Basic Electronics 1, Chapter 1: Challenge 3 solution (build it and measure it)

WIRING CHECKLIST (breadboard)
 [ ] Battery disconnected while wiring
 [ ] LED long leg (anode) on the resistor side, short leg (cathode, flat edge) to the - rail
 [ ] R1 = 820 ohm (grey-red-brown), one end to the + rail, other end in the LED anode's column
 [ ] Battery red lead to + rail, black lead to - rail, connected last
 [ ] LED lights; if not: swap the LED, check rails and fresh battery

EXPECTED READINGS (9 V battery, 820 ohm, red LED)
   Battery terminals      9.0 V calculated; 9.3 to 9.6 V real (fresh); the set value in a simulator
   Across R1              7.0 V calculated; typically 7.2 to 7.6 V real
   Across LED             2.0 V calculated; 1.8 to 2.1 V real
   Loop current           8.5 mA calculated (7.0 V / 820 ohm); about 8 to 9 mA real
 Check: V(R1) + V(LED) should equal the battery voltage within meter accuracy.
 Check: V(R1) / 820 should equal the measured current. Example: 7.4 V / 820 = 9.0 mA.
 To read current, break the loop and insert the meter in series (mA socket). Never across the battery.

SWAP TO 220 OHM (predict first)
 1. V_R = 9.0 - 2.0 = 7.0 V (the LED drop barely changes)
 2. I = 7.0 / 220 = 0.0318 A = 31.8 mA
 3. Power in R1:  P = 7.0 x 0.0318 = 0.222 W = 222 mW
 4. Verdict: NOT safe.
    - 31.8 mA is above the 20 mA rated maximum of a typical 5 mm LED (and at or above the
      absolute maximum of 20 to 30 mA); it will be very bright and shorten its life or burn out.
    - 222 mW is about 89 % of a 250 mW resistor's rating, so it runs hot (aim for 50 % or less).
    - With a fresh battery at 9.4 V the figures are slightly worse (I about 33.6 mA).
 5. Lesson: a smaller series resistor means more current. Run the 220 ohm case briefly in the
    simulator only (Falstad will show the LED current exceeding its limit), not on the
    breadboard for any length of time.