Emf, Heating & Internal Resistance
Batteries pump charge; resistors turn the work into heat.
Lesson at a glanceShow the key points
- Emf ℰ = W/q by the non-electrostatic battery force (32.10); equals open-circuit voltage.
- Heat in a resistor U = i²Rt; P = i²R = V²/R = Vi (32.11–32.12).
- Real battery: ideal emf plus internal resistance r; V = ℰ − ir; heat i²rt inside.
- Discharging: current leaves the + terminal; charging: forced in.
A battery is a charge pump. Its emf measures how much work it does on each coulomb, and a resistor turns that work into heat.
How a battery keeps its voltage
Inside a battery some non-electrostatic mechanism (chemistry, in a cell) pushes positive charge toward terminal A and negative toward B with a force F⃗b. As charge piles up, an electrostatic field builds between the terminals and pushes back with F⃗e = qE⃗. Accumulation stops when Fe = Fb.
With nothing connected, Fbd = qEd = qV, so the emf equals the terminal voltage of an open battery. The two are different quantities: emf comes from the non-electrostatic battery force, potential difference from the electrostatic field of the piled-up charges.
Connect a wire across the terminals and charge flows round; the battery mechanism keeps restoring the charge on A and B, maintaining a steady current. Normally positive charge leaves from the + terminal (discharging); driving current into the + terminal instead is charging.
Heat in a resistor
A resistor R joined to a battery by wires of negligible resistance has potential difference V = iR across it. In time t a charge q = it passes through and loses potential energy qV, which turns into thermal energy:
A resistor gives out 600 J in 15 s with 2.0 A through it. (a) Find R. (b) How much heat does it produce in 15 s at 3.0 A?
Internal resistance
As charge q = it passes through a battery, the battery force does work qℰ, but the charge gains only qV of electric potential energy. The difference warms the battery. An ideal battery wastes nothing, so V = ℰ always. A real battery behaves like an ideal emf in series with an internal resistance r:
A 9.0 V battery with internal resistance 1.5 Ω delivers 0.40 A. Find (a) its terminal voltage, (b) the heat produced inside it in one minute.
- 1Emf ℰ = work per unit charge by the battery force (32.10); equals open-circuit terminal voltage.
- 2Heat in a resistor: U = i²Rt, P = i²R = V²/R = Vi (32.11–32.12).
- 3Real battery: ideal ℰ plus internal r; V = ℰ − ir while discharging.
Check yourself
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