Everything so far has analysed a circuit somebody else drew. This lesson designs one, and the answer is not a number — it is a component you could order.

An LED is a diode, not a resistor. Above its threshold it holds a nearly fixed forward voltage $V_f$, in volts — around 1.8 V for a red one, past 3 V for blue and white — almost regardless of how much current you push through it. Which means it cannot limit its own current: connect one straight across a supply and the current is set by nothing but the wiring, and the part lasts about as long as it takes to say so. Every LED needs ballast, and the cheapest ballast is one resistor in series.

Size it with KVL and Ohm's law, in that order. The loop rule says the resistor gets whatever volts the LED does not take, $V_s - V_f$; Ohm's law turns that into ohms at the current you want. $R = \dfrac{V_s - V_f}{I}$ — read aloud R equals V-s minus V-f, over I. Here $V_s$ is the supply in volts, $V_f$ is the LED's forward voltage in volts from its datasheet, $I$ is the design current in amperes — and datasheets quote it in milliamps, so 20 mA goes into the formula as 0.02 A. Forget that conversion and your answer is a thousand times too small.

Then the part. Resistors are not stocked at every value: the E12 series holds twelve per decade — 100, 120, 150, 180, 220, 270, 330, 390, 470, 560, 680, 820 — and your computed figure almost never lands on one. The rule is always round UP. More resistance means slightly less current, which means a slightly dimmer LED that outlives the equipment. Round down and you are over the datasheet current, running the part hot, for the sake of a difference nobody can see. Rounding to the nearest value is a habit from arithmetic class; in design it is a defect.