Node by Node — Untangling a Network of Junctions
A small network of brass posts and wires sits on a circuit board, part of a larger circuit. Six currents cross its edge and are read on LED ammeters — 1 A and 2 A leave it; 2 A, 3 A, 4 A and 2 A enter it — but the meters on five other wires are covered with masking tape. Press play to work out the hidden currents one junction at a time: each node glows while it is being solved, its equation appears beside the scene, and the tape peels off the meter whose current has just been found, so you can check the answer. In the last two seconds a dashed boundary is drawn round the whole network to show that it, too, obeys the junction rule.
Cara memakai simulasi ini
- Out via a, to the left — 0 to 8 A in steps of 0.5 A, starting at 1 A (a plain wire joins a and b, so this current leaves node b).
- Out at b, to the right — 0 to 8 A, starting at 2 A.
- Into c, from the left — 0 to 8 A, starting at 2 A.
- Into c, from the front — 0 to 8 A, starting at 3 A.
- Into e, from the front — 0 to 8 A, starting at 4 A.
- Into f, from the right — 0 to 8 A, starting at 2 A. A change solves the network again 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.
Yang perlu diamati
- The beads flow in every wire from the very first moment — the currents are real from the start; only our knowledge grows as the tapes come off.
- At the starting values the five hidden meters read 3.000, 5.000, 2.000, 6.000 and 8.000 A — exactly the answers of the five node equations.
- Set both currents into c to zero: I_d comes out as −3 A. Once d is solved its arrow points from e up to d, its beads run that way, and its meter shows −3.000.
- Whatever the slider settings, the check in the last two seconds always balances: the total current into the network equals the total current out of it.
- The order matters: d cannot be solved first, because three of its wires are unknown at the start.
Fisika di baliknya
Kirchhoff's current law, ΣI_in = ΣI_out, holds at every node — every point where three or more wires meet. Points joined by a plain wire are the same node, so a and b count as one. To untangle a network, start at a node with only one unknown current, solve it and move on: each answer leaves the next node with one unknown fewer. At the start b (with a) and c each have a single unknown; d has three; e and f have two each. So: at b, I_b = 1 + 2 = 3 A; at c, 2 + 3 = I_c = 5 A; at d, 5 = 3 + I_d, so I_d = 2 A; at e, 2 + 4 = I_e = 6 A; at f, 6 + 2 = I = 8 A. As a check, the whole network behaves like one big junction: 2 + 3 + 4 + 2 = 11 A flows in and 1 + 2 + 8 = 11 A flows out. If a result comes out negative, the current flows against the drawn direction — set both currents into c to zero and I_d becomes −3 A, flowing from e up to d.