RL Time Constant Calculator

Calculator

Provide any two of \( R \), \( L \), \( \tau \) — the third is solved. Optionally provide a time \( t \) and a supply voltage or initial current to calculate the energizing or de-energizing current at that time.

• Resistance (R):
• Ω kΩ MΩ

• Inductance (L):
• H mH µH nH

• Time constant (τ):
• s ms µs min

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• Time (t) — optional:
• s ms µs min

• Supply voltage V_s (energizing) — optional:
• V mV kV

• Initial current I₀ (de-energizing) — optional:
• A mA

Provide any two of R, L, τ to calculate.

Calculate

RL Time Constant — Explanation

The RL time constant \( \tau \) characterises how quickly the current in an RL circuit rises or falls in response to a voltage step: \[ \tau = \frac{L}{R} \] Where \( L \) is in Henrys and \( R \) is in Ohms, giving \( \tau \) in seconds. A larger inductance stores more energy and takes longer to change its current; a larger resistance dissipates energy faster and shortens the transition. The time constant is the single number that governs every transient in the circuit.

The unit check is worth doing once: \( [L/R] = [\text{H}/\Omega] = [\text{H} \cdot \text{A} / \text{V}] = [\text{Wb} / \text{V}] = [\text{V} \cdot \text{s} / \text{V}] = [\text{s}] \). This is the RL counterpart of the RC time constant \( \tau = RC \); see the RC time constant calculator for the capacitive version.

Energizing

When a voltage \( V_s \) is applied to a series RL circuit with zero initial current, the current rises exponentially toward its steady-state value \( I_{\max} = V_s / R \): \[ I(t) = \frac{V_s}{R} \left(1 - e^{-t/\tau}\right) \] At \( t = 0 \) the inductor looks like an open circuit — it carries no current yet and all of \( V_s \) appears across it. As current builds up, the back-EMF of the inductor falls and the resistor takes an increasing share of the voltage. After five time constants the current is within 1% of \( I_{\max} \) and the inductor looks like a plain wire carrying DC: \[ \begin{align} t = 1\tau &\Rightarrow I = 63.2\%\ I_{\max} \\ t = 2\tau &\Rightarrow I = 86.5\%\ I_{\max} \\ t = 3\tau &\Rightarrow I = 95.0\%\ I_{\max} \\ t = 4\tau &\Rightarrow I = 98.2\%\ I_{\max} \\ t = 5\tau &\Rightarrow I = 99.3\%\ I_{\max} \end{align} \]

De-energizing

When the supply is removed from an inductor carrying an initial current \( I_0 \), the current decays exponentially through the discharge path (typically a freewheeling diode or the source resistance): \[ I(t) = I_0 \cdot e^{-t/\tau} \] After one time constant the current has fallen to \( 1/e \approx 36.8\% \) of \( I_0 \). The inductor resists the drop and will develop whatever voltage across itself is needed to keep current flowing — this is the source of the inductive voltage spike that appears when an inductive load is switched off without a suppression path. The energy stored at \( I_0 \) is \( E = \frac{1}{2}LI_0^2 \); see the inductor energy calculator to quantify it.

Purpose of the Calculator

Given any two of \( R \), \( L \), \( \tau \), the calculator solves for the third:

• To solve for \( \tau \): \[ \tau = \frac{L}{R} \]
• To solve for \( R \): \[ R = \frac{L}{\tau} \]
• To solve for \( L \): \[ L = \tau \cdot R \]

Optionally, providing \( t \) and \( V_s \) calculates the energizing current at that time, and providing \( t \) and \( I_0 \) calculates the de-energizing current. Both can be provided together to compare the two in the same circuit. If L is a combination of inductors, compute the equivalent value first with the inductors in series or inductors in parallel calculator.

Related Tools

• RC Time Constant Calculator — the capacitive equivalent: \( \tau = RC \), voltage across capacitor as the state variable.
• Inductor Impedance Calculator — frequency-domain view of the same RL circuit.
• Inductor Energy Calculator — quantify the energy stored at a given current.
• Inductors in Series / Inductors in Parallel — compute the equivalent inductance of a network before using it here.

More calculators: blog.hirnschall.net/tools/.