The Unbalanced Bridge — Equivalent Resistance by Kirchhoff
Five power resistors are wired as a bridge between two posts, a and d, with a fifth resistor crossing the middle from b to c. A 13 V battery feeds the bridge through a knife switch and a panel ammeter, and a multimeter is clipped across a and d. Because current can cross the middle, the five resistors are neither in series nor in parallel, so the usual rules cannot combine them. Press play: the blade swings down, the current starts, and the two meters give the voltage across the bridge and the current the battery delivers. Their ratio is the equivalent resistance R′, the single resistor that would draw the same current from the battery.
Cara memakai simulasi ini
- Battery V_B — 1 to 20 V, starting at 13 V. It changes every current in the same proportion and leaves R′ as it is.
- Top left R_ab (a to b) — 0.5 to 5 Ω, starting at 1 Ω.
- Top right R_ac (a to c) — 0.5 to 5 Ω, starting at 1 Ω.
- Across the middle R_bc — 0.5 to 5 Ω, starting at 1 Ω; its current I₃ passes through a second ammeter.
- Bottom left R_bd (b to d) — 0.5 to 5 Ω, starting at 1 Ω.
- Bottom right R_cd (c to d) — 0.5 to 5 Ω, starting at 2 Ω.
- Play/pause, back to the start, a timeline you can drag to any moment, and ½×, ¼×, ⅛× slow motion. The switch closes 1 s into the 6 s timeline.
- Drag anywhere on the scene to look around.
Yang perlu diamati
- Before the switch closes nothing flows: both ammeters and the multimeter read 0, and V_B / I has nothing to divide by.
- At the start values the battery ammeter reads 11.00 A and the multimeter 13.00 V: R′ = 13/11 = 1.18 Ω. The middle ammeter reads +1.000 A along its arrow, from c to b.
- At a the 11 A splits into 6 A towards b and 5 A towards c; at b the 1 A from the middle joins in, so 7 A goes on to d, while only 4 A leaves c for d.
- Set the bottom right to 1 Ω: the ratios match, the middle ammeter falls to 0, and R′ = (1 + 1) ∥ (1 + 1) = 1 Ω, exactly what series and parallel give. At 0.5 Ω the middle current reverses and the meter reads a negative value.
- Changing V_B changes every current in proportion but not R′: at 20 V the battery delivers 16.92 A and R′ is still 1.18 Ω.
- R′ always lies between 1.167 Ω (the middle branch replaced by a plain wire) and 1.200 Ω (the middle branch cut); a larger R_bc moves it towards the second value.
Fisika di baliknya
Kirchhoff's junction rule is built into the names of the currents: I₂ leaves a towards b and I₁ towards c, I₃ crosses the middle from c to b, so b → d carries I₂ + I₃ and c → d carries I₁ − I₃. Three loops then give three equations (ΣV_B = ΣIR). Round the top triangle: I₁ − I₂ + I₃ = 0. Round the bottom triangle: 2I₁ − I₂ − 4I₃ = 0. Round the battery loop: 2I₂ + I₃ = 13. The solution is I₁ = 5 A, I₂ = 6 A and I₃ = 1 A, all positive, so every assumed direction is right. The battery delivers I = I₁ + I₂ = 11 A, so R′ = V_B / I = 13/11 = 1.18 Ω. Energy confirms the answer: the battery gives out V_B I = 143 W, and the five resistors turn 25 + 36 + 1 + 49 + 32 = 143 W into heat, the same as I²R′. The middle current flows because c sits 1 V above b. If R_ab/R_bd = R_ac/R_cd, b and c sit at the same potential, nothing crosses the middle and the bridge is balanced: then series and parallel work again.