Voltage divider output, or R1 or R2 for a target voltage; the current-limiting resistor for LEDs with its nearest E12 value and power rating.
R1 = 10 kΩ over R2 = 4.7 kΩ passes 31.97% of the input: 3.837 V from 12 V, drawing 816.3 µA.
A voltage divider of two resistors gives Vout = Vin × R2/(R1 + R2). The calculator solves for the output, or for R1 or R2 when you know the output you want, and shows how a load across R2 pulls the output down. An LED needs a series resistor R = (Vs − n·Vf)/I, where Vf is the forward voltage of each of n LEDs in series and I is the current you want.
The divider default, 10 kΩ over 4.7 kΩ on 12 V, gives 3.84 V and draws 0.82 mA. A red LED (Vf ≈ 2.0 V) at 20 mA on 5 V needs exactly 150 Ω, an E12 value, and the resistor dissipates 60 mW, so a ⅛ W part covers it with a 2× margin.
Exact values are rounded up to the next E12 value (IEC 60063), which keeps the LED current at or below target. Forward voltages are typical figures; datasheets vary by a few tenths of a volt, and the current is most sensitive when little voltage is left across the resistor.
Checked against: Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA
Checked against: Python 3.8 fractions: R2 = R1·Vout/(Vin − Vout) = 33000/1.7 Ω = 19.411765 kΩ
Checked against: Python 3.8 fractions: R2‖RL = 5 kΩ, 12 × 5/15 = 4 V (6 V unloaded)
Checked against: Python 3.8 fractions: (5 − 2.0)/0.02 = 150 Ω (in E12); P = 0.02² × 150 = 0.06 W; 2 × 0.06 ≤ 0.125
For one red LED at 20 mA, 150 Ω: (5 − 2.0 V)/0.020 A. A blue or white LED with a 3.2 V forward drop needs (5 − 3.2)/0.020 = 90 Ω, rounded up to 100 Ω in the E12 series, which gives 18 mA. Many indicator LEDs are bright enough at 5–10 mA, which doubles to quadruples the resistance.
For a single red LED at 20 mA, (12 − 2.0)/0.020 = 500 Ω, rounded up to 560 Ω (E12). That gives 17.9 mA and 0.18 W in the resistor, so use a ½ W part to keep a 2× margin. Three white LEDs in series instead (3 × 3.2 V = 9.6 V) need only 120 Ω, and the resistor wastes 48 mW.
Vout = Vin × R2/(R1 + R2), where R2 is the resistor between the output and ground. 10 kΩ over 4.7 kΩ on 12 V gives 12 × 4.7/14.7 = 3.84 V. For a target output, fix one resistor and solve for the other: R2 = R1 × Vout/(Vin − Vout), so 3.3 V from 5 V with R1 = 10 kΩ needs R2 = 19.4 kΩ.
The load sits in parallel with R2 and lowers the bottom resistance. Two 10 kΩ resistors on 12 V give 6 V unloaded, but a 10 kΩ load makes the bottom 5 kΩ and the output falls to 4 V. Keep the load at least 10 times R2 to stay within about 10%, or buffer the output; dividers suit reference and sensing inputs, not powering circuits.
Work out P = I²R and choose a standard rating at least twice that. A 150 Ω resistor passing 20 mA dissipates 0.02² × 150 = 0.06 W, so a ⅛ W (0.125 W) resistor meets the 2× margin. With one red LED on 12 V the resistor dissipates 0.18 W and needs ½ W. Common through-hole ratings are ⅛, ¼, ½, 1 and 2 W.
Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “12 V with 10 kΩ over 4.7 kΩ” is checked against Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA.
Horowitz & Hill, The Art of Electronics (3rd ed.), §1.2.3 Voltage dividers and §2.1 (LED current limiting); IEC 60063:2015 — Preferred number series (E12); OpenStax University Physics Volume 2, §10.2 Resistors in series and parallel.
7 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.
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