Two Loops, Three Currents — Both Laws at Once
A slate circuit board on a lab bench carries two sealed lead-acid batteries, three gold power resistors and three LED ammeters, wired as two loops that share a middle branch. Each ammeter's printed arrow is the direction guessed for its current. Sliders set both batteries and all three resistors, and the circuit is solved exactly the moment anything changes. Press ▶ and a test charge walks loop 1 and then loop 2 while a fence above it rises and falls with the potential — the walk plays once and stops, and the timeline takes you back to any moment.
How to use this simulation
- Left battery (V_B)₁ — 0 to 12 V, starting at 6 V.
- Middle battery (V_B)₂ — 0 to 12 V, starting at 2 V.
- Left resistor R₁ — 0.5 to 10 Ω, starting at 2 Ω.
- Middle resistor R₂ — 0.5 to 10 Ω, starting at 3 Ω.
- Right resistor R₃ — 0.5 to 10 Ω, starting at 5 Ω. Every change re-solves the circuit and keeps the moment on screen; nothing moves until you press ▶.
- 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.
What to look for
- At the starting values the middle ammeter reads −0.516 A: its printed arrow points up, but the current flows down.
- Follow the fence round loop 1: it falls 2.45 V across R₁, climbs 6 V inside the left battery, falls 2 V through the middle battery from + to − and 1.55 V across R₂ — and ends exactly where it began.
- Raise the middle battery above 4.29 V: the minus sign disappears and that battery starts giving out energy instead of taking it in.
- Make both batteries 6 V and the left and middle resistors equal: the two batteries share the load equally.
- The voltmeter across a and b always equals I₃ × R₃: all three branches see the same potential difference.
The physics behind it
Kirchhoff's first law, the junction rule, is conservation of charge: at junction c the currents arriving equal the currents leaving, I₁ + I₂ = I₃. The second law, the loop rule, is conservation of energy: round any closed loop the rises in potential across batteries equal the falls across resistors, ΣV_B = ΣIR. With a guessed direction for every current, loop abcfa gives 6 − 2 = 2I₁ − 3I₂ and loop cdefc gives 2 = 3I₂ + 5I₃. Three equations for three unknowns: I₁ = 1.226 A, I₂ = −0.516 A and I₃ = 0.710 A. The minus sign is information, not a mistake: the middle current really flows down, from + to − inside the 2 V battery, so that battery takes in energy — it is being charged. Energy checks the answer: the 6 V battery gives out 6 × 1.226 = 7.36 W, and the heat in the three resistors plus the charging power 2 × 0.516 W add up to the same 7.36 W. A voltmeter across a and b reads 3.548 V by every route: 6 − 1.226 × 2, or 2 + 0.516 × 3, or 0.710 × 5. A third loop round the outside gives nothing new, because it is the sum of the other two. The same sliders solve related circuits too: 12 V and 5 V batteries with 3 Ω, 2 Ω and 4 Ω give 2 A, −0.5 A and 1.5 A.