The Formula for Current in an Inductor: Derivations, Tables, and Worked Examples

The Core Formula for Current in an Inductor

The fundamental relationship governing an inductor is that it opposes changes in current by inducing a voltage. When we flip this relationship to solve for current, we get the time-domain integral formula. The instantaneous current i(t) through an inductor is the integral of the voltage across it over time, scaled by its inductance, plus any initial current present at the start of the observation window.

The primary time-domain formula is:

i(t) = (1 / L) ∫[t₀ to t] v(τ) dτ + i(t₀)

For the highly common practical case of a DC step voltage applied to a resistor-inductor (RL) series circuit, the integral resolves into the exponential step-response formula:

i(t) = (V_s / R) * (1 - e^(-(R/L)*t))

Symbol Definition Table

Symbol	Parameter	Standard SI Unit	Practical Unit Variants
i(t)	Instantaneous current at time t	Amperes (A)	mA, µA
L	Inductance	Henries (H)	mH, µH, nH
v(τ)	Instantaneous voltage across the inductor	Volts (V)	mV
t, t₀, τ	Time, initial time, and integration dummy variable	Seconds (s)	ms, µs
i(t₀)	Initial current at time t₀	Amperes (A)	mA
V_s	DC source voltage (for RL step response)	Volts (V)	-
R	Series resistance (including inductor DCR)	Ohms (Ω)	mΩ

Real-World Inductor Parameters & Magnitudes

A common mistake on the bench is calculating a current that is physically impossible for the chosen component. Before running the math, you must know the realistic magnitude of your answer. A 10 nH RF choke will never carry 5 amps, and a 10 mH toroidal line filter will saturate instantly if you push 40 amps through it. Below is a data-dense reference table of real-world inductor classes to calibrate your expectations.

Component Class	Typical Inductance (L)	Max RMS / Saturation Current	Typical DCR	Realistic Answer Magnitude
0603 SMD RF Choke	10 nH - 100 nH	200 mA - 500 mA	0.1 Ω - 0.5 Ω	Microamps to low milliamps
SMD Shielded Power Inductor	1 µH - 10 µH	2 A - 8 A	10 mΩ - 30 mΩ	1 A to 5 A (Buck/Boost converters)
Toroidal Line Filter (Mains)	5 mH - 20 mH	1 A - 3 A	100 mΩ - 250 mΩ	< 2 A (AC mains filtering)
High-Current Buck Choke (Through-hole)	50 µH - 200 µH	20 A - 50 A	1 mΩ - 3 mΩ	10 A to 40 A (Motor drives, welders)

Source reference for magnetic component characteristics: Georgia State University HyperPhysics: Inductors.

Rearranged Forms & Unit Traps

You will rarely use the integral form exactly as written unless you are simulating a circuit in SPICE. On the bench, you usually need to solve for a missing component value or a specific timing threshold. Here are the algebraically rearranged forms derived from the RL step response and the fundamental derivative relationship v(t) = L * (di/dt).

Rearranged Forms List

• Solving for Inductance (L): L = (V_s * t) / (-R * ln(1 - (i(t)*R / V_s))) — Use this when designing a buck converter and you know your target ripple current and switching time.
• Solving for Time (t): t = -(L / R) * ln(1 - (i(t) * R / V_s)) — Use this to calculate how long it takes for a relay coil (which is essentially an inductor) to reach its pull-in current threshold.
• Solving for Required Voltage (V_s): V_s = (i(t) * R) / (1 - e^(-(R/L)*t)) — Use this to size the drive voltage for an injector solenoid or actuator to guarantee it hits a specific current within a strict time window.
• Solving for Peak Voltage (Derivative form): V_peak = L * (Δi / Δt) — Use this to calculate the flyback voltage spike when a switch opens, forcing Δt to approach zero.

Unit Mistakes That Break the Math

Worked Examples with Unit Tracking

Let's run two practical scenarios. We will track the units through every intermediate step to prove the math holds up.

Example 1: Constant Voltage Integration (Ideal Inductor)

Scenario: An ideal 10 mH inductor has an initial current of 0 A. A constant 5 V DC is applied across it for 2 ms. What is the final current?

Formula: i(t) = (1 / L) ∫ v(t) dt + i(0)

1. Convert to SI units:
   L = 10 mH = 0.01 H
t = 2 ms = 0.002 s
v(t) = 5 V (constant)
2. Evaluate the integral:
   Since voltage is constant, the integral of v(t) from 0 to 0.002 is simply V * Δt.
   ∫ 5 dt = 5 V * 0.002 s = 0.01 V·s (Volt-seconds, which is equivalent to Webers, the unit of magnetic flux).
3. Divide by Inductance:
   (1 / 0.01 H) * 0.01 V·s = 1 A.
   Unit check: Volt-seconds divided by Henrys equals Amperes. (Since H = V·s / A, the units cancel perfectly).
4. Add initial current:
   1 A + 0 A = 1 A.

Final Answer: The current ramps linearly to 1 A in 2 ms.

Example 2: RL Transient Step Response (Real-World Circuit)

Scenario: A 12V DC source is connected via a switch to a series circuit containing a 4 Ω power resistor and a 50 mH inductor. The switch closes at t=0. Find the current at t=5 ms.

Formula: i(t) = (V_s / R) * (1 - e^(-(R/L)*t))

1. Convert to SI units:
   V_s = 12 V
R = 4 Ω
L = 50 mH = 0.050 H
t = 5 ms = 0.005 s
2. Calculate Final Steady-State Current (V_s / R):
   12 V / 4 Ω = 3 A. (This is the maximum current if we wait forever).
3. Calculate the Time Constant (τ = L / R):
   τ = 0.050 H / 4 Ω = 0.0125 s (or 12.5 ms).
   Unit check: Henrys divided by Ohms equals seconds.
4. Calculate the Exponent (-t / τ):
   -0.005 s / 0.0125 s = -0.4. (Dimensionless, as required).
5. Evaluate the Exponential Term:
   e^(-0.4) ≈ 0.6703.
6. Compute Final Current:
   i(5ms) = 3 A * (1 - 0.6703) = 3 A * 0.3297 = 0.9891 A.

Final Answer: At 5 ms, the current is approximately 0.99 A. (Note: 5 ms is less than half of one time constant (12.5 ms), so it makes physical sense that the current hasn't even reached 1/3 of its 3A maximum yet).

When This Formula Applies (and When It Fails)

The formulas above are mathematically rigorous, but they rely on physical assumptions that break down on a real workbench if you aren't paying attention.

Core Assumptions

• Linearity: The formulas assume L is a constant value. This is strictly true only for air-core inductors or magnetic-core inductors operating well below their saturation limit.
• Ideal Components: The basic integral form ignores parasitic elements. In reality, every inductor has DC resistance (DCR), parallel parasitic capacitance (which creates a self-resonant frequency), and core losses.

The Core Saturation Failure Mode

If you apply the formula to a ferrite-core inductor and calculate a current that exceeds its saturation rating (I_sat), the math is wrong. When a magnetic core saturates, the permeability of the core drops toward that of air. The inductance L violently collapses—sometimes dropping to 10% of its nominal value.

Because di/dt = V/L, a sudden drop in L causes the rate of current change to spike. In a switching power supply, this manifests as a massive, sharp spike in current at the end of the switching cycle, which will invariably blow the drain of your MOSFET or trigger overcurrent protection. Always check the datasheet for I_sat (saturation current) alongside I_rms (thermal current limit). For deeper analysis on magnetic saturation limits in power design, refer to MIT OpenCourseWare's Circuits and Electronics materials.

High-Frequency AC Limitations

If you are applying an AC waveform at high frequencies (e.g., RF circuits above 1 MHz), the simple time-domain formulas must be supplemented with AC impedance calculations. The skin effect forces current to the outer edge of the wire, drastically increasing the effective AC resistance beyond the datasheet's DCR value. At these frequencies, you must transition from time-domain integrals to phasor-domain complex impedance: Z_L = jωL, and use RMS magnitudes rather than instantaneous time-domain integration.