Basic Electronics 2, Chapter 2: Challenge 2 solution (Kirchhoff's laws)

Given: V1 = 10 V feeds node A through R1 = 1 k ohm. V2 = 4 V feeds node A through
R2 = 2 k ohm. R3 = 2 k ohm connects node A to ground. All source negatives are ground.

Assume currents I1 and I2 flow INTO node A, and I3 flows OUT of A through R3.
(A negative answer just means the real current runs the other way.)

PART A: node method (KCL at A, one unknown: the voltage Va)
  I1 = (10 - Va) / 1000
  I2 = (4 - Va) / 2000
  I3 = Va / 2000
  KCL: I1 + I2 = I3
  (10 - Va)/1000 + (4 - Va)/2000 = Va/2000
  Multiply every term by 2000:
  2(10 - Va) + (4 - Va) = Va
  20 - 2Va + 4 - Va = Va
  24 = 4 Va
  Va = 6.0 V

Currents:
  I1 = (10 - 6)/1000 = 4.0 mA
  I2 = (4 - 6)/2000  = -1.0 mA   (negative: current actually flows from A back into V2)
  I3 = 6/2000        = 3.0 mA
  KCL check: 4.0 + (-1.0) = 3.0 mA. Good.

PART B: KVL check around the two loops
  Loop V1 -> R1 -> R3:  10 = I1 x 1k + I3 x 2k = 4 + 6 = 10 V. Good.
  Loop V2 -> R2 -> R3:   4 = I2 x 2k + I3 x 2k = -2 + 6 = 4 V.  Good.

PART C: power
  V1 delivers   10 x 4.0 mA  = 40 mW
  V2 "delivers" 4 x (-1.0) mA = -4 mW (it ABSORBS 4 mW: it is being charged)
  R1 = I1^2 x R1 = (0.004)^2 x 1000 = 16 mW
  R2 = (0.001)^2 x 2000          = 2 mW
  R3 = (0.003)^2 x 2000          = 18 mW
  Dissipated total = 16 + 2 + 18 = 36 mW
  Supplied total   = 40 - 4     = 36 mW. Balanced.

Meaning: V2 is the smaller source, so the bigger source V1 pushes current backwards
through it. This is why you never connect two different supplies to the same node
without resistors, and why a battery must not be charged by accident.

Checklist
 [ ] choose a reference (ground), name the node voltage
 [ ] write each branch current as (voltage difference) / R, with a consistent direction
 [ ] KCL: currents in = currents out
 [ ] solve; confirm with KVL around each loop and with a power balance