Cells, Internal Resistance and EMF
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Direct answer
The terminal voltage of a cell of emf epsilon and internal resistance r is V = epsilon − I r while discharging and V = epsilon + I r while being charged — the emf is the open-circuit potential difference. Driving an external resistance R, the current is I = epsilon/(R + r) and the useful delivered power P = I^2 R peaks exactly when R = r, at P(max) = epsilon^2/(4 r), where the efficiency is only 50%. Grouping multiplies the options: series cells add emfs and internal resistances; parallel cells keep one emf with internal resistance r/n; mixed grouping is optimised when external resistance equals effective internal resistance.
What you must remember
- Definitions: emf = work done per unit charge by the source's non-electrostatic forces (chemical, in a cell); terminal voltage V = epsilon − I r while discharging, exceeding epsilon only when current is forced backward (charging).
- Current and terminal drop: I = epsilon/(R + r); V = I R = epsilon R/(R + r); plotting V against I gives a straight line with intercept epsilon and slope −r — the standard experiment.
- Maximum power transfer: P(out) = epsilon^2 R/(R + r)^2 is maximum at R = r, with P(max) = epsilon^2/(4 r); efficiency = R/(R + r) = 50% at that point, so power-optimal is not efficiency-optimal.
- Series grouping: n cells in series drive I = n epsilon/(R + n r); best for high external resistance.
- Parallel grouping: n identical cells in parallel drive I = epsilon/(R + r/n); best for low external resistance.
- Mixed grouping: m rows of n cells each gives I = n epsilon/(R + n r/m), maximised when R = n r/m — the general matching rule.
- Potentiometer comparison: emfs compare without drawing current (null method), epsilon(1)/epsilon(2) = l(1)/l(2).
- Pattern note: Main tests V–I line readings and single-cell power; Advanced builds battery groups with unequal cells and asks which cell charges which.
Reading the V–I line, then matching the load
Plot terminal voltage against current for a battery and the data fall on a straight line: intercept V = 12 V at I = 0 (that is the emf) and a drop to 10 V at 4 A (slope = −0.5, so r = 0.5 ohm). Load R = 5.5 ohm: I = 12/6 = 2 A, terminal 11 V, useful power 22 W, heat inside the battery 2 W. Ask for maximum useful power instead: set R = r = 0.5 ohm, I = 12 A, P(max) = 144/2 = 72 W — and the terminal voltage falls to 6 V, half the emf, the signature of matched load.
Grouping repeats the matching theme. Eight cells of 1.5 V, 0.5 ohm each feeding R = 0.25 ohm: all in series gives about 2.8 A, all in parallel about 4.8 A. The best grouping is the one whose effective internal resistance lands nearest R — check that equality before computing any current.
Where students slip
Charging versus discharging decides the sign of the I r correction, and candidates who memorise only V = epsilon − I r misstate the terminal voltage of a battery being charged (it reads above emf). Second, maximum power does not mean maximum efficiency: at R = r, half the energy cooks inside the cell. Third, in parallel grouping, only identical cells are safe to combine — unequal emfs in parallel drive circulating currents that drain the stronger cell, a fact Advanced turns into "which cell is being charged" questions: the one with lower emf, current forced into its positive terminal. Finally, a voltmeter across a cell reads V = epsilon R(V)/(R(V) + r), approaching epsilon only for enormous R(V) — the potentiometer's null method is the honest emf measurement.
Frequently asked questions
What is the emf of a cell and how does it differ from terminal voltage?
Emf is the open-circuit work per unit charge supplied by the cell; terminal voltage is the p.d. across its terminals under load, V = epsilon − I r, equal to emf only when no current flows.
When is power delivered to an external resistor maximum?
When R = r, giving P(max) = epsilon^2/(4 r), with exactly half the power dissipated internally.
How do series and parallel grouping of identical cells differ?
Series adds both emfs and internal resistances (n epsilon, n r); parallel keeps one cell's emf but divides internal resistance by n (epsilon, r/n).
Why does a battery's terminal voltage rise above emf while charging?
The charger pushes current backward through the cell, so the internal drop adds: V = epsilon + I r, with energy being stored chemically rather than delivered.
Why is a potentiometer preferred over a voltmeter for comparing emfs?
At null point the potentiometer draws no current from the cell, so no I r drop exists, while a voltmeter's finite resistance always lowers the reading below true emf.