Combinations of Cells
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Direct answer
n identical cells of emf ε and internal resistance r combine like batteries with rules worth stating precisely: in series, emf nε and internal resistance nr (voltages add, so this arrangement feeds high-resistance loads); in parallel, emf ε and internal resistance r/n (currents add, suiting low-resistance loads). The circuit current follows I = nε/(R + nr) for series and I = ε/(R + r/n) for parallel, and for m rows of n cells each (mixed grouping), I = nε/(R + nr/m), maximised when the external resistance equals the total internal resistance, R = nr/m — the maximum power transfer condition. Non-identical cells in parallel are handled by the general equivalent: ε_eq = (Σε_i/r_i)/(Σ1/r_i), the current-weighted mean of the emfs.
What you must remember
- Series pack: emf = nε, internal resistance = nr; current I = nε/(R + nr); advantageous when R >> r.
- Parallel pack: emf = ε, internal resistance = r/n; current I = nε/(nR + r); advantageous when R << r.
- Mixed grouping (m rows of n): I = nε/(R + nr/m); maximum current when R = nr/m, i.e. total internal resistance equals external.
- Matching rule: series for high-resistance circuits (torch with several cells), parallel for low-resistance draws — and identical cells in parallel do not raise emf, only sustain larger current longer.
- Maximum power theorem: the load receives greatest power P = ε^2R/(R + r)^2 when R = r, delivering P_max = ε^2/4r; efficiency then is only 50 per cent — power versus efficiency is a conceptual favourite.
- Terminal voltage: V = ε − Ir while discharging, V = ε + Ir while being charged; a cell's emf is its open-circuit potential difference.
- Non-identical parallel cells: ε_eq = (ε_1r_2 + ε_2r_1)/(r_1 + r_2) for two cells — a weighted average, never the arithmetic mean.
A worked grouping comparison
Four cells, each ε = 2 V and r = 0.5 Ω, feed a resistor R = 0.5 Ω. Series: I = nε/(R + nr) = 8/(0.5 + 2) = 3.2 A. Parallel: I = ε/(R + r/n) = 2/(0.5 + 0.125) = 3.2 A. Identical answers — and not by accident: setting the two expressions equal, the n in the numerator cancels against cross-multiplying, and the currents match exactly when R = r, which these numbers satisfy. The clean general rule: series wins when R > r, parallel wins when R < r, and at R = r the grouping does not matter. Swap the load to R = 10 Ω and series delivers 8/12 ≈ 0.67 A against parallel's 2/10.125 ≈ 0.20 A — a threefold victory for series; swap it to R = 0.1 Ω and parallel returns 2/0.225 ≈ 8.9 A while series manages only 8/2.1 ≈ 3.8 A. The lesson: when a question offers a choice of grouping, do not trust the reflex that "more emf is better"; compare nε/(R + nr) with ε/(R + r/n) using the actual R.
Where students slip
The reflex error is assuming series is always better because emf adds; with a low-resistance load, most of a series pack's emf drops across its own internal resistance as heat. Second, the maximum power point is often quoted as maximum efficiency — at R = r the cell delivers only half its power to the load, the rest warming itself; the exam separates these deliberately. Third, in mixed-grouping questions, count carefully: "24 cells, 4 per row" means n = 4 in series and m = 6 rows in parallel, and swapping m and n in nr/m flips the answer. A final discipline point: terminal voltage V = ε − Ir shrinks as current grows, which is why a car battery reads about 12 V open-circuit but sags during cranking — a daily-life framing NEET has used.
Frequently asked questions
What are the equivalent emf and internal resistance of n cells in series?
Emf nε and internal resistance nr, giving current I = nε/(R + nr); this suits external resistances much larger than r.
Why does connecting cells in parallel not increase the emf?
Identical parallel cells all maintain the same potential difference, so the emf stays ε while the internal resistances divide (r/n), allowing larger total current for low-resistance loads.
When does a mixed grouping deliver maximum current?
When the external resistance equals the total internal resistance, R = nr/m for m rows of n cells — the matched condition underlying maximum power transfer.
At what load resistance is the power delivered by a cell maximum, and what is that power?
At R = r, the power peaks at P_max = ε^2/4r, with exactly half the cell's output dissipated internally.
What is the terminal voltage of a cell during discharge?
V = ε − Ir, less than the emf by the internal drop; when the cell is being charged, the terminal voltage exceeds ε by the same amount.