# RC Time Constant

> The RC time constant tells you how long a capacitor takes to fill up through a resistor. You get it by multiplying the resistance in ohms by the capacitance in farads, and the answer comes out in seconds.

Page: https://formulabro.yunusemrevurgun.com/electricity/rc-time-constant/  
Category: Electricity  
Published: 2026-10-22  
Updated: 2026-10-22

## Formula

$$
\tau = R C
$$

Read it out loud: "Tau equals R times C. Tau is the Greek letter used for the time constant, measured in seconds. R is the resistance, measured in ohms. C is the capacitance, measured in farads. So the time constant equals resistance times capacitance."

## Before you start

- **Powers of ten and the prefixes micro and kilo**: A **power of ten** is 10 multiplied by itself a set number of times. $10^{3}$ is $10 \times 10 \times 10 = 1000$, and $10^{-6}$ is one millionth, a decimal point followed by five zeros and then a 1. The prefix **kilo** (k) means $10^{3}$, so 10 kilohms is $10 \times 10^{3} = 10\,000$ ohms. The prefix **micro** ($\mu$) means $10^{-6}$, so 100 microfarads is $100 \times 10^{-6}$ farads, which is 0.0001 farads.
- **Capacitor, capacitance and the farad (F)**: A **capacitor** is a part that stores electric charge, usually between two metal plates separated by an insulator. Its **capacitance**, written $C$, is how much charge it holds for each volt across it. Capacitance is measured in **farads** (F), and one farad is one coulomb of charge per volt. A whole farad is very large, so real capacitors are often rated in microfarads.
- **Resistance and the ohm ($\Omega$)**: **Resistance**, written $R$, is how strongly a part opposes the flow of current, and it is measured in **ohms**, written $\Omega$. One ohm means one volt drives one ampere through the part. A resistor placed between a supply and a capacitor limits how fast charge can flow into the capacitor, so a bigger resistance means slower charging.
- **The Greek letter tau, $\tau$**: Tau, written $\tau$, is a letter of the Greek alphabet. Greek letters are often used as names for quantities, and each page says what its letters stand for. Here $\tau$ is the name of the **time constant**, which is measured in seconds. It is said as 'tau', rhyming with 'how' or 'taw'.
- **The number e and exponentials, like $e^{-t/\tau}$**: The number $e$ is a fixed number, about 2.718, that appears whenever something changes at a rate proportional to its own size. $e^{x}$ is read 'e to the power x'. A negative power such as $e^{-1}$ means $\frac{1}{e} \approx 0.3679$, and $e^{-5}$ is much smaller, about 0.0067. So $e^{-t/\tau}$ starts at 1 and shrinks toward zero as time passes.

More notation help: https://formulabro.yunusemrevurgun.com/math-basics/

## What each symbol means

- $\tau$: The **time constant**, the Greek letter tau, measured in seconds (s). It sets how fast the capacitor charges or discharges. After one time constant, the capacitor has covered about 63.2% of the way to the supply voltage.
- $R$: **Resistance** of the resistor in the circuit, measured in ohms ($\Omega$). A small resistor such as 100 $\Omega$ lets charge flow in quickly, and a large one such as 10 k$\Omega$ lets it flow in slowly.
- $C$: **Capacitance**, how much charge the capacitor holds for each volt across it, measured in farads (F). One farad is one coulomb per volt. A practical capacitor is often 100 microfarads, written $100\ \mu\text{F}$, which is $100 \times 10^{-6}$ F.

## The formula piece by piece

- $C$: Start with the capacitor. $C$ is its capacitance in farads, which is the charge it stores for each volt. A bigger $C$ holds more charge, so it takes longer to fill up.
- $R$: $R$ is the resistance in ohms, which limits how fast charge can flow in. A bigger $R$ lets less current through, so the charge arrives more slowly.
- $R C$: Multiply the resistance by the capacitance. The units give seconds, because ohms times farads equals seconds, $\Omega \cdot \text{F} = \text{s}$. A bigger resistor or a bigger capacitor makes the product larger, and the time longer.
- $\tau = R C$: The whole formula in one sentence: **the time constant equals the resistance times the capacitance**, and it tells you how many seconds the capacitor needs to charge up to most of the supply voltage.
- $V_C(t) = V_s \left(1 - e^{-t/\tau}\right)$: A standard second form, the charging curve. $V_C(t)$ is the capacitor voltage at time $t$, starting at 0 V, and $V_s$ is the supply voltage. The factor $e^{-t/\tau}$ shrinks as time passes, so the voltage climbs toward the supply voltage.

## What the formula is doing, and why

A capacitor stores electric charge, a bit like a small reservoir. When a capacitor is connected through a resistor to a battery, it does not fill up at once. The resistor limits how fast charge can flow in, so the voltage across the capacitor starts at 0 V and rises gradually toward the supply voltage $V_s$.

**Why multiply?** Here is a rough reasoning. To fill a capacitor to a voltage $V$, it must receive a charge $Q = C V$. At the start the resistor limits the current to about $\frac{V}{R}$. Time is charge divided by current, so the time is about $\frac{C V}{V / R} = R C$. This gives the scale of the time. A bigger $R$ or a bigger $C$ makes charging slower by the same factor, so doubling either one doubles $\tau$. The exact curve needs more careful mathematics, because the current falls as the capacitor fills, but the scale is the same.

**Why the curve bends.** As the capacitor fills, the gap between the supply voltage and the capacitor voltage shrinks, and the current shrinks with it. So the curve is steep at the start and flattens out as it nears the supply voltage. After one time constant the capacitor has reached about 63.2% of the supply voltage. After two it is at about 86.5%, after three about 95.0%, and after five about 99.3%.

After about five time constants the capacitor is more than 99% charged, which is why five time constants is a common rule of thumb for full charge. The same time constant applies when the capacitor discharges through the resistor. Its voltage then falls as $V_C(t) = V_0 e^{-t/\tau}$, and after one time constant it has dropped to about 36.8% of its starting voltage.

## An everyday picture

Think of filling a bucket through a hose. The resistor is the hose: a narrower hose fills the bucket more slowly. The capacitor is the bucket: a bigger bucket takes longer to fill. The time constant is the time it takes the water level to rise most of the way up. Doubling the hose's narrowness or the bucket's size doubles the time, which is the same as doubling $\tau$.

## Worked example: Charging a 100 microfarad capacitor through 10 kiloohm

A 5 V supply charges a capacitor with $C = 100\ \mu\text{F}$ through a resistor with $R = 10\ \text{k}\Omega$. The k means kilo, so $R$ is 10 000 ohms. The capacitor starts at 0 V. Find the time constant, then the capacitor voltage after 1 s and after 5 s.

1. **Write each input in plain units.** Kilo means one thousand, so $10\ \text{k}\Omega = 10 \times 10^{3}\ \Omega = 10\,000\ \Omega$. Micro means one millionth, $10^{-6}$, so $100\ \mu\text{F} = 100 \times 10^{-6}\ \text{F} = 0.0001\ \text{F}$. Converting first avoids the common mistake of mixing scales.
2. **Multiply to get the time constant.** $\tau = R C = 10\,000\ \Omega \times 0.0001\ \text{F} = 1\ \text{s}$. In powers of ten, $10 \times 10^{3} \times 100 \times 10^{-6}$: multiply the front numbers, $10 \times 100 = 1000$, and add the exponents, $3 + (-6) = -3$, so the result is $1000 \times 10^{-3} = 1$. The unit is ohm times farad, which is a second.
3. **Find the fraction charged after 1 s.** The ratio $\frac{t}{\tau} = \frac{1\ \text{s}}{1\ \text{s}} = 1$, so the exponent is $-1$. The fraction charged is $1 - e^{-1}$, and $e^{-1} = \frac{1}{e} \approx 0.3679$. So $1 - 0.3679 = 0.632$, which means the capacitor is 63.2% charged.
4. **Find the voltage after 1 s.** Multiply the fraction by the supply voltage: $V_C = 5\ \text{V} \times 0.632 \approx 3.16\ \text{V}$. The capacitor started at 0 V, so in one time constant it has risen by 3.16 V.
5. **Find the fraction charged after 5 s.** Now $\frac{t}{\tau} = \frac{5\ \text{s}}{1\ \text{s}} = 5$, so the exponent is $-5$. $e^{-5} \approx 0.00674$, so the fraction charged is $1 - 0.00674 = 0.993$, which is 99.3%. Five seconds is five time constants here.
6. **Find the voltage after 5 s.** $V_C = 5\ \text{V} \times 0.993 \approx 4.97\ \text{V}$. The capacitor is almost at the full supply voltage, and the remaining gap is less than 0.04 V.
7. **Check against the five time constant rule.** Five time constants give about 99% charged, which is the 99.3% we found. The rule of thumb and the calculation agree, so the answer is consistent.

**Answer:** The time constant is **1 s**. After 1 s the capacitor is at **3.16 V**, which is 63.2% of 5 V. After 5 s it is at **4.97 V**, which is 99.3% of the supply, so it is almost fully charged.

## Common mistakes

- Forgetting the prefixes. 100 microfarad is $100 \times 10^{-6}$ F, and 10 kiloohm is 10 000 ohm. Skipping either one changes the answer by a factor of a thousand or a million.
- Thinking the capacitor is full after one time constant. After one $\tau$ it is only at about 63.2%, and it takes about five $\tau$ to get above 99%.
- Using the supply voltage for the capacitor voltage at every moment. The capacitor starts at 0 V and only reaches the supply voltage after a long time.
- Using $\tau = R C$ with the capacitance in farads and the resistance in kiloohms without converting. Convert to ohms and farads first, or the time comes out on the wrong scale.

## Where you will meet it

- Simple timing circuits, such as the delay in a timed switch or a short pause before a light turns off.
- Power supply smoothing and audio filters, where a resistor and capacitor set the frequency at which the signal starts to be cut.
- Button debouncing, where the slow charging of a capacitor smooths out the rapid switching when mechanical contacts bounce.
- Sensors that measure a resistance by timing how long a capacitor takes to charge.

## Questions people ask

### What is a farad?

A farad (F) is one coulomb of charge stored per volt. Real capacitors are often rated in microfarads ($\mu\text{F}$) or picofarads, because one whole farad is a very large capacitor.

### When is the capacitor fully charged?

Not after one time constant. After about five time constants it is more than 99% charged, which is the usual rule of thumb for fully charged.

### Does the capacitor discharge with the same time constant?

Yes. Its voltage falls as $V_C(t) = V_0 e^{-t/\tau}$, and after one time constant it has dropped to about 36.8% of its starting voltage.

### Why does the formula have an e in it?

The charging speed is proportional to how much voltage is still missing. A rule like that gives an exponential curve, and $e$ is the natural base for exponential curves, the one that appears when growth or decay speed depends on the current size.

## Related formulas

- [Ohm's Law](https://formulabro.yunusemrevurgun.com/electricity/ohms-law/)
- [Resistors in Series and Parallel](https://formulabro.yunusemrevurgun.com/electricity/resistors-in-series-and-parallel/)
- [Kirchhoff's Current Law](https://formulabro.yunusemrevurgun.com/electricity/kirchhoffs-current-law/)

## Sources

- [RC circuit, Wikipedia](https://en.wikipedia.org/wiki/RC_circuit)
- [Capacitor, Wikipedia](https://en.wikipedia.org/wiki/Capacitor)
- [Exponential decay, Wikipedia](https://en.wikipedia.org/wiki/Exponential_decay)

From Formula Bro (https://formulabro.yunusemrevurgun.com/), a side project developed by yunusemrevurgun.com.