Three Batteries Side by Side — Internal Resistance in Every Branch
Three sealed lead-acid batteries stand on a circuit board, side by side in three branches: 15 V on the back rail, 10 V in the middle and 3 V at the front, each with its internal resistance written on a sticker. Two load resistors on the right, 9.5 Ω and 1.4 Ω, complete two loops. A panel ammeter in every battery branch reads the current along its printed arrow, and a multimeter is clipped across the back battery. Press play and a test charge walks loop abcfa and then loop fcdef; the fence above the wires rises and falls with the potential, so you can watch each battery lift it and its own internal resistance take a little back. Eight sliders set every emf, every internal resistance and both loads.
How to use this simulation
- (V_B)₁, the back battery's emf — 0 to 20 V, starting at 15 V.
- r₁, the back battery's internal resistance — 0 to 2 Ω, starting at 1 Ω.
- (V_B)₂, the middle battery's emf — 0 to 20 V, starting at 10 V.
- r₂, the middle battery's internal resistance — 0 to 2 Ω, starting at 0.5 Ω.
- (V_B)₃, the front battery's emf — 0 to 20 V, starting at 3 V.
- r₃, the front battery's internal resistance — 0 to 2 Ω, starting at 0.1 Ω.
- R_bc, the upper load resistor — 0.5 to 20 Ω, starting at 9.5 Ω.
- R_dc, the lower load resistor — 0.5 to 20 Ω, starting at 1.4 Ω. Every 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 meters read 2.000 A, 8.000 A and 6.000 A, and the junction rule holds at f: 8 = 2 + 6.
- Inside each battery the fence first climbs by the emf and then slopes down by I r: 2 V in the 15 V battery, 4 V in the 10 V battery, only 0.6 V in the 3 V battery.
- The multimeter across the back battery reads 13.00 V, not 15 V: the missing 2 V is I₁ r₁, lost inside the battery itself.
- Each loop walk ends exactly where it began: the steps in the walk card add up to 0.00 V.
- Set (V_B)₂ to 0 and R_bc to 0.5 Ω: I₃ turns negative (−0.4 A). Current now enters the 3 V battery at its + terminal and charges it, 1.2 W going in while the 15 V battery gives out 114 W.
- Raise r₂ from 0.5 to 2 Ω and the middle current drops from 8 A to 4.38 A — internal resistance matters as much as any load.
The physics behind it
A real battery behaves like an ideal emf V_B in series with an internal resistance r. Walking through it from − to + in the direction of the current, the potential rises by V_B and then falls by I r, so between its terminals you measure V = V_B − I r. With the assumed directions (I₁ from a to b, I₂ from c to f, I₃ from e to d), the junction rule at f gives I₂ = I₁ + I₃. Walking loop abcfa clockwise: 15 + 10 = (1 + 9.5) I₁ + 0.5 I₂. Walking loop fcdef clockwise, both batteries are met from + to − and both currents run against the walk: −10 − 3 = −0.5 I₂ − (0.1 + 1.4) I₃. Solved together: I₁ = 2 A, I₂ = 8 A and I₃ = 6 A, all positive, so every assumed direction was right and all three batteries give out energy. Check with energy: 15 × 2 + 10 × 8 + 3 × 6 = 128 W produced, and 2² × (1 + 9.5) + 8² × 0.5 + 6² × (0.1 + 1.4) = 128 W turned into heat. The terminal voltages are 13 V, 6 V and 2.4 V: the 10 V battery loses 4 V inside itself because it carries the largest current.