=2\tau$ .
Find: Find time, voltage, plate charge and current.

Approach: Use the general response with actual initial voltage.
$\begin{aligned}\tau&=RC\\{}u_C&=U_f+(U_i-U_f)e^{-\frac{t}{\tau}}\end{aligned}$
Convert prefixes to SI.
$\begin{aligned}\tau&=2200\cdot47\cdot10^{-6}=0.1034\,\mathrm s\\{}t&=0.2068\,\mathrm s\end{aligned}$
Use the actual initial deviation −4 V.
$u_C=5-4e^{-2}=4.458658867\,\mathrm V$
Find charge and decreasing current from the same response.
$\begin{aligned}q&=Cu_C=209.5569668\,\mu\mathrm C\\{}i&=\frac{4e^{-2}}{2200}=0.2460642\,\mathrm{mA}\end{aligned}$

Result: At 206.8 ms voltage is about 4.459 V and charge 209.56 μC; current 0.246 mA.

Check: Remaining deviation $5\,\mathrm V-U_C\approx 0.54134\,\mathrm V$ equals $4 e^{-2}$ . $iR+U_C$ equals 5 V; at t=0 the formula gives exactly 1 V.

Discharge to a specified threshold

Given: $U_i=8.0\,\mathrm V$ , $U_f=0$ , $R=4.7\,\mathrm{k\Omega}$ , $C=100\,\mathrm{\mu F}$ .
Find: Find time to 2.0 V, and voltage and remaining energy at 0.70 s.

Approach: Solve discharge toward zero logarithmically for the threshold, then find energy from voltage.
$\begin{aligned}u_C&=U_i e^{-\frac{t}{\tau}}\\{}t&=\tau\ln\frac{U_i}{u_C}\\{}E_C&=\frac{1}{2}Cu_C^2\end{aligned}$
Find time constant and logarithmic threshold time.
$\begin{aligned}\tau&=0.47\,\mathrm s\\{}t_{2\mathrm V}&=0.47\ln(\frac{8}{2})=0.65155835\,\mathrm s\end{aligned}$
Evaluate the exponential at the second instant.
$u_C(0.70)=8e^{-\frac{0.70}{0.47}}=1.80413245\,\mathrm V$
Energy follows the square of voltage.
$\begin{aligned}E_0&=\frac{1}{2}C8^2=3.20\,\mathrm{mJ}\\{}E(0.70)&=\frac{1}{2}Cu_C^2=0.162744695\,\mathrm{mJ}\end{aligned}$

Result: 2 V is reached at about 651.6 ms; at 700 ms 1.804 V and 0.1627 mJ remain.

Check: Voltage decays as $e^{-\frac{t}{\tau}}$ , energy as $e^{-2\frac{t}{\tau}}$ ; about 3.0373 mJ has been dissipated in the ideal resistor.

A new switch position requires a new final value and equivalent resistance. Capacitor voltage just before switching supplies the next interval’s initial condition. The following example uses separate time origins for the two phases.[1][2]

A second switching event starts from the existing voltage

Given: A 20 μF capacitor starts at 0 V and charges through 1.0 kΩ from 10 V. After 20 ms it is disconnected from that source and connected to reference through 2.0 kΩ. The capacitor is ideal and linear; there are no impulse currents.
Find: Find voltage at switching and 40 ms later.

Approach: each switch position has its own time constant. The voltage reached at the end of the first interval becomes the initial voltage of the second.
$\begin{aligned}\tau_1&=R_1C\\u_1(t)&=10(1-e^{-\frac t{\tau_1}})\\\tau_2&=R_2C\\u_2(s)&=u_1(20\,\mathrm{ms})e^{-\frac s{\tau_2}}\end{aligned}$
The first phase lasts exactly one of its time constants.
$\begin{aligned}\tau_1&=20\,\mathrm{ms}\\u_1(20\,\mathrm{ms})&=10(1-e^{-1})\approx6.32120559\,\mathrm V\end{aligned}$
In the second phase, time is measured since switching.
$\begin{aligned}\tau_2&=40\,\mathrm{ms}\\u_2(40\,\mathrm{ms})&=10(1-e^{-1})e^{-1}\\&\approx2.32544158\,\mathrm V\end{aligned}$

Result: Voltage is 6.321 V at total time 20 ms and 2.325 V at total time 60 ms. Capacitor voltage remains continuous at switching.

Check: The second phase reduces existing voltage by $e^{-1}$ . Current may jump: just before switching about +3.679 mA enters the positive terminal, immediately afterwards −3.161 mA. Charge and its rate of change are different quantities.

Convert target accuracy into time

\begin{aligned}\frac{|u_C-U_f|}{|U_i-U_f|}&=\varepsilon\\{}t&=-\tau\ln\varepsilon\\{}0&<\varepsilon<1\end{aligned}

Relative remaining deviation, provided Ui≠Uf; exactly zero error cannot be reached in finite time.[1][2]

10% remaining error requires ln 10≈2.303 time constants; 1% requires ln 100≈4.605. At 5τ, zero-start charging reaches 99.326% of the final value, not exactly 100%. Real R/C and threshold tolerances affect timing. One nominal number is therefore not a guaranteed switching instant.[2]

For a specified voltage threshold U*, first check whether it is reachable. If Ui ≠ Uf, a threshold reached at a positive finite time lies strictly between Ui and Uf. U* = Ui corresponds to t = 0; U* = Uf is approached only asymptotically. If Ui = Uf, voltage stays constant: the same threshold has no unique crossing time, while any other threshold is never reached.[1][2]

t=R_{\mathrm{eq}}C\ln\left|\frac{U_i-U_f}{U_*-U_f}\right|

Threshold time in the constant one-capacitor model; a positive finite time requires U* strictly between Ui and Uf.[1][2]

Check farads, coulombs and seconds

\begin{aligned}[RC]&=\frac{\mathrm V}{\mathrm A}\frac{\mathrm C}{\mathrm V}=\mathrm s\\{}1\,\mu\mathrm F&=10^{-6}\,\mathrm F\end{aligned}

The exponent $\frac{t}{\tau}$ is dimensionless; microfarads cannot be entered as whole farads.[2]

The introductory 10 kΩ/100 μF example remains τ=1.0 s and $\frac{q}{Q}=1-e^{-1}\approx0.632$ after one second, for an initially uncharged plate. With initial voltage this fraction is not simply $\frac{q}{Q}$ ; use relative deviation. Charge Q=CUf requires the same constant capacitance and plate orientation.[2]

Five time-response exercises

Calculate · Foundation

R=1 kΩ, C=20 μF, zero-start charging toward 3 V. Voltage at τ?

Show hint

Do not confuse it with full source voltage.

Learning goal: Find voltage after one time constant.

Show solution and calculation

Approach: for zero initial voltage, use the achieved fraction of the whole voltage change.
The time constant is 20 ms.
Capacitor voltage then is approximately 1.89636 V; the remaining source difference is 1.10364 V.
Together they give 3 V.

u_C(\tau)=3(1-e^{-1})\approx1.89636\,\mathrm V

Transfer · Application

How does doubling R at fixed C, Ui, Uf affect the response?

Show hint

Separate time scale and initial current.

Learning goal: Transfer resistance changes to time and current.

Show solution and calculation

Approach: compare time constant and initial current at identical initial and final voltages. τ doubles and initial current halves; time to the same relative deviation doubles.
Final voltage remains Uf.

\begin{aligned}\tau_2&=2\tau_1\\{}i_2(0)&=\frac{i_1(0)}{2}\end{aligned}

Rearrange · Application

Discharge should reach 10% of initial voltage at 30 ms. C=10 μF. Find model resistance.

Show hint

Use $\tau=\frac{t}{\ln10}$ , then $R=\frac{\tau}{C}$ .

Learning goal: Rearrange a discharge target for resistance.

Show solution and calculation

Approach: the specified residual is a fraction of initial voltage; solve the exponent logarithmically. τ≈13.0288 ms; R≈1302.88 Ω.
Substitution into the discharge response restores 10%.

R=\frac{0.030}{10^{-5}\ln10}\approx1302.88\,\Omega

Interpret · Application

An ideal capacitor approaches 10 V. At t=0, 1 s and 2 s voltages are exactly 2 V, 6 V and 8 V. Is this consistent with a constant-parameter RC model? Find τ from residual voltages.

Show hint

Subtract each reading from the final voltage.

Learning goal: Recognize an exponential model from equal residual ratios.

Show solution and calculation

Approach: Residual voltages are 8 V, 4 V and 2 V.
They halve over equal time intervals, consistent with an exponential.
The half-time is one second rather than the time constant.
The time constant is approximately 1.4427 s; substitution reproduces both later readings.
Voltage itself does not grow by a constant factor.

\begin{aligned}\frac{U_f-U(1\,\mathrm s)}{U_f-U(0)}&=\frac12=e^{-\frac{1\,\mathrm s}\tau}\\{}\tau&=\frac{1\,\mathrm s}{\ln2}\approx1.4427\,\mathrm s\end{aligned}

[1]

Error analysis · Application

For a voltage transition from 4 V to 10 V someone claims: after τ voltage is 63.2% of 10 V, namely 6.32 V. Correct the value and identify the reference for 63.2%; assume an ideal RC circuit.

Show hint

Initial voltage is not zero.

Learning goal: Apply a percentage to the correct voltage reference.

Show solution and calculation

Approach: Add the exponentially achieved fraction of the total 6 V change to the initial voltage.
After one time constant voltage is approximately 7.79272 V.
The 63.2% refers to the removed fraction of the initial deviation from the endpoint, not the fraction of final voltage.
The remaining gap is about 2.20728 V; together they sum to 10 V.

\begin{aligned}U(\tau)&=U_i+(U_f-U_i)(1-e^{-1})\\{}&=4+6(1-e^{-1})\approx7.79272\,\mathrm V\end{aligned}

[1]

One storage pole is not enough for every network

Several independent storage elements may produce several time constants; inductance may permit overshoot. Leakage, ESR, voltage-dependent capacitance and a current-limited driver alter the simple RC model. R=0 in an ideal switching circuit is a singular limit, not a finite zero-time charging calculation. The response here requires positive Req, constant C and a constant equivalent after switching.[1][2]

In brief

The time constant tracks decay of the initial deviation. State initial and final voltage and use the resistance actually seen after switching. A percentage target gives a logarithmic time, not the claim that charging is complete after one time constant.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added reachable voltage thresholds and their logarithmic timing, linked two prerequisites in context and clarified the English statement about finite time.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. RC time constant. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rc-zeitkonstante

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Frequently asked questions about RC time constant

What does τ=RC mean in practice?+

A time constant is the interval after which initial deviation from the final value falls by a factor of one over e. Charging from zero toward a constant source then reaches about 63.2% of the final value. Discharging toward zero leaves 36.8% of initial voltage. After five time constants the response is very close but mathematically not exactly complete. The resistance used is the equivalent seen by the capacitor after switching. Different switch positions can therefore have different charge and discharge times. Existing initial voltage is included in the general response rather than assuming zero-start charging.

How do I convert microfarads correctly?+

Micro means a factor of one millionth, so 47 microfarads must be entered as 47 times ten to the minus six farads. Together with 2.2 kilohms, the original example gives a time constant of 0.1034 seconds. Kilohms additionally mean one thousand ohms. Convert both values consistently to SI or use a checked compatible prefix combination; do not mix raw numbers from different scales. The unit check is ohms times farads equals seconds. The exponent, time divided by time constant, must also be dimensionless. An incorrectly converted capacitance changes the whole response by orders of magnitude.

Which function applies during discharge?+

For ideal discharge toward zero, voltage equals initial voltage times the decaying exponential factor. Signed plate charge has the same relative response when C is constant. Energy instead follows voltage squared, so its exponential has twice the exponent magnitude. In the original 8 V/4.7 kΩ/100 μF case, the time constant is 0.47 seconds. The 2 V threshold is reached at about 0.6516 seconds; at 0.70 seconds about 1.804 V remains. A different final value requires the general deviation form. Removing the source alone does not guarantee discharge: the model needs an actual conducting discharge path.

What happens with a larger resistor?+

At unchanged capacitance and identical initial and final voltages, a larger positive equivalent resistance increases time constant proportionally. Doubling it doubles time to the same relative remaining deviation while halving initial current. Final voltage remains unchanged in the same ideal source model. The relevant resistance is the full equivalent seen by the capacitor, not necessarily one resistor changed within a network. Loading a divider may alter both final voltage and equivalent resistance. Leakage, driver current limiting and temperature-dependent values can also limit simple scaling. Check time scale separately from final value.

How is an RC circuit related to a differential equation?+

The node equation equates source current through equivalent resistance with capacitor current, capacitance times voltage derivative. Rearrangement gives a first-order linear differential equation. The constant final voltage is a particular solution; deviation from it satisfies a homogeneous equation with exponential decay. Initial voltage fixes the coefficient. This produces charging and discharging from the same approach rather than two unrelated memorised formulas. The original example begins at 1 V and approaches 5 V; after two time constants it gives 4.459 V. Several independent storage elements can create additional poles and require an extended equation system rather than one time constant.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §10.5: RC Circuits

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RC time constant

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Symbol

Meaning

Unit

τ

Time constant

R

Resistance

C

Capacitance

Q

Final charge during charging

t

Time since switching

Derivation & background

Exam blueprint

Validity range

Derivation steps

Resistance limits charging current while capacitor voltage reduces the remaining driving voltage difference.

1After switching, consider one ideal capacitor connected to a time-invariant resistive network.
2Replace the network at the capacitor by its Thévenin voltage and effective resistance; dependent sources remain active.
3The loop gives $R_{\mathrm{eq}}i+U_C=U_f$ , while the capacitor gives $i=C\frac{\mathrm dU_C}{\mathrm dt}$ .
4Set $x=U_C-U_f$ : then $\frac{\mathrm dx}{\mathrm dt}=-\frac{x}{R_{\mathrm{eq}}C}$ .
5With initial voltage $U_i$ , $U_C(t)=U_f+(U_i-U_f)e^{-\frac t\tau}$ , where $\tau=R_{\mathrm{eq}}C$ .
6After one time constant, the reciprocal of e of the initial deviation remains. The familiar 63.2 percent of final voltage specifically applies to charging from zero volts.

How do you calculate with RC time constant?

Here is how to work through a typical RC time constant (τ = RC; q(t)=Q(1−e^(−t/τ))) task step by step:

Task

R=10 kΩ and C=100 μF. Find the time constant τ.

Solution path

With $R=10\,\mathrm{k\Omega}$ and $C=100\,\mathrm{\mu F}$ , $\tau=RC=1.0\,\mathrm s$ .
Only for the explicitly assumed zero start toward constant nonzero final voltage does one second give $\frac q{Q_f}=1-e^{-1}\approx0.632$ .
In general, 36.8% of initial deviation remains.

Check: $\frac{1\,\mathrm s}{RC}=1$ , so the exponent is dimensionless.

Understand and apply the equation

An exponential approach, not a completion instant

\begin{aligned}\tau&=R_{\mathrm{eq}}C\\{}u_C(t)&=U_f+(U_i-U_f)e^{-\frac{t}{\tau}}\end{aligned}

One ideal linear capacitor and a constant stable resistive equivalent after switching; t=0 at the switching instant.[1][2]

Voltage and energy decay at different rates

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Both curves describe a linear capacitor discharging toward zero:

\frac{u_C}{U_i}=e^{-\frac t\tau}

\frac{W_C}{W_i}=e^{-\frac{2t}\tau}

. This normalization requires nonzero initial voltage.[2][4]

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Combine the component relation and node balance

\begin{aligned}i&=C\frac{\mathrm du_C}{\mathrm dt}=\frac{U_f-u_C}{R_{\mathrm{eq}}}\\{}\tau\frac{\mathrm du_C}{\mathrm dt}+u_C&=U_f\end{aligned}

Current is positive into the positive capacitor terminal; capacitance is constant.[1]

\begin{aligned}w&=u_C-U_f\\{}\frac{\mathrm dw}{w}&=-\frac{\mathrm dt}{\tau}\\{}w(t)&=(U_i-U_f)e^{-\frac{t}{\tau}}\end{aligned}

Solve the deviation from steady state; the special case Ui=Uf remains constant.[1]

Capacitance of a capacitor connects plate charge with voltage. For constant C its time derivative gives capacitor current in the balance above.[1]

Find the correct resistance after switching

Initial charge, threshold time and switching

Charge from 1 V to 5 V rather than from zero

Given: $R=2.2\,\mathrm{k\Omega}$ , $C=47\,\mathrm{\mu F}$ , $U_i=1.0\,\mathrm V$ , $U_f=5.0\,\mathrm V$ , $t=2\tau$ .
Find: Find time, voltage, plate charge and current.

Approach: Use the general response with actual initial voltage.
$\begin{aligned}\tau&=RC\\{}u_C&=U_f+(U_i-U_f)e^{-\frac{t}{\tau}}\end{aligned}$
Convert prefixes to SI.
$\begin{aligned}\tau&=2200\cdot47\cdot10^{-6}=0.1034\,\mathrm s\\{}t&=0.2068\,\mathrm s\end{aligned}$
Use the actual initial deviation −4 V.
$u_C=5-4e^{-2}=4.458658867\,\mathrm V$
Find charge and decreasing current from the same response.
$\begin{aligned}q&=Cu_C=209.5569668\,\mu\mathrm C\\{}i&=\frac{4e^{-2}}{2200}=0.2460642\,\mathrm{mA}\end{aligned}$

Result: At 206.8 ms voltage is about 4.459 V and charge 209.56 μC; current 0.246 mA.

Check: Remaining deviation $5\,\mathrm V-U_C\approx 0.54134\,\mathrm V$ equals $4 e^{-2}$ . $iR+U_C$ equals 5 V; at t=0 the formula gives exactly 1 V.

Discharge to a specified threshold

Given: $U_i=8.0\,\mathrm V$ , $U_f=0$ , $R=4.7\,\mathrm{k\Omega}$ , $C=100\,\mathrm{\mu F}$ .
Find: Find time to 2.0 V, and voltage and remaining energy at 0.70 s.

Approach: Solve discharge toward zero logarithmically for the threshold, then find energy from voltage.
$\begin{aligned}u_C&=U_i e^{-\frac{t}{\tau}}\\{}t&=\tau\ln\frac{U_i}{u_C}\\{}E_C&=\frac{1}{2}Cu_C^2\end{aligned}$
Find time constant and logarithmic threshold time.
$\begin{aligned}\tau&=0.47\,\mathrm s\\{}t_{2\mathrm V}&=0.47\ln(\frac{8}{2})=0.65155835\,\mathrm s\end{aligned}$
Evaluate the exponential at the second instant.
$u_C(0.70)=8e^{-\frac{0.70}{0.47}}=1.80413245\,\mathrm V$
Energy follows the square of voltage.
$\begin{aligned}E_0&=\frac{1}{2}C8^2=3.20\,\mathrm{mJ}\\{}E(0.70)&=\frac{1}{2}Cu_C^2=0.162744695\,\mathrm{mJ}\end{aligned}$

Result: 2 V is reached at about 651.6 ms; at 700 ms 1.804 V and 0.1627 mJ remain.

Check: Voltage decays as $e^{-\frac{t}{\tau}}$ , energy as $e^{-2\frac{t}{\tau}}$ ; about 3.0373 mJ has been dissipated in the ideal resistor.

A second switching event starts from the existing voltage

Given: A 20 μF capacitor starts at 0 V and charges through 1.0 kΩ from 10 V. After 20 ms it is disconnected from that source and connected to reference through 2.0 kΩ. The capacitor is ideal and linear; there are no impulse currents.
Find: Find voltage at switching and 40 ms later.

Approach: each switch position has its own time constant. The voltage reached at the end of the first interval becomes the initial voltage of the second.
$\begin{aligned}\tau_1&=R_1C\\u_1(t)&=10(1-e^{-\frac t{\tau_1}})\\\tau_2&=R_2C\\u_2(s)&=u_1(20\,\mathrm{ms})e^{-\frac s{\tau_2}}\end{aligned}$
The first phase lasts exactly one of its time constants.
$\begin{aligned}\tau_1&=20\,\mathrm{ms}\\u_1(20\,\mathrm{ms})&=10(1-e^{-1})\approx6.32120559\,\mathrm V\end{aligned}$
In the second phase, time is measured since switching.
$\begin{aligned}\tau_2&=40\,\mathrm{ms}\\u_2(40\,\mathrm{ms})&=10(1-e^{-1})e^{-1}\\&\approx2.32544158\,\mathrm V\end{aligned}$

Result: Voltage is 6.321 V at total time 20 ms and 2.325 V at total time 60 ms. Capacitor voltage remains continuous at switching.

Convert target accuracy into time

\begin{aligned}\frac{|u_C-U_f|}{|U_i-U_f|}&=\varepsilon\\{}t&=-\tau\ln\varepsilon\\{}0&<\varepsilon<1\end{aligned}

Relative remaining deviation, provided Ui≠Uf; exactly zero error cannot be reached in finite time.[1][2]

t=R_{\mathrm{eq}}C\ln\left|\frac{U_i-U_f}{U_*-U_f}\right|

Threshold time in the constant one-capacitor model; a positive finite time requires U* strictly between Ui and Uf.[1][2]

Check farads, coulombs and seconds

\begin{aligned}[RC]&=\frac{\mathrm V}{\mathrm A}\frac{\mathrm C}{\mathrm V}=\mathrm s\\{}1\,\mu\mathrm F&=10^{-6}\,\mathrm F\end{aligned}

The exponent $\frac{t}{\tau}$ is dimensionless; microfarads cannot be entered as whole farads.[2]

Five time-response exercises

Calculate · Foundation

R=1 kΩ, C=20 μF, zero-start charging toward 3 V. Voltage at τ?

Show hint

Do not confuse it with full source voltage.

Learning goal: Find voltage after one time constant.

Show solution and calculation

Approach: for zero initial voltage, use the achieved fraction of the whole voltage change.
The time constant is 20 ms.
Capacitor voltage then is approximately 1.89636 V; the remaining source difference is 1.10364 V.
Together they give 3 V.

u_C(\tau)=3(1-e^{-1})\approx1.89636\,\mathrm V

Transfer · Application

How does doubling R at fixed C, Ui, Uf affect the response?

Show hint

Separate time scale and initial current.

Learning goal: Transfer resistance changes to time and current.

Show solution and calculation

Approach: compare time constant and initial current at identical initial and final voltages. τ doubles and initial current halves; time to the same relative deviation doubles.
Final voltage remains Uf.

\begin{aligned}\tau_2&=2\tau_1\\{}i_2(0)&=\frac{i_1(0)}{2}\end{aligned}

Rearrange · Application

Discharge should reach 10% of initial voltage at 30 ms. C=10 μF. Find model resistance.

Show hint

Use $\tau=\frac{t}{\ln10}$ , then $R=\frac{\tau}{C}$ .

Learning goal: Rearrange a discharge target for resistance.

Show solution and calculation

Approach: the specified residual is a fraction of initial voltage; solve the exponent logarithmically. τ≈13.0288 ms; R≈1302.88 Ω.
Substitution into the discharge response restores 10%.

R=\frac{0.030}{10^{-5}\ln10}\approx1302.88\,\Omega

Interpret · Application

An ideal capacitor approaches 10 V. At t=0, 1 s and 2 s voltages are exactly 2 V, 6 V and 8 V. Is this consistent with a constant-parameter RC model? Find τ from residual voltages.

Show hint

Subtract each reading from the final voltage.

Learning goal: Recognize an exponential model from equal residual ratios.

Show solution and calculation

Approach: Residual voltages are 8 V, 4 V and 2 V.
They halve over equal time intervals, consistent with an exponential.
The half-time is one second rather than the time constant.
The time constant is approximately 1.4427 s; substitution reproduces both later readings.
Voltage itself does not grow by a constant factor.

\begin{aligned}\frac{U_f-U(1\,\mathrm s)}{U_f-U(0)}&=\frac12=e^{-\frac{1\,\mathrm s}\tau}\\{}\tau&=\frac{1\,\mathrm s}{\ln2}\approx1.4427\,\mathrm s\end{aligned}

[1]

Error analysis · Application

For a voltage transition from 4 V to 10 V someone claims: after τ voltage is 63.2% of 10 V, namely 6.32 V. Correct the value and identify the reference for 63.2%; assume an ideal RC circuit.

Show hint

Initial voltage is not zero.

Learning goal: Apply a percentage to the correct voltage reference.

Show solution and calculation

Approach: Add the exponentially achieved fraction of the total 6 V change to the initial voltage.
After one time constant voltage is approximately 7.79272 V.
The 63.2% refers to the removed fraction of the initial deviation from the endpoint, not the fraction of final voltage.
The remaining gap is about 2.20728 V; together they sum to 10 V.

\begin{aligned}U(\tau)&=U_i+(U_f-U_i)(1-e^{-1})\\{}&=4+6(1-e^{-1})\approx7.79272\,\mathrm V\end{aligned}

[1]

One storage pole is not enough for every network

In brief

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Quanta · Updated October 2, 2026

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Change history

October 2, 2026Added reachable voltage thresholds and their logarithmic timing, linked two prerequisites in context and clarified the English statement about finite time.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. RC time constant. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rc-zeitkonstante

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