Basic Electronics 2, Chapter 2: Challenge 1 solution (series-parallel network)

Given: 12 V supply, R1 = 470 ohm in series with (R2 = 680 ohm in parallel with R3 = 1.2 k ohm).

Step 1: reduce the parallel pair (product over sum)
  R23 = (680 x 1200) / (680 + 1200) = 816,000 / 1,880 = 434.0 ohm
  Sanity check: 434 is below the smaller branch (680). Good.

Step 2: total resistance (series)
  R_total = 470 + 434.0 = 904.0 ohm

Step 3: total current (Ohm's law)
  I = 12 / 904.0 = 0.013274 A = 13.27 mA

Step 4: voltage across each part
  V1  = I x R1 = 0.013274 x 470 = 6.239 V
  V23 = I x R23 = 0.013274 x 434.0 = 5.761 V   (or 12 - 6.239 = 5.761 V: matches)

Step 5: branch currents (same voltage, 5.761 V, across both)
  I2 = 5.761 / 680  = 0.008472 A = 8.47 mA
  I3 = 5.761 / 1200 = 0.004801 A = 4.80 mA
  Check (KCL at the junction): 8.47 + 4.80 = 13.27 mA = I. Good.
  Current-divider check: I2 = I x R3/(R2+R3) = 13.27 x 1200/1880 = 8.47 mA.

Step 6: power
  P1 = V1 x I  = 6.239 x 0.013274 = 0.0828 W = 82.8 mW
  P2 = V23 x I2 = 5.761 x 0.008472 = 0.0488 W = 48.8 mW
  P3 = V23 x I3 = 5.761 x 0.004801 = 0.0277 W = 27.7 mW
  Total = 82.8 + 48.8 + 27.7 = 159.3 mW
  Supply: 12 x 0.013274 = 0.1593 W = 159.3 mW. Energy balance holds.
  All three resistors are far below a 250 mW rating (R1 is the hottest, about one third).

Summary table
  Part   Voltage    Current    Power
  R1     6.24 V     13.27 mA   82.8 mW
  R2     5.76 V      8.47 mA   48.8 mW
  R3     5.76 V      4.80 mA   27.7 mW

Checklist
 [ ] reduce parallel first, then add the series parts
 [ ] total current from V / R_total
 [ ] voltages from I x R, and they must add to the supply
 [ ] branch currents must add to the total current
 [ ] powers must add to the supply power