Grouping of Cells

PHYCells & Circuits

This subtopic treats how cells combine: in series their EMFs add (and so do internal resistances), in parallel identical cells keep the EMF but lower the effective internal resistance, and unequal cells in parallel are handled with an equivalent-EMF formula. The key idea is to reduce any battery network to a single equivalent EMF and internal resistance before applying Ohm's law.

ISC tests series strings, identical parallel cells, and the unequal-parallel case.

Cells in series

E_{eq} = \sum E_i, \qquad r_{eq} = \sum r_i, \qquad I = \dfrac{n E}{R + n r}

n

identical cells, each EMF

E

and internal resistance

r

, external load

R

. Series adds EMFs (if same polarity) and internal resistances.

Identical cells in parallel

E_{eq} = E, \qquad r_{eq} = \dfrac{r}{m}, \qquad I = \dfrac{E}{R + r/m}

m

identical cells in parallel. EMF is unchanged; effective internal resistance drops to

r/m

, boosting current when

r

dominates.

Equivalent EMF of unequal parallel cells

E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}

E_1, E_2

EMFs and

r_1, r_2

internal resistances of two cells in parallel with like terminals joined. Weighted by the opposite internal resistance.

Derivation

Let $I_1, I_2$ be the currents delivered by the two cells into the common node, and $I$ the current through $R$ . The junction rule gives:
$I = I_1 + I_2$
The node voltage $V$ is common to both cell branches, so each cell's loop equation can be solved for its own current:
$V = E_1 - I_1 r_1 = E_2 - I_2 r_2 \;\Rightarrow\; I_1=\dfrac{E_1-V}{r_1}, \;\; I_2=\dfrac{E_2-V}{r_2}$
Substitute into the junction rule and collect terms in $V$ :
$I = \dfrac{E_1-V}{r_1}+\dfrac{E_2-V}{r_2} = \dfrac{E_1r_2+E_2r_1}{r_1r_2} - V\left(\dfrac{1}{r_1}+\dfrac{1}{r_2}\right)$
Matching this to the single-cell form $I=(E_{eq}-V)/r_{eq}$ identifies the equivalent emf and internal resistance:
$E_{eq} = \dfrac{E_1r_2+E_2r_1}{r_1+r_2}, \qquad r_{eq} = \dfrac{r_1r_2}{r_1+r_2}$

Equivalent internal resistance (parallel) and current

r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}, \qquad I = \dfrac{E_{eq}}{R + r_{eq}}

r_{eq}

is the parallel combination of the two internal resistances;

R

external load. Then

I=E_{eq}/(R+r_{eq})

Series grouping increases total EMF and is best when external resistance $R \gg r$ .
Parallel grouping of identical cells keeps EMF fixed but reduces internal resistance to $r/m$ , best when $R \ll r$ .
For unequal cells in parallel, reduce to a single equivalent EMF $E_{eq}$ and internal resistance $r_{eq}$ first, then apply $I=E_{eq}/(R+r_{eq})$ .
The equivalent EMF of unequal parallel cells lies between $E_1$ and $E_2$ , weighted by the opposite internal resistance.
In series, if a cell is connected with reversed polarity its EMF subtracts from the total.
Mixed (series-parallel) grouping of $mn$ cells gives maximum current when the load $R$ equals the total internal resistance.
Always check polarity before adding EMFs in series; like terminals joined in parallel is the standard ISC convention.

Where the marks go

Adding internal resistances in parallel cells the same way as series — parallel uses $r/m$ or $r_1r_2/(r_1+r_2)$ , not a sum.
Assuming the EMF rises when identical cells are placed in parallel; the EMF stays the same.
Forgetting opposite-resistance weighting in $E_{eq}=(E_1r_2+E_2r_1)/(r_1+r_2)$ and just averaging the EMFs.
Ignoring polarity in series strings, so a reversed cell that should subtract is wrongly added.

How the board asks it

Numericalequivalent emf and internal resistance of unequal parallel cells
Two cells of emf $2\,V$ and $4\,V$ with internal resistances $1\,\Omega$ and $2\,\Omega$ respectively are connected in parallel and joined to an external resistance of $10\,\Omega$ . Calculate the equivalent emf, the equivalent internal resistance, and the current drawn from the combination.
Derive / proveequivalent emf of unequal parallel cells
Two cells of emf $E_1$ and $E_2$ having internal resistances $r_1$ and $r_2$ are connected in parallel across an external resistance $R$ . Derive an expression for the equivalent emf $E_{eq}$ and the equivalent internal resistance $r_{eq}$ of the combination.
Give reasonsseries best when external resistance is large, parallel best when it is small
Account for the following: when several identical cells are required to supply current to a very small external resistance, they should be connected in parallel rather than in series.
Numericalnet emf and current in a series string with one reversed cell
Three cells each of emf $1.5\,V$ and internal resistance $0.5\,\Omega$ are joined in series, but one cell is connected with reversed polarity. Find the net emf of the combination and the current through an external resistance of $3\,\Omega$ .
Define / statecondition for maximum current from a mixed grouping
State the condition under which a mixed (series-parallel) grouping of identical cells delivers maximum current to an external resistance $R$ , expressing it in terms of the total internal resistance of the combination.
Assertion–Reasonemf of identical cells in parallel equals that of a single cell
Assertion: connecting $m$ identical cells in parallel does not change the net emf of the combination. Reason: parallel grouping reduces the effective internal resistance of the cells to $r/m$ . Select the correct option: (a) both assertion and reason are true and the reason is the correct explanation; (b) both are true but the reason is not the correct explanation; (c) assertion true, reason false; (d) assertion false, reason true.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.