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What you will learn
- Terminal voltage falls as current through internal resistance rises.
- Useful power is εI − I²r for this battery model.
- Maximum current is not the same as maximum useful performance.
Electricity & Magnetism · reward round
Drive the rover to each flag on one battery. Hold to draw more current — the wheels get P = εI − I²r, the internal resistance burns the rest, and the cell drains at exactly I. Five stages, ageing cells.
Score 0 · stage 1/5 · fresh cell · r = 1.0 Ω
The cell delivers power εI, but its own internal resistance dissipates I²r of it as heat before anything reaches the motor. The wheels get the difference, P = εI − I²r — a downward parabola in I. It is zero at I = 0, zero again at I = ε/r where every watt burns inside the cell, and greatest exactly half-way between, at I = ε/2r. The game never prints that value: you find it by nudging the draw and watching the power readout peak.
The battery is a charge budget, not an energy budget. Capacity in ampere-hours drains at exactly the current you draw, so drawing softly always travels further per ampere-hour — but the stage clock makes pure thrift lose too. As the cells age from stage to stage, r rises, the parabola flattens, and the sweet spot slides to lower currents: an old cell simply cannot deliver the power a fresh one could.
The dashed ghost line runs the same update law forward with your present draw held fixed — position gaining v = P/drag each second while the charge falls at I — and marks where you would halt. At the maximum-power draw exactly half the cell’s energy is lost inside it: real engineering trades speed against efficiency on the same curve you are playing.
Game 20 · Electricity & Magnetism learning guide
Learning objectiveChoose a useful current draw that delivers enough power to a rover without wasting the battery’s energy internally.
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Why does very large current increase the power wasted inside a cell?
Suitable forSecondary and upper-secondary physics · cells and internal resistance
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