Three Loops — Kirchhoff on a Star of Resistors
Two 80 V lead-acid batteries stand in line along the left-hand side of a circuit board, both pushing the same way. From them run six 150 W power resistors: 20 Ω along the top, 20 Ω along the bottom, 20 Ω across the middle to a star point O, 30 Ω in each arm of the star and 30 Ω down the right-hand side. Ammeters measure the top, right and bottom currents, and a multimeter is clipped across the middle resistor. Press play and a test charge walks the three loops in turn while a fence above the wires rises and falls with the potential. Four sliders set the two batteries, the middle resistor and the right-hand resistor.
Bu simülasyon nasıl kullanılır
- (V_B)₁, the top battery — 0 to 100 V, starting at 80 V.
- (V_B)₂, the bottom battery — 0 to 100 V, starting at 80 V.
- R_eO, the middle resistor — 5 to 40 Ω, starting at 20 Ω.
- R_bc, the right-hand resistor — 10 to 60 Ω, starting at 30 Ω. The other four resistors keep their marked values; every change re-solves the circuit and keeps the timeline where it is.
- Play/pause, back to the start, a timeline you can drag to any moment, and ½×, ¼×, ⅛× slow motion.
- Drag anywhere on the scene to look around.
Nelere dikkat etmeli
- At the start values the ammeters read 2.667 A, 1.778 A and 2.667 A, the fractions 8/3, 16/9 and 8/3; the middle branch shows no arrow, no moving charge and 0.000 V on the multimeter.
- On the walk round loop 1 the fence stays perfectly level across the middle resistor: e and O are at the same potential, 80 V above d.
- Keep the two batteries equal and move R_eO or R_bc anywhere on their sliders: the middle current stays at zero.
- Turn the top battery down to 0 V: 8/9 A now flows from e to O and the multimeter reads −17.78 V. Make the top battery the stronger one and that current reverses.
- Loop 3 has no battery: the two rises of 26.67 V in the star arms are paid back exactly by the 53.33 V drop across R_bc.
- Power: 80 × 8/3 + 80 × 8/3 = 426.7 W supplied, and exactly the same turned into heat in the six resistors.
Arkasındaki fizik
Only three currents need names: I₁ along the top (a to b), I₂ down the right (b to c) and I₃ along the bottom (c to d). The junction rule then gives every other branch: I₃ − I₁ through the middle from e to O, I₁ − I₂ down the upper star arm from b to O, and I₂ − I₃ along the lower arm from c to O. One equation per loop, walking clockwise: 80 = 20(I₃ − I₁) − 30(I₂ − I₃) + 20 I₃ for the lower-left loop, 80 = 20 I₁ + 30(I₁ − I₂) − 20(I₃ − I₁) for the upper loop, and 0 = −30(I₁ − I₂) + 30 I₂ + 30(I₂ − I₃) for the star triangle, which contains no battery. The solution is I₁ = 8/3 A, I₂ = 16/9 A and I₃ = 8/3 A, so the middle current I₃ − I₁ is zero. The reason is symmetry: with equal batteries the circuit is its own mirror image top to bottom, so the star point O sits at exactly the potential of e (80 V above d) and nothing pushes charge through the middle resistor, whatever its value. In the lower arm I₂ − I₃ = −8/9 A: the minus sign only means that this current really flows from O to c. The energy balances: 80 × 8/3 + 80 × 8/3 = 1280/3 ≈ 426.7 W supplied and dissipated.