# RL Time Constant > The RL time constant tells you how long a coil takes to let its current build up through a resistor: divide the coil's inductance by the resistor's resistance. Page: https://formulabro.yunusemrevurgun.com/electricity/rl-time-constant/ Category: Electricity Published: 2026-10-22 Updated: 2026-10-22 ## Formula $$ \tau = \frac{L}{R} $$ Read it out loud: "Tau equals L over R. Tau is the Greek letter t, said to rhyme with wow. The time constant equals L, the inductance in henries, divided by R, the resistance in ohms. The answer is a time in seconds." ## Before you start - **Resistance and the ohm ($\Omega$)**: **Resistance** is how strongly a component limits the flow of current. It is measured in **ohms**, written with the Greek capital letter omega, $\Omega$. One ohm means that one volt of push drives one ampere of current through it, so $1\ \Omega = 1$ V/A. A 10 $\Omega$ resistor lets 0.1 A flow when 1 V is across it, because 1 divided by 10 is 0.1. - **Inductance and the henry (H)**: An **inductor** is a coil, and its **inductance**, written $L$, measures how strongly it opposes changes in current. The unit is the **henry**, written H, which is one volt-second per ampere. The example coil has 2 H of inductance. Smaller coils are measured in millihenries, written mH, which are thousandths of a henry. - **Time constant and the Greek letter tau ($\tau$)**: A **time constant** is a length of time that sets the speed of a change. After one time constant, a change heading toward a fixed final value has covered about 63.2 percent of the distance to that value. The Greek letter tau, written $\tau$ and said 'tow', is the usual name for the time constant. In an RL circuit it is measured in seconds. - **The number $e$ and powers like $e^{-x}$**: The number $e$ is about 2.71828. It is a fixed number that appears wherever something grows or shrinks at a rate proportional to its own size. $e^{x}$ means the number $e$ raised to the power $x$, and for whole numbers it means multiplying $e$ by itself $x$ times. A negative power, $e^{-x}$, means one divided by $e^{x}$, so it is a small number that shrinks as $x$ grows. For example, $e^{-1} \approx 0.368$ and $e^{-5} \approx 0.0067$. - **Percent, written %, and approximately equal ($\approx$)**: **Percent** means parts out of 100, so 63.2% means 63.2 parts out of every 100, or 0.632 as a decimal. The symbol $\approx$ means approximately equal, and it is used when a result has been rounded. The value 0.632 is a rounded form of $1 - e^{-1}$, which is about 0.6321. More notation help: https://formulabro.yunusemrevurgun.com/math-basics/ ## What each symbol means - $\tau$: The **time constant**, measured in seconds (s). It tells you how quickly the current rises or falls. In the worked example $\tau = 0.2$ s, which is a fifth of a second, so the current reaches about 63 percent of its final value in that time. - $L$: The **inductance** of the coil, measured in henries (H). A bigger coil opposes changes in current more strongly, so it makes the current rise more slowly and makes $\tau$ bigger. The worked example uses a 2 H coil. - $R$: The **resistance** of the resistor connected in series with the coil, measured in ohms ($\Omega$). A bigger resistance limits the current more strongly and lets the current settle sooner, so it makes $\tau$ smaller. The worked example uses a 10 $\Omega$ resistor. ## The formula piece by piece - $R$: The resistance in ohms, which is how much the resistor limits the current. Dividing by R means a bigger resistance gives a smaller answer, because the current reaches its final value sooner when the resistor is larger. - $L$: The inductance in henries. It measures how strongly the coil opposes changes in its current, and it is the part that makes the current rise slowly. A bigger inductance gives a bigger time constant. - $\frac{L}{R}$: Henries divided by ohms. A henry is a volt-second per ampere and an ohm is a volt per ampere, so dividing one by the other cancels the volts and the amperes and leaves seconds. The result is a time, which is exactly what a time constant must be. - $\tau = \frac{L}{R}$: The whole formula in one sentence: the RL time constant, tau, equals the inductance divided by the resistance, and it is the time in seconds that the current needs to reach about 63.2 percent of its final value. - $I(t) = \frac{V}{R}\left(1 - e^{-t/\tau}\right)$: A standard form for the rising current. It gives the current $I(t)$ at time $t$ after switch-on, where $V$ is the supply voltage. The current starts at zero and approaches the final value $\frac{V}{R}$ as time goes on. ## What the formula is doing, and why A coil connected through a resistor to a battery does not let its current jump straight to the final value. The inputs are the coil's inductance $L$ and the resistor's resistance $R$. The output is the time constant $\tau$, in seconds, which tells you how fast the current rises toward its final value of $\frac{V}{R}$. The current is zero at the moment the switch closes, and the coil's opposition to change is the reason it rises slowly. **Why divide?** A henry is a volt-second per ampere, and an ohm is a volt per ampere. Dividing henries by ohms cancels the volts and the amperes and leaves seconds. A bigger coil has more inductance, so it opposes change for longer, and $\tau$ grows. A bigger resistor lets the current settle sooner, so $\tau$ shrinks. Doubling $L$ doubles $\tau$, and doubling $R$ halves it. **Why the curve has this shape?** The current follows $I(t) = \frac{V}{R}\left(1 - e^{-t/\tau}\right)$. At $t = 0$ the exponent is zero, and $e^{0} = 1$, so the current is zero. After one time constant, $t/\tau = 1$, and the current is $\frac{V}{R}(1 - e^{-1})$, about 63.2 percent of its final value. After about five time constants the current is more than 99 percent of its final value, because $e^{-5}$ is only about 0.0067. The curve is steep at first and flattens as it nears the final value, because each part of the rise is proportional to how far the current still has to go. **Why the coil slows the rise.** The voltages around the loop must add up to the supply voltage: $V = IR + L\frac{dI}{dt}$. At the start the current is zero, so the resistor has no voltage and the coil carries the full supply voltage. That makes the current rise fastest at the start. As the current grows, the resistor takes more of the voltage, the coil's share falls, and the rise slows until the coil has no voltage left and the current stops changing. **When it applies.** The formula describes a simple series circuit with an ideal coil and a fixed resistor, switched on to a steady supply. The same time constant also describes the fall of the current after the supply is switched off. The current then drops to about 36.8 percent of its starting value after one time constant. ## An everyday picture Think of pushing a heavy cart across a floor with friction. The mass of the cart plays the part of the inductance, because a heavier cart takes longer to speed up. The friction plays the part of the resistance, because more friction lowers the top speed and lets the cart settle sooner. The steady push plays the part of the supply voltage, and the cart's speed plays the part of the current. The cart's speed follows the same curve as the current, and its time constant is mass divided by friction, the same pattern as $L / R$. ## Worked example: A 2 henry coil and a 10 ohm resistor on a 5 volt supply Imagine a made-up example with easy numbers: a coil with $L = 2\ \text{H}$ and a resistor with $R = 10\ \Omega$ are connected in series to a 5 V supply, switched on at $t = 0$. Find the time constant, the final current, and the current at $t = 0.2$ s and at $t = 1$ s. 1. **Write down the inputs.** The coil has $L = 2\ \text{H}$, the resistor has $R = 10\ \Omega$, and the supply is $V = 5\ \text{V}$. The questions are the time constant, the final current, and the current at two times. All the values are already in base units, so no prefix conversion is needed. 2. **Find the time constant.** Divide the inductance by the resistance: $\tau = \frac{L}{R} = \frac{2\ \text{H}}{10\ \Omega} = 0.2\ \text{s}$. The units are henries divided by ohms, which give seconds, so the answer is a time. A time constant of 0.2 s, a fifth of a second, means the current reaches about 63 percent of its final value after that time. 3. **Find the final current.** After a long time the current stops changing, and the coil adds no voltage. Only the resistor limits the current, so $I_\infty = \frac{V}{R} = \frac{5\ \text{V}}{10\ \Omega} = 0.5\ \text{A}$. The symbol $\infty$ means 'after a very long time'. The unit works out because volts divided by ohms gives amperes. 4. **Find the current after one time constant, 0.2 s.** Here $t = 0.2$ s, so $\frac{t}{\tau} = \frac{0.2}{0.2} = 1$. The formula gives $I = 0.5\ \text{A} \times (1 - e^{-1})$. The number $e^{-1}$ is one divided by $e$, about 0.3679, so $1 - 0.3679 = 0.6321$, and $0.5 \times 0.6321 \approx 0.316\ \text{A}$. The current has covered about 63 percent of its final value. 5. **Find the current after five time constants, 1 s.** Here $t = 1$ s, so $\frac{t}{\tau} = \frac{1}{0.2} = 5$. The formula gives $I = 0.5\ \text{A} \times (1 - e^{-5})$. The number $e^{-5}$ is about 0.0067, a tiny number, so $1 - 0.0067 = 0.9933$ and $I \approx 0.5 \times 0.9933 \approx 0.497\ \text{A}$. That is more than 99 percent of the final current. 6. **Check the start.** At $t = 0$ the exponent is zero, and $e^{0} = 1$, so $I = 0.5 \times (1 - 1) = 0$ A. This matches the fact that the current starts at zero when the switch closes. As time passes, the current gets closer and closer to 0.5 A without quite reaching it. 7. **Check the units of the answer.** Volts divided by ohms gives amperes, and the bracket is a plain number with no unit. So every current in this example comes out in amperes, which is the unit a current should have. **Answer:** The time constant is **0.2 s**. The current reaches **0.316 A** after 0.2 s, which is about 63 percent of the final **0.5 A**. After 1 s it is **about 0.497 A**, more than 99 percent of the way to the final value. ## Common mistakes - Assuming the current starts at its final value. Right after the switch closes, the current is zero, not $\frac{V}{R}$. - Thinking the current is final after one time constant. After one $\tau$ it is only about 63 percent of the final value, and it takes about five $\tau$ to get above 99 percent. - Writing the formula upside down as $\tau = R / L$. A bigger coil should be slower and a bigger resistor should be faster, and only $L / R$ shows both effects. - Entering millihenries as henries. A 2 mH coil has $L = 0.002$ H, and skipping that conversion makes $\tau$ 1000 times too big. ## Where you will meet it - Relay and solenoid circuits, where the coil current rises over a short, known time after switch-on. - Motor windings, where the slow build-up of current after switch-on sets how quickly the motor reaches its running current. - Switching power supplies and DC-to-DC converters, which store energy in an inductor while its current rises. - Laboratory experiments that measure a coil's inductance by timing how fast its current rises. ## Questions people ask ### Is the RL time constant the same as the RC time constant? No. The formulas differ: $\tau = L / R$ for an RL circuit, and $\tau = RC$ for an RC circuit, where $C$ is the capacitance. In both circuits, though, the change covers about 63 percent of its final value after one time constant. ### Why does the coil slow the current down? A changing current makes a voltage across the coil that opposes the change. The effect is strongest right after switch-on, when the current is changing fastest, which is why the current starts at zero and rises slowly. ### What happens when the supply is switched off? The current decays with the same time constant. After one $\tau$ it has fallen to about 37 percent of its starting value. The coil's stored energy has to go somewhere, which is why a switch can spark when it opens, unless a diode or resistor gives the current another path. ### Why is the time constant L over R? A bigger inductance means the coil opposes change for longer, so the time constant grows. A bigger resistance lets the current settle sooner, so the time constant shrinks. Dividing $L$ by $R$ captures both effects in one formula. ## Related formulas - [RC Time Constant](https://formulabro.yunusemrevurgun.com/electricity/rc-time-constant/) - [Ohm's Law](https://formulabro.yunusemrevurgun.com/electricity/ohms-law/) - [Inductive Reactance](https://formulabro.yunusemrevurgun.com/electricity/inductive-reactance/) ## Sources - [RL circuit, Wikipedia](https://en.wikipedia.org/wiki/RL_circuit) - [Inductance, Wikipedia](https://en.wikipedia.org/wiki/Inductance) - [Exponential decay, Wikipedia](https://en.wikipedia.org/wiki/Exponential_decay) From Formula Bro (https://formulabro.yunusemrevurgun.com/), a side project developed by yunusemrevurgun.com.