Basic Electronics 2, Chapter 2: Challenge 3 solution (build, measure, meter loading)

PART A: expected readings for the series-parallel circuit (9 V; R1 = 1 k; R2 = 2.2 k || R3 = 3.3 k)
  R23 = (2200 x 3300)/(2200 + 3300) = 7,260,000 / 5,500 = 1,320 ohm
  R_total = 1000 + 1320 = 2,320 ohm
  I = 9 / 2320 = 3.879 mA
  V(R1) = 3.879 V,  V(X to ground) = 9 - 3.879 = 5.121 V
  I(R2) = 5.121 / 2200 = 2.328 mA,  I(R3) = 5.121 / 3300 = 1.552 mA  (sum 3.880 mA)

  Meter checklist (DC V, probes across the part; DC mA, meter in the gap):
   Battery           9.0 V   (real fresh battery: 9.3 to 9.6 V, everything scales up about 4 %)
   Across R1         3.88 V
   Node X to ground  5.12 V
   Total current     3.88 mA (meter in series with R1)
   Current in R2     2.33 mA (open the R2 branch at column 10, meter in the gap)
   Current in R3     1.55 mA
  Resistance with the battery DISCONNECTED, probes at the + rail and the - rail: about 2.32 k
  (this works only because nothing else is on the board; see the in-circuit warning in section 3).

  What to conclude: R1's voltage plus node X's voltage equals the supply (KVL); the two branch
  currents add up to the total (KCL). Real readings differ by 2 to 5 %: resistor tolerance,
  battery voltage, meter accuracy.

PART B: meter loading on a high-resistance divider
  Circuit: 9 V across two 470 k resistors in series. Ideal midpoint = 4.5 V.
  A voltmeter adds its own input resistance in parallel with the lower resistor.

  Meter input 10 M ohm:
   470k || 10M = (470 x 10,000)/(470 + 10,000) = 4,700,000 / 10,470 = 448.9 k
   Vmid = 9 x 448.9 / (470 + 448.9) = 9 x 0.4885 = 4.397 V
   Error = (4.397 - 4.5) / 4.5 = -2.3 %

  Meter input 1 M ohm (a cheap or analogue meter):
   470k || 1M = (470 x 1000)/(470 + 1000) = 470,000 / 1,470 = 319.7 k
   Vmid = 9 x 319.7 / (470 + 319.7) = 9 x 0.4048 = 3.644 V
   Error = (3.644 - 4.5) / 4.5 = -19 %

  Rule of thumb: the meter resistance should be at least 10 to 100 times larger than the
  resistance of the circuit at the point you measure. With 1 k resistors instead of 470 k
  the same 10 M meter reads 4.4998 V: no visible error.

PART C: ammeter burden (optional extension)
  A meter on its mA range may have about 10 ohm of internal resistance. In a 5 V, 100 ohm
  circuit the true current is 5 / 100 = 50 mA. With the meter in series:
   I = 5 / (100 + 10) = 45.5 mA, error -9 %.
  In the 9 V, 2.32 k circuit above the same 10 ohm changes the current by only 0.4 %: harmless.

Wiring checklist (breadboard)
 [ ] battery clip disconnected while wiring
 [ ] R1: column 5 to column 10; R2 and R3: column 10 to column 20 (rows b and c)
 [ ] jumper from + rail to column 5; jumper from column 20 to - rail
 [ ] connect the battery last; measure voltages across parts, currents in series