LED Current Limiter - Electricity and Electronic Circuits Step by Step

This LED-current-limiter lesson shows the resistor taking the voltage left after the ideal LED forward drop, then setting branch current by Ohm's law. The 12 V source, 2 V drop, 10 V resistor budget, and 5 ohm limiter give 2 A only for the stated schematic training values. Real LED circuits need rated currents, nonlinear forward-voltage curves, temperature effects, resistor tolerances, supply limits, finite precision, and measurement uncertainty.

A fixed LED drop leaves the rest of the source voltage for the current-limiting resistor. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

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The resistor gets the remaining voltage

The source voltage minus the LED forward drop leaves 10 volts across the limiter.

V_R = 12\ \text{V} - 2\ \text{V} = 10\ \text{V}

Ohm's law gives the branch current

The limiter is 5 ohm, so 10 volts gives 2 amperes.

I = \frac{V_R}{R} = \frac{10\ \text{V}}{5\ \text{ohm}} = 2\ \text{A}

The resistor limits the LED current

In this ideal LED circuit, the resistor is the element that turns the remaining voltage into a checked current. These values are schematic training values, not component recommendations.

I_{\text{LED}} = 2\ \text{A}

More resistance means less current

Keep the source and LED drop fixed. The remaining voltage is the same, so the resistor controls the branch current.

\begin{array}{c|c|c|c|c}V_s&V_f&R&V_R&I\\12\ \text{V}&2\ \text{V}&10\ \text{ohm}&10\ \text{V}&1\ \text{A}\\12\ \text{V}&2\ \text{V}&5\ \text{ohm}&10\ \text{V}&2\ \text{A}\\12\ \text{V}&2\ \text{V}&2\ \text{ohm}&10\ \text{V}&5\ \text{A}\\\end{array}