Topic 7 — Cells in Series and in Parallel

Part 1 · The idea, told simply

A single cell has a fixed push and a fixed internal rustiness. Combining cells lets you trade one for the other.

Story track · series is a ladder

Stand two people on each other's shoulders and you can reach twice as high. Stack two cells nose to tail — positive of one joined to negative of the next — and their pushes add up.

But there is a cost. The current now has to fight its way through both sets of rusty internal pipes, one after the other. So the internal resistances add too.

SERIES: ε_eq = ε₁ + ε₂ r_eq = r₁ + r₂ More voltage — but also more internal resistance

Story track · parallel is a wider door

Now put two identical cells side by side instead, positives joined together and negatives joined together. Nobody is standing on anybody's shoulders, so the height — the voltage — is unchanged.

But now the current has two internal paths to choose from instead of one. Two doors are easier to get through than one. So the internal resistance halves.

PARALLEL (identical cells): ε_eq = ε r_eq = r/n Same voltage — but much less internal resistance

That is the whole trade. Series buys voltage at the price of internal resistance. Parallel buys low internal resistance at no gain in voltage.

Animation 1 · Stacking cells in series

Add cells to the ladder. Watch both bars grow together — that is the bargain.

number of cells n =

Animation 2 · Banking cells in parallel

Add identical cells side by side. The voltage bar refuses to move; the resistance bar collapses.

number of cells n =

Part 2 · Which arrangement gives more current?

This is the question NEET actually asks. With n identical cells of emf ε and internal resistance r driving an external resistance R:

All in SERIES: I = nε/(R + nr) All in PARALLEL: I = ε/(R + r/n) = nε/(nR + r)

The decision rule — compare R against r, nothing else

Put the two expressions side by side. Series wins when nR + r > R + nr, which simplifies to R > r.

R ≫ r → use SERIES (external resistance dominates; extra voltage is worth it) R ≪ r → use PARALLEL (internal resistance dominates; you must reduce it) R = r → both give exactly the same current

The physical reading: if the external circuit is already the main obstacle, adding internal resistance barely matters, so take the voltage — series. If the cells' own internal resistance is the main obstacle, adding more of it would be self-defeating, so spread the load — parallel.

Worked check with four cells of 1.5 V and r = 0.5 Ω:

External R	Series current	Parallel current	Winner
R = 2 Ω (R > r)	1.50 A	0.71 A	series
R = 0.5 Ω (R = r)	1.20 A	1.20 A	tie
R = 0.1 Ω (R < r)	2.86 A	6.67 A	parallel

Animation 3 · The crossover

Slide the external resistance. The two curves cross at exactly R = r — and swap places.

external R =

Part 3 · The general formulas

Maths track · series, from the loop rule

Two cells joined so that the negative of the first meets the positive of the second. From Topic 2, the potential difference across each is

V_AB = ε₁ − I r₁ V_BC = ε₂ − I r₂

Adding, since the same current I flows through both:

V_AC = (ε₁ + ε₂) − I(r₁ + r₂)

Comparing with V_AC = ε_eq − I r_eq gives the two series results at once. NCERT extends this to n cells: the equivalent emf is the sum of the individual emfs, and the equivalent internal resistance is the sum of the internal resistances.

Maths track · parallel, from the junction rule

Two cells with positives joined and negatives joined, carrying currents I₁ and I₂ into a common node, with I = I₁ + I₂. The same V appears across both, so

V = ε₁ − I₁r₁ ⇒ I₁ = (ε₁ − V)/r₁ V = ε₂ − I₂r₂ ⇒ I₂ = (ε₂ − V)/r₂

Adding and rearranging for V:

I = (ε₁/r₁ + ε₂/r₂) − V(1/r₁ + 1/r₂)

Matching against V = ε_eq − I r_eq gives NCERT's compact pair:

1/r_eq = 1/r₁ + 1/r₂ ε_eq/r_eq = ε₁/r₁ + ε₂/r₂

These are far easier to remember than the expanded forms, and they extend to n cells by simply adding more terms. The expanded versions are

ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) r_eq = r₁r₂/(r₁ + r₂)

Note carefully: ε₁ is paired with r₂ in the numerator, not with r₁. Getting this cross-pairing backwards is the standard slip.

When a cell is connected the wrong way round

Story track

One person in the human ladder is standing on their head. They do not add their height — they subtract it.

In a series chain, a reversed cell contributes −ε instead of +ε. But its internal resistance still adds, because resistance never cancels. NCERT states it precisely: if the current leaves a cell from its negative electrode, that cell's emf enters ε_eq with a negative sign.

n cells in series, one reversed: ε_eq = (n − 2)ε r_eq = n r

Why n − 2 and not n − 1? Because that cell not only fails to contribute its +ε, it actively cancels one other cell's +ε. You lose it twice over.

Animation 4 · One cell turned around

Click any cell to flip it. Watch ε_eq fall by 2ε each time — while r_eq never budges.

Part 4 · Mixed grouping

Beyond NCERT's text but regularly examined: N identical cells arranged as m rows in parallel, each row containing n cells in series, so that N = mn.

Each row: emf nε, internal resistance nr m such rows in parallel: ε_eq = nε, r_eq = nr/m Current: I = nε / (R + nr/m) = mnε / (mR + nr)

The current is largest when the equivalent internal resistance matches the external resistance:

Maximum current when R = nr/m (external = internal) Then I_max = mnε/(2mR) = nε/(2R)

This is the same matching condition as maximum power transfer in Topic 2, arrived at from a different direction.

Animation 5 · Arranging twelve cells

Twelve identical cells, arranged as m rows of n. Only some arrangements are any good for a given load.

rows m = external R =

Trap · the six that cost marks

Subtracting internal resistances when emfs oppose. Resistances always add, never cancel.
Writing ε_eq = (ε₁r₁ + ε₂r₂)/(r₁+r₂) for parallel cells. The pairing is crossed: ε₁ goes with r₂.
Taking a reversed cell as costing you one ε. It costs you 2ε.
Assuming series always gives more current. It only does when R > r.
Thinking parallel cells give more voltage. Identical cells in parallel give exactly ε.
Forgetting that in parallel the cells must have their like terminals joined, or one will simply charge the other.

Part 5 · Formula sheet

Series

ε_eq = ε₁ + ε₂ + … + ε_n r_eq = r₁ + r₂ + … + r_n n identical cells: ε_eq = nε, r_eq = nr Current: I = nε/(R + nr) One cell reversed: ε_eq = (n − 2)ε, r_eq = nr (unchanged) k cells reversed: ε_eq = (n − 2k)ε

Parallel

1/r_eq = 1/r₁ + 1/r₂ + … ε_eq/r_eq = ε₁/r₁ + ε₂/r₂ + … ← easiest form to remember Expanded (two cells): ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) ← note the CROSSED pairing r_eq = r₁r₂/(r₁ + r₂) n identical cells: ε_eq = ε, r_eq = r/n Current: I = ε/(R + r/n) = nε/(nR + r) If one cell is reversed, put ε₂ → −ε₂ in the formulas

Which grouping wins

Series current I_s = nε/(R + nr) Parallel current I_p = nε/(nR + r) R > r → series gives more current R < r → parallel gives more current R = r → identical currents Rule of thumb: R ≫ r use series · R ≪ r use parallel

Mixed grouping (m rows of n, N = mn cells)

ε_eq = nε r_eq = nr/m I = nε/(R + nr/m) = mnε/(mR + nr) Maximum current when R = nr/m Then I_max = nε/(2R) = mnε/(2mR)

Sanity checks

Identical cells in parallel NEVER raise the voltage above ε Series always raises emf but also raises r r_eq in parallel is always SMALLER than the smallest r A reversed cell lowers ε_eq by 2ε but leaves r_eq alone Two identical cells in parallel: ε_eq = ε, r_eq = r/2

Part 6 · 50 NEET-pattern questions with full solutions

Includes 4 graph-based questions, 3 assertion–reason questions, and 9 questions tagged Grouping covering the series-versus-parallel decision and mixed arrangements. Year labels are not attached: the patterns are authentic, the wording is mine.