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

Given: 12 V supply, R1 = 1 k (1000 ohm), R2 = 2.2 k (2200 ohm).

SERIES
 1. Total resistance:  R_total = 1000 + 2200 = 3200 ohm
 2. Current (same everywhere):  I = 12 / 3200 = 0.00375 A = 3.75 mA
 3. Voltage across each:
      V1 = I x R1 = 0.00375 x 1000 = 3.75 V
      V2 = I x R2 = 0.00375 x 2200 = 8.25 V
      Check: 3.75 + 8.25 = 12.00 V   (voltages add to the supply)
 4. Power in each:
      P1 = V1 x I = 3.75 x 0.00375 = 0.01406 W = 14.1 mW
      P2 = V2 x I = 8.25 x 0.00375 = 0.03094 W = 30.9 mW
      Total = 45.0 mW = 12 V x 3.75 mA   (matches the supply)

PARALLEL (both across 12 V)
 1. Branch currents (same voltage):
      I1 = 12 / 1000 = 12.0 mA
      I2 = 12 / 2200 = 5.45 mA
 2. Total current:  I = 12.0 + 5.45 = 17.45 mA
 3. Equivalent resistance:
      R_eq = (1000 x 2200) / (1000 + 2200) = 2,200,000 / 3200 = 687.5 ohm
      Check: 12 / 687.5 = 17.45 mA, which matches.
 4. Power: P1 = 12 x 0.012 = 144 mW; P2 = 12 x 0.00545 = 65.5 mW.
    The 1 k resistor is above half of a 1/4 W (250 mW) rating, so it runs warm; use a
    1/2 W part for a permanent circuit.

SANITY CHECKS
 - Series total (3200) is larger than either resistor.
 - Parallel total (687.5) is smaller than the smaller resistor (1000).
 - Parallel draws more current (17.45 mA) than series (3.75 mA) from the same supply.
 - Total power in the parts equals the power from the supply in both cases.

Simulator check: in Falstad, build each version and click the parts; the readings should
match the figures above to within rounding.