The fundamental relationship governing an inductor is that it opposes changes in current by inducing a voltage. While the derivative form (v = L di/dt) is commonly used to find voltage, finding the current through an inductor requires the integral form of the equation. The direct equation for instantaneous current is i(t) = (1/L) ∫ v(τ) dτ + i(t₀). This formula dictates that the current at any given moment is the accumulated integral of the voltage applied over time, scaled by the inverse of the inductance, plus whatever initial current was already flowing.
Below, we break down the derivation, define every variable, rearrange the formula for different bench scenarios, and walk through worked problems with strict unit tracking to ensure your calculations translate to real-world circuits.
The Core Current Through Inductor Equation
To calculate the current i(t) flowing through an inductor at a specific time t, we use the integral form of Faraday's law applied to inductance. This equation is mandatory when dealing with time-varying voltages, switching power supplies, or transient DC step responses.
i(t) = (1/L) ∫t₀t v(τ) dτ + i(t₀)
| Symbol | Parameter | SI Unit | Practical Bench Notes |
|---|---|---|---|
| i(t) | Instantaneous current at time t | Amperes (A) | The target variable. Must be measured with a current probe or shunt resistor on an oscilloscope for transient verification. |
| L | Inductance | Henries (H) | Assumed constant in this ideal equation. In reality, L drops as the core approaches magnetic saturation. |
| v(τ) | Instantaneous voltage across the inductor | Volts (V) | The dummy variable τ (tau) represents time during the integration window. |
| t₀ | Initial time | Seconds (s) | Usually set to 0 in switching transient problems. |
| t | Current evaluation time | Seconds (s) | Must strictly be in seconds to satisfy the Henry unit cancellation. |
| i(t₀) | Initial current at time t₀ | Amperes (A) | Crucial for continuous conduction mode (CCM) power supplies where current never hits zero. |
Rearranged Forms for Circuit Analysis
On the bench, you rarely have all variables neatly isolated. Depending on whether you are selecting a component, measuring a transient, or debugging a waveform, you will need to rearrange the current through inductor equation. Here are the algebraically isolated forms:
- Solving for Inductance (L):
L = [ ∫t₀t v(τ) dτ ] / [ i(t) - i(t₀) ]
Use case: You know the voltage profile and the desired current ripple (Δi), and need to specify an inductor part number for a buck converter. - Solving for Voltage (v(t)) - The Derivative Form:
v(t) = L * (di(t) / dt)
Use case: You have a measured current waveform on your scope and need to calculate the voltage spike (flyback) generated when the switch opens. - Solving for Initial Current (i(t₀)):
i(t₀) = i(t) - (1/L) ∫t₀t v(τ) dτ
Use case: Debugging a fault condition where you know the final current and the applied voltage, but need to determine what the baseline current was before the fault. - Solving for Time (t) under Constant DC Voltage:
t = t₀ + [ L * (i(t) - i(t₀)) ] / V
Use case: Calculating the exact on-time required for a switch to ramp an inductor's current to a specific trip threshold.
Worked Examples with Unit Tracking
Abstract math fails on the workbench if you drop a milli- prefix. The following examples explicitly track units through every step to demonstrate how the Henry (H) resolves into Amperes (A).
Problem 1: DC Step Response (Constant Voltage)
Scenario: A 50 mH inductor has an initial current of 0 A. A constant 12 V DC source is switched across it at t = 0. Calculate the current at t = 5 ms.
- Identify and convert variables to base SI units:
L = 50 × 10⁻³ H
v(τ) = 12 V (constant)
t₀ = 0 s
t = 5 × 10⁻³ s
i(t₀) = 0 A - Substitute into the integral equation:
i(t) = (1 / 50×10⁻³ H) ∫05×10⁻³ (12 V) dτ + 0 A - Evaluate the integral:
The integral of a constant V over time t is simply V × t.
∫ (12 V) dτ = 12 V × (5×10⁻³ s - 0 s) = 0.06 V·s - Divide by Inductance and track units:
i(t) = (0.06 V·s) / (0.05 H)
Unit check: By definition, 1 Henry = 1 Volt-second per Ampere (H = V·s/A). Therefore, (V·s) / (V·s/A) = A.
i(t) = 1.2 A
Answer: The current through the inductor at 5 ms is exactly 1.2 Amperes.
Problem 2: Time-Varying Linear Voltage Ramp
Scenario: A 100 mH inductor is subjected to a linearly increasing voltage defined by v(t) = 0.5t Volts (where t is in seconds). The initial current i(0) is 0 A. Find the current at t = 100 ms.
- Identify and convert variables:
L = 100 × 10⁻³ H = 0.1 H
v(τ) = 0.5τ V
t₀ = 0 s
t = 100 × 10⁻³ s = 0.1 s
i(t₀) = 0 A - Set up the integral:
i(0.1) = (1 / 0.1 H) ∫00.1 (0.5τ) dτ + 0 A - Perform the calculus integration:
The integral of 0.5τ with respect to τ is 0.25τ².
Evaluate from 0 to 0.1: [0.25(0.1)²] - [0.25(0)²] = 0.25(0.01) = 0.0025 V·s - Final calculation:
i(0.1) = (1 / 0.1 H) × 0.0025 V·s
i(0.1) = 10 H⁻¹ × 0.0025 V·s = 0.025 A
Answer: The current at 100 ms is 25 milliamperes (mA).
Assumptions, Limitations, and Unit Traps
The integral equation for inductor current is mathematically pure, but physical components are not. Understanding when this formula breaks down is what separates textbook theory from working bench diagnostics.
When the Formula Applies (and its Assumptions)
This equation assumes an ideal inductor. It assumes the component has zero DC resistance (DCR), zero parasitic parallel capacitance, and a perfectly linear magnetic core. In low-frequency or slow-switching DC circuits, these assumptions hold well enough for hand calculations. However, in high-frequency switching converters (like a 500 kHz buck regulator), parasitic capacitance causes ringing, and DCR causes an exponential decay curve that the pure integral equation ignores. For high-precision work, you must model the inductor as an L-R-C network.
Realistic Answer Magnitudes
If your calculation yields an absurd number, trust your physics intuition over your math.
- Signal/RF Inductors (nH to μH): Typically handle 10 mA to 500 mA. If your math says 5 A, you have the wrong component.
- Power Inductors (μH to low mH): Designed for switching supplies, handling 1 A to 30 A continuously.
- Chokes/Filter Inductors (high mH to H): Used in audio or line-filtering, often handling 50 mA to 2 A.
Unit Mistakes That Break the Math
The most common failure mode in student and junior engineering calculations is the prefix mismatch. The unit cancellation (V·s / H = A) only works if time is strictly in base seconds. If you integrate voltage over milliseconds but forget to convert the time bounds to seconds (e.g., using '5' instead of '0.005'), your calculated current will be off by a factor of 1,000. Always convert μH, mH, ms, and μs into base H and s before plugging them into the integral.
For deeper reading on inductor behavior and magnetic saturation limits, refer to the All About Circuits guide on inductor saturation and the foundational calculus concepts outlined by Georgia State University's HyperPhysics.
Frequently Asked Questions
Why does the current through an inductor equation use an integral instead of a simple multiplication like Ohm's Law?
Ohm's Law (V = IR) describes a static, instantaneous relationship where voltage and current scale linearly at any exact moment. An inductor, however, stores energy in a magnetic field. The current is not determined by the instantaneous voltage, but by the history of the voltage applied over time. The integral is simply the mathematical tool for summing up that continuous accumulation of volt-seconds. You cannot know the current without knowing how long the voltage has been pushing it.
How do I calculate the current through an inductor in an AC circuit using this equation?
For steady-state sinusoidal AC, you technically can use the integral equation by integrating v(t) = V_peak * sin(ωt), which yields a negative cosine function (showing the 90-degree phase lag). However, in practice, electrical engineers abandon the time-domain integral for AC analysis and use phasor domain impedance: I = V / (jωL). This algebraic approach is vastly faster for AC circuits, reserving the integral equation primarily for DC transients, switching waveforms, and non-sinusoidal pulses.
What happens to the current equation when the inductor core saturates?
When the magnetic core saturates, it cannot support any additional magnetic flux. Physically, the permeability of the core drops toward that of air, causing the inductance value (L) to plummet rapidly. Because L is in the denominator of the current equation (i = 1/L ∫ v dt), a collapsing L causes the calculated current to spike exponentially. On an oscilloscope, this looks like a current waveform that starts as a neat linear ramp and suddenly curves sharply upward. At this point, the ideal equation fails, and the current is limited only by the winding's DC resistance (DCR) and the source's internal resistance.
