Whatever Flows In Must Flow Out — the Junction Rule
Five wires meet at a brass binding post, c, on a lab circuit board. Four of them carry currents you set with sliders — 4 A in from the left, 5 A in from the right, 2 A in from the upper left and 8 A out towards the back — and an LED ammeter on each wire shows its reading. The fifth wire's current is the unknown: choose the direction you guess, and its meter turns so that the arrow printed on it shows your guess. Press play to watch four seconds of steady flow, with every glowing bead standing for 1 C of charge, while two counters add up the charge crossing a dashed ring round the junction.
Cara memakai simulasi ini
- From the left, into c — I₁, 0 to 10 A in steps of 0.5 A, starting at 4 A.
- From the right, into c — I₂, 0 to 10 A, starting at 5 A.
- Upper left, into c — I₃, 0 to 10 A, starting at 2 A.
- Upwards, out of c — I₄, 0 to 20 A, starting at 8 A.
- I assumed: into c / out of c — a two-way switch for the direction you guess for the unknown current. The fifth meter turns round so that its printed arrow shows the guess, and it reads negative when the guess is wrong. A change 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.
Yang perlu diamati
- With the starting values the fifth meter reads −3.000 A while the guess is "into c" and 3.000 A when it is "out of c": the size never changes, only the sign. The beads and the solid arrow always show the real direction — out of c.
- During the 4 s the charge that came in and the charge that went out grow together, 11 C every second, and the charge piled up at c stays at 0.0 C.
- Count the beads: each one is 1 C, so 44 beads cross the ring inwards and 44 outwards in the 4 s.
- Raise I₄ to 15 A: the unknown now flows into c, I = 4 A, and the beads on the fifth wire turn round.
- Make I₁ + I₂ + I₃ equal to I₄ and the fifth wire carries no current at all — the four known currents already balance.
Fisika di baliknya
A junction cannot store charge. Charge is conserved, so in any stretch of time the charge arriving at c equals the charge leaving it, and dividing by the time gives Kirchhoff's current law: ΣI_in = ΣI_out. Counting a current flowing in as positive and one flowing out as negative, the same rule reads ΣI = 0. With the unknown I assumed to flow into c: 4 + 5 + 2 + I = 8, so I = −3 A. Assumed to flow out: 4 + 5 + 2 = 8 + I, so I = +3 A. Both guesses give the same physical answer — 3 A flowing out of c — because a negative result only means that the current flows opposite to the assumed direction. In 4 s, (4 + 5 + 2) A × 4 s = 44 C enters through the ring and (8 + 3) A × 4 s = 44 C leaves; nothing piles up. The currents are conventional currents: electrons drift the other way, so an arrow that marks the motion of electrons must be reversed before the rule is applied.