Two Batteries Head to Head — Which One Is Charging?
Two sealed lead-acid batteries, 6 V at the back and 12 V at the front, sit in one loop on a circuit board with a 10 Ω and an 8 Ω power resistor. They are connected so that they push against each other: the 6 V battery pushes charge clockwise, the 12 V battery anticlockwise. A small monitor clipped to each battery reports whether it is charging or discharging, and an ammeter whose printed arrow points clockwise measures the current. Pick an assumed direction for the current, then press play: a test charge walks the loop in that direction, a fence above it follows the potential, and a chart adds up the energy given out by one battery, turned into heat in the resistors and stored in the other battery.
Como usar esta simulação
- Back battery (V_B)₁ — 0 to 12 V, starting at 6 V; it pushes clockwise.
- Front battery (V_B)₂ — 0 to 12 V, starting at 12 V; it pushes anticlockwise.
- Left resistor R₁ — 1 to 20 Ω, starting at 10 Ω.
- Right resistor R₂ — 1 to 20 Ω, starting at 8 Ω.
- Assumed current — clockwise or anticlockwise. It turns the ring on the board, the equation and the walk; the ammeter stays wired clockwise, so it reads −0.333 A either way.
- 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.
O que observar
- With clockwise assumed, the equation gives I = −0.333 A, the ammeter reads −0.333 and the glowing beads flow anticlockwise: the bigger push wins.
- The monitor on the 6 V battery reads CHARGING ▲ and the one on the 12 V battery DISCHARGING ▼: the current enters the 6 V battery at its + terminal.
- The energy chart always stacks the heat and the stored energy exactly up to the energy given out: 40 J = 20 J + 20 J after 10 s.
- Switch to anticlockwise: the computed current becomes +0.333 A and the ring turns round, but the beads, the ammeter and the monitors do not change.
- Make the two emfs equal and the current stops; both monitors read IDLE. Make the back battery the bigger one and the current turns clockwise: now the front battery is the one being charged.
A física por trás
A single loop has no junctions, so Kirchhoff's first law has nothing to say; the second law, ΣV_B = ΣIR, gives one equation for the one unknown current. Choose a loop direction and an assumed current direction (here both clockwise), and count each battery as positive if it pushes the way you go round and negative if it pushes against it: (V_B)₁ − (V_B)₂ = I(R₂ + R₁), so 6 − 12 = I(8 + 10) and I = −6/18 = −0.333 A. The minus sign is the answer, not a mistake: the current really flows anticlockwise, 0.333 A. It passes through the 12 V battery from − to +, so that battery gives out energy at (V_B)₂I = 12 × 0.333 = 4 W, and through the 6 V battery from + to −, against its push, so that battery takes in energy at (V_B)₁I = 6 × 0.333 = 2 W: it is being charged. The resistors turn I²(R₁ + R₂) = 0.333² × 18 = 2 W into heat. Power produced equals power consumed, 4 W = 2 W + 2 W, so in 10 s 40 J are given out, 20 J become heat and 20 J are stored. Assume the anticlockwise direction instead and the same equation gives +0.333 A: the physics does not depend on the guess.