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Inductor Current Growth
Describes the exponential rise of current through an inductor when a DC voltage is first applied.
Interpretation
Inductor current growth when a DC voltage is applied: I(t) = (V/R)(1 − e^(−tR/L)).
The current rises exponentially toward its final value V/R.
Example: V=5V, R=2Ω, L=4H, t=1s → I = (5/2)(1 − e^(−1×2/4)) = 2.5(1 − e^(−0.5)) ≈ 0.983A.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| I | Current | A |
| V | Voltage | V |
| R | Resistance | Ω |
| L | Inductance | H |
| t | Time | s |
What it means
When a DC voltage is suddenly applied to a series RL circuit, the current grows exponentially according to I(t) = (V/R) * (1 − e^(−tR/L)). Here V/R is the final steady‑state current, and the time constant τ = L/R. The current cannot change instantaneously in an inductor, so it starts at 0 and asymptotically approaches the final value. This equation describes the build‑up of current in inductors and is used to analyse the transient behaviour of inductive circuits, such as relay coils, motor windings, and inductor‑based filters. The time to reach a certain percentage of the final current is similar to the RC case. Understanding this growth is essential for protecting circuits from inductive spikes and for designing switching power supplies. Example: V=5V, R=2Ω, L=4H. τ = L/R = 2s. At t=2s (one τ), I = (5/2)*(1−e^(−2/2)) = 2.5*(1−e^(−1)) = 2.5*0.632 = 1.58A. At t=4s (2τ), I = 2.5*(1−e^(−2)) = 2.5*0.865 = 2.16A, approaching 2.5A.
Worked example
Inductor Current Growth – Practical Example
Real‑World| Parameter | Value |
|---|---|
| V | 10 V |
| R | 50 Ω |
| L | 10 mH |
| Formula | I(t) = (V/R)(1 − e−t/τ), τ = L/R |
Common mistakes
- Final current: V/R is the steady‑state current (after a long time).
- Exponent: The term is −tR/L – the time constant is L/R.
- Initial condition: At t=0, the current is zero (assuming no initial current).
- Voltage across inductor: At t=0, the inductor acts as an open circuit (all voltage drop).
- AC: This formula is for a DC step applied to an RL circuit.
Applications
Inductor current growth when a DC voltage is applied is given by I(t) = (V/R)(1 − e^(−tR/L)). The current rises exponentially to its final value V/R with a time constant L/R. This is essential for analysing circuits with inductors, such as motors, transformers, and magnetic actuators. Engineers use it to calculate the inrush current, to design soft‑start circuits, and to determine the energy stored in an inductor. In power electronics, it helps predict the current waveform in boost and buck converters. By understanding the current growth, professionals can avoid excessive currents that might trip protection devices. This formula is a key part of transient analysis in circuits.
- Motor start‑up current and inrush control
- Design of inductor‑based filters and converters
- Relay and solenoid coil energisation timing
- Transformer magnetising current analysis
- Educational understanding of inductive transients
Frequently Asked Questions
The current through an inductor when a DC voltage is applied is I(t) = (V/R)(1 − e^(−tR/L)).
As t → ∞, I(t) → V/R, which is the steady-state current (inductor behaves as a short).
I(τ) = (V/R)(1 − e⁻¹) ≈ 0.632 (V/R).
Larger τ means slower current rise; smaller τ means faster.
V_L(t) = V e^(−tR/L). It starts at V and decays to zero as current reaches steady state.
The form is similar but with L/R instead of RC; the current grows, not voltage.
Inrush current limiting, inductor charging in power supplies, and relay coil energization.
The formula becomes I(t) = I_final + (I_initial − I_final) e^(−t/τ).
t = −(L/R) ln(1 − I/(V/R)).
Common errors include: 1) using the capacitor charging formula, 2) forgetting the initial current, 3) using the wrong time constant, 4) applying the formula to AC, and 5) ignoring the inductor's resistance.