Walking the Loop — Where the Potential Rises and Falls
A 12 V sealed lead-acid battery drives a power resistor round a single loop on a circuit board. A test charge sets off from point a, at the battery's negative terminal, and walks once round the loop: through the battery, past b, through the resistor R and back to a. A translucent fence above its path rises and falls with the electric potential, and a graph draws the same story as the walk goes on. Three sliders set the battery's emf, its internal resistance and the resistor. An ammeter in the loop shows the current, and a multimeter clipped across the battery's terminals shows the voltage the battery actually delivers.
How to use this simulation
- Battery emf V_B — 1 to 12 V, starting at 12 V.
- Internal resistance r — 0 to 2 Ω, starting at 0.5 Ω; it is written on the battery's sticker, which disappears at r = 0.
- Resistor R — 0.5 to 20 Ω, starting at 5.5 Ω. A 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.
What to look for
- At the start values the ammeter reads 2.000 A and the multimeter 11.00 V, not 12 V: the missing volt is spent inside the battery.
- Inside the battery the fence first jumps up by the full 12 V, then slopes down by Ir = 1 V before the charge reaches the + terminal.
- Along every wire the fence stays level; through R it slopes down by 11 V, and the walk ends exactly where it began: +12 − 1 − 11 = 0.
- Set r to 0 and the multimeter reads the full emf. Raise r to 2 Ω and lower R to 0.5 Ω: 4.8 A flows, Ir is 9.6 V and only 2.4 V is left at the terminals.
- Whatever the settings, the graph always returns to 0 V at a: the battery gives each coulomb exactly the energy the circuit takes from it.
The physics behind it
Kirchhoff's second law is conservation of energy for charge going round a closed loop: the potential changes add up to zero, ΣV = 0, or equivalently the emfs balance the IR drops, ΣV_B = ΣIR. Going through the battery from − to +, the potential rises by its emf V_B; the battery's own resistance r then takes back Ir; going through R with the current, the potential falls by IR; the connecting wires change nothing. So V_B − Ir − IR = 0, which gives I = V_B/(R + r). With V_B = 12 V, r = 0.5 Ω and R = 5.5 Ω the current is 12 ÷ 6 = 2 A: the battery lifts every coulomb by 12 J, 1 J of it is spent inside the battery and 11 J in the resistor. A voltmeter across the terminals reads the terminal voltage V = V_B − Ir = 11 V, less than the emf whenever a current flows. Multiply by the current for power: the battery converts 24 W, of which 2 W heats the battery itself and 22 W heats R.