Inductor RL Charge Time Calculator

RL Circuit Charging Time

Drive a series RL circuit from a DC source and the current builds along an exponential curve, i(t) = (V/R)(1 - e^(-tR/L)). V is the source voltage, R the resistance in ohms, L the inductance in henries. The time constant τ = L/R tells you how many seconds the current needs to reach about 63.2% of its steady-state value V/R. With L = 10 mH and R = 100 Ω you get τ = 100 μs, so by roughly 500 μs (5τ) the current has all but settled at its final value.

An inductor fights any change in current (Lenz's law), so the current ramps up slowly while the voltage across the inductor follows V₄(t) = V e^(-tR/L). That voltage starts at V, since the inductor looks like an open circuit at t = 0⁺, and falls toward zero as the resistor takes over. Switch the supply off and the stored energy E = ½LI² has nowhere to go but out, which is what produces the familiar inductive kickback. Boylestad and Sedra/Smith work through these first-order responses in detail.

Applications

RL behavior shapes how motors and solenoids start up, how relay coils and contactors pick up, how ignition coils fire, and how chokes in switching supplies and current-limiting reactors respond. Standards like IEC 60076 (transformers) and IEC 60947 (low-voltage switchgear) lay out inrush and time-constant behavior so you can size protection correctly. To soak up the energy released when an inductive load is switched off, designers usually add free-wheeling diodes, RC snubbers, or TVS clamps.

FAQ

Why does current rise slower in larger inductors? The time constant τ = L/R climbs as L grows. A bigger inductor holds more magnetic energy per ampere, so with the same R it takes longer to drive the current up to its steady-state value.

What happens at the instant the switch closes? Right at t = 0⁺ the current is still zero and the inductor takes the entire source voltage across it, behaving mathematically like an open circuit. It only starts to act like a short for DC after several τ have passed.

Why use a free-wheeling diode across a relay coil? The moment the drive transistor turns off, the coil insists on keeping its current going and throws out a high reverse voltage spike (-L·di/dt). A diode gives that current somewhere to circulate, dissipating the stored energy safely rather than punching through the transistor.