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In Lesson 4 we carefully distinguished EMF (energy per unit charge into the circuit from the source) from potential difference (energy per unit charge out to a component). The "distinction" matters most when the source has a significant internal resistance, which is practically always.
Every real battery, cell or generator has some resistance inside it — due to the materials, the electrodes and the chemistry. When current flows, some of the EMF is dissipated inside the source, so the terminal pd (what you would measure across the battery's terminals) is less than the EMF.
This lesson derives the equation ε = I(R + r), discusses the concept of "lost volts", and shows how to measure internal resistance experimentally — another OCR H556 required practical.
Module 4.2.2 / 4.3.1.
A real cell can be modelled as an ideal EMF source of strength ε in series with a fixed internal resistance r. This is not a physical resistor you can remove — it is a mathematical representation of all the internal losses.
flowchart LR
subgraph Cell
EMF[EMF ε] --- r[Internal resistance r]
end
Cell --> R[External resistor R] --> Cell
When the cell drives a current I through an external load R, the circuit is a simple series loop, and by Kirchhoff's second law (which we formalise in Lesson 11, but can state intuitively here):
ε = I R + I r
or equivalently:
ε = I (R + r)
The quantity Ir is the voltage dropped inside the cell and is called the "lost volts". It represents the electrical energy per coulomb that is dissipated inside the battery as heat, rather than being delivered to the external load.
The terminal pd (what a voltmeter across the battery's terminals would read) is:
V = ε − I r
So the terminal pd is the EMF minus the lost volts. When I = 0 (no current drawn), V = ε exactly.
A cell of EMF 1.50 V and internal resistance 0.50 Ω is connected to an external resistor of 4.5 Ω. Calculate:
(a) The current in the circuit. (b) The terminal pd. (c) The lost volts. (d) The power dissipated in the external resistor and in the cell.
(a) Total resistance = R + r = 4.5 + 0.50 = 5.0 Ω.
(b) V = ε − Ir = 1.50 − (0.30 × 0.50) = 1.50 − 0.15 = 1.35 V (Alternatively: V = IR = 0.30 × 4.5 = 1.35 V ✓)
(c) Lost volts = Ir = 0.30 × 0.50 = 0.15 V
(d) P_ext = I²R = 0.09 × 4.5 = 0.405 W
Notice the energy conservation check: P_ext + P_int = εI.
As the load R changes, the current and terminal pd change too:
| R (Ω) | I = ε/(R+r) | V = ε − Ir |
|---|---|---|
| ∞ (open circuit) | 0 | 1.50 V (= ε) |
| 100 | 0.0149 A | 1.493 V |
| 10 | 0.143 A | 1.429 V |
| 1 | 1.00 A | 1.00 V |
| 0.5 | 1.50 A | 0.75 V |
| 0 (short circuit) | 3.00 A | 0 V |
Observations:
Exam Tip: "Why does the terminal pd of a battery fall when the headlights are switched on?" Answer: more current is drawn, so the lost volts Ir increase, and V = ε − Ir drops.
A surprisingly deep result: the external power P_ext = I²R is maximised when R = r. This is called the maximum power transfer theorem. You can derive it with calculus, but the intuition is:
At maximum power transfer, half the total power goes to the load and half is wasted inside the cell (P_int = I²r = I²R). So the maximum-power condition is also the 50% efficient condition. This is why batteries designed to deliver significant power have deliberately low internal resistance — they can run efficient loads and deliver lots of current.
For R > r, efficiency rises above 50% but the raw power delivered is lower. For a torch battery you typically want efficiency rather than maximum power.
OCR H556 includes the determination of the EMF and internal resistance of a cell as a required practical.
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