Inductor Voltage Calculator: Formula and Examples

An inductor voltage calculator finds the voltage across an inductor from its inductance and the rate at which current changes. Enter inductance in henries and current change in amperes per second to get voltage in volts; the result is signed, so it also indicates polarity relative to the chosen reference direction.

Formula

For an ideal inductor with constant inductance, the voltage magnitude is calculated using:

v = L × (Δi / Δt)

• v is the voltage across the inductor, in volts (V).
• L is inductance, in henries (H).
• Δi is the change in current, in amperes (A).
• Δt is the time interval over which that current change occurs, in seconds (s).

The calculator uses the signed current change: final current minus initial current. A negative result means the voltage has the opposite polarity from the reference direction implied by the entered current change and passive sign convention. The time interval must be positive. Inductance may be zero in the input, which gives zero voltage in this idealized calculation, although a zero-inductance component is not a practical inductor model.

Worked Example

Using the calculator’s default values, inductance is 0.01 H, current changes by 2 A, and the interval is 0.001 s.

1. Find the rate of change: Δi / Δt = 2 A / 0.001 s = 2,000 A/s.
2. Multiply by inductance: v = 0.01 H × 2,000 A/s = 20 V.
3. Convert inductance for the helper result: 0.01 H × 1,000 = 10 mH.

The calculated inductor voltage is 20.00 V, the inductance is 10.00 mH, and the current rate is 2,000.00 A/s. The sign is positive because the entered current change is positive.

What the Result Means

An inductor stores energy in a magnetic field when current flows through it. Its voltage is related to how quickly that current changes, not simply to the current’s size. A steady current has a rate of change of zero, so an ideal inductor has zero voltage across it after the current has settled. In a real circuit, resistance and other effects can still produce voltage drops.

A rapid current change can create a large voltage even when the inductance is modest. For instance, the example’s 2 A change occurs in just 1 millisecond, producing a rate of 2,000 A/s. If the same change took 10 milliseconds instead, the ideal voltage would be 2 V. This is why switching events and abrupt interruptions can produce voltage spikes.

The equation is based on the passive sign convention: current is considered to enter the terminal marked positive for voltage. With that reference, the constitutive relation is v = L di/dt. Lenz’s law describes the physical response: the induced voltage opposes the change that produces it. “Opposes” does not mean the numerical result must always be negative; the sign depends on how current direction and voltage polarity were defined.

When to Use the Calculator

Use this calculation when you know an inductor’s inductance and a current change over a measured or specified interval. It is useful for introductory circuit analysis, estimating an ideal voltage during a current ramp, or checking whether a stated change and voltage are consistent. It calculates the average voltage over the entered interval if the current change is summarized by its endpoints.

• Use henries for inductance. For conversions, 1 H = 1,000 mH = 1,000,000 μH.
• Use amperes for current change. A change from 0.5 A to 1.5 A is +1 A; a change from 1.5 A to 0.5 A is −1 A.
• Use seconds for time. Convert milliseconds by dividing by 1,000 and microseconds by dividing by 1,000,000.
• Keep the direction of current change if voltage polarity matters. Use its magnitude only when you need voltage magnitude.

For example, 10 mH is 0.01 H, and a 2 A increase over 1 ms gives 20 V. Mixing milliseconds directly with seconds would make the result wrong by a factor of 1,000. The calculator accepts seconds only, so convert before entering a different time unit.

Limits and Common Mistakes

The simple formula assumes constant inductance and a current change that can be represented by a single rate. If current varies nonlinearly, v = L di/dt still applies instantaneously for a constant ideal inductance, but an endpoint calculation gives the average rate across the interval, not every moment’s voltage. For a changing inductance, the more general flux relation is v = dλ/dt, where λ is flux linkage; under a simple linear model this can include a term associated with changing L.

• Confusing current with current change: the formula needs Δi, not the final current alone. If current rises from 0 A to 2 A, Δi is 2 A.
• Using the wrong time unit: convert milliseconds or microseconds to seconds before calculating.
• Dropping the sign: a negative Δi produces a negative voltage in the chosen reference convention. It does not mean the physical voltage has no magnitude.
• Treating an estimate as a component rating: real circuits include winding resistance, parasitic capacitance, magnetic-core effects, and switching behavior. This calculation does not establish a safe operating voltage or component rating.

For an ideal inductor, the stored magnetic energy is E = ½Li², measured in joules when L is in henries and current in amperes. This is a separate calculation from voltage: energy depends on current magnitude, while voltage depends on the current’s rate of change. A voltage estimate alone therefore does not tell you how much energy is stored.

Frequently Asked Questions

How do you calculate voltage across an inductor?
Multiply inductance in henries by the current change in amperes divided by the elapsed time in seconds: v = LΔi/Δt. The result is in volts.

What happens to inductor voltage when current is constant?
For an ideal inductor, constant current means di/dt is zero, so its voltage is zero. Real circuit resistance and other non-ideal effects may still create a voltage drop.

Can inductor voltage be negative?
Yes. A negative result indicates polarity opposite to the chosen reference direction. The magnitude is the absolute value of that result.

Does a faster current change create more voltage?
Yes, for the same inductance and current change, reducing the time interval increases Δi/Δt and therefore increases the calculated voltage.

Should I enter current or change in current?
Enter the difference between final and initial current, including its sign. For example, a rise from 1 A to 3 A is +2 A; a fall from 3 A to 1 A is −2 A.

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in l | Inductance | H | 0.01 | 0
in di | Current change | A | 2 | -
in dt | Time interval | s | 0.001 | 0
out v | Inductor voltage | V | l * di / dt | 2
out vl | Inductance | mH | l * 1000 | 2
out rate | Current rate of change | A/s | di / dt | 2