Basic Electronics 1, Chapter 1: Challenge 1 solution (size a resistor)

Given: green LED, VF = 2.1 V, maximum 20 mA, target about 10 mA.
Formulas: R = (V_supply - VF) / I      P_R = (V_supply - VF) x I

PART A: 5 V (USB) supply
 1. Voltage across the resistor:  V_R = 5.0 - 2.1 = 2.9 V
 2. Ideal resistor:  R = 2.9 / 0.010 = 290 ohm
 3. E12 neighbours: 270 and 330 ohm. Choose 330 (rounding up is the safe direction).
 4. Actual current:  I = 2.9 / 330 = 0.00879 A = 8.8 mA   (below the 20 mA limit)
 5. Power in the resistor:  P = 2.9 x 0.00879 = 0.0255 W = 25.5 mW
    A 1/4 W (250 mW) resistor is used at about 10 % of its rating: fine.
 (270 ohm would also be acceptable: I = 10.7 mA, P = 31 mW.)

PART B: 4.5 V (3 x AA) pack
 1. V_R = 4.5 - 2.1 = 2.4 V
 2. Ideal resistor:  R = 2.4 / 0.010 = 240 ohm
 3. E12 neighbours: 220 and 270 ohm.
 4. Is 220 ohm acceptable?  I = 2.4 / 220 = 0.0109 A = 10.9 mA
    That is well below the 20 mA maximum, so YES, 220 ohm is acceptable.
    P_R = 2.4 x 0.0109 = 26 mW: no heat problem.
 5. 270 ohm gives 8.9 mA (slightly dimmer): also fine.

Note: a fresh 3 x AA pack reads about 4.8 V and falls as it ages, so current changes a little.

Checklist for any LED resistor:
 [ ] V_R = V_supply - VF        [ ] R = V_R / I_target
 [ ] round to an E12 value      [ ] recompute actual I (must be below the LED maximum)
 [ ] check P_R is below half the resistor rating