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Cells Connected in Series, Parallel and Mixed Grouping (Class 12 Physics)

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eSaral Academic and Editorial Team

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Summary

Cells connected in series add their EMFs and internal resistances: $E_{eq} = nE$ and $r_{eq} = nr$. In parallel, the EMF stays $E$ and the internal resistance falls to $r/m$. In mixed grouping, with $n$ cells per row and $m$ rows, $E_{eq} = nE$ and $r_{eq} = nr/m$, and the current is maximum when $R = nr/m$. This page gives the formulas, derivations, a comparison table, solved examples and FAQs for Class 12, JEE and NEET.

Cells connected in series, parallel and mixed grouping with the equivalent EMF and internal resistance

A single cell often cannot give the voltage or the current that a circuit needs, so cells are grouped. Cells connected in series raise the voltage, cells connected in parallel lower the internal resistance, and a mixed grouping does both. This topic is part of the Current Electricity chapter of Class 12 Physics, and the JEE Main syllabus lists the combination of cells in series and in parallel. For the chapter questions, see eSaral's NCERT Solutions for Class 12 Physics.

What Is Grouping of Cells?

Grouping of cells means connecting two or more cells so that they act as one equivalent cell with its own EMF and internal resistance. It is similar to the combination of resistances: a group of resistors is replaced by one equivalent resistor, and a group of cells is replaced by one equivalent cell. Before you start, revise the EMF and internal resistance of a cell.

Take $n$ identical cells in a row and $m$ rows, where each cell has EMF $E$ and internal resistance $r$, and the external resistance is $R$.

Arrangement

Equivalent EMF

Equivalent internal resistance

Current through $R$

Series ($n$ cells)

$nE$

$nr$

$I = \dfrac{nE}{R + nr}$

Parallel ($m$ cells)

$E$

$\dfrac{r}{m}$

$I = \dfrac{E}{R + r/m}$

Mixed ($n$ cells per row, $m$ rows)

$nE$

$\dfrac{nr}{m}$

$I = \dfrac{nE}{R + nr/m}$

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Cells Connected in Series: Formula and Derivation

Cells connected in series have the negative terminal of the first cell joined to the positive terminal of the second, and so on. The external resistance is connected between the free terminals of the first and the last cell.

Cells connected in series: n cells joined end to end with an external resistance R

The same current $I$ flows through every cell. The EMFs add, and so do the internal resistances, so the whole group acts as one cell of EMF $nE$ and internal resistance $nr$.

Equivalent circuit for cells connected in series: one cell of EMF nE and internal resistance nr

$$E_{eq} = nE, \qquad r_{eq} = nr$$ $$I = \frac{nE}{R + nr}$$

The potential difference across the external resistance is $V = IR = \dfrac{nER}{R + nr}$.

Special Cases for Cells Connected in Series

The current in cells connected in series depends on how the total internal resistance $nr$ compares with the external resistance $R$.

Condition

Current

Meaning

$nr \ll R$

$I \approx \dfrac{nE}{R}$

The current is $n$ times the current from one cell

$nr \gg R$

$I \approx \dfrac{E}{r}$

The current is the same as the short-circuit current of one cell

So cells connected in series are useful when the external resistance is large compared with the total internal resistance. A common example is a 12 V lead-acid car battery, which has six cells of about 2 V each connected in series.

Cells Connected in Series With Some Cells Reversed

If $p$ cells out of $n$ are connected the wrong way round, their EMFs subtract, but their internal resistances still add:

$$E_{eq} = (n - p)E - pE = (n - 2p)E, \qquad r_{eq} = nr$$ $$I = \frac{(n - 2p)E}{R + nr}$$

Non-Identical Cells Connected in Series

$$E_{eq} = E_1 + E_2 + \ldots = \sum E_i, \qquad r_{eq} = r_1 + r_2 + \ldots = \sum r_i$$ $$I = \frac{\sum E_i}{R + \sum r_i}$$

If a cell is connected in the opposite direction, take its EMF with a negative sign, but keep its internal resistance positive.

Cells Connected in Parallel: Formula and Derivation

Cells are connected in parallel when all the positive terminals are joined at one point and all the negative terminals at another point. The external resistance is connected between these two points.

Cells connected in parallel: m cells with all positive terminals joined and all negative terminals joined

Let $m$ identical cells, each of EMF $E$ and internal resistance $r$, be connected in parallel. The current $I$ divides equally, so each cell gives $I/m$. The terminal potential difference is

$$V = E - \frac{I}{m}\,r$$

Since $V = IR$,

$$IR = E - \frac{Ir}{m} \quad \Rightarrow \quad I = \frac{E}{R + r/m}$$

Equivalent circuit for cells connected in parallel: one cell of EMF E and internal resistance r/m

So the group acts as one cell of EMF $E$ and internal resistance $r/m$.

$$E_{eq} = E, \qquad r_{eq} = \frac{r}{m}$$

Special Cases for Cells Connected in Parallel

Condition

Current

Meaning

$r/m \ll R$

$I \approx \dfrac{E}{R}$

The current is the same as from a single cell

$r/m \gg R$

$I \approx \dfrac{mE}{r}$

The current is $m$ times the short-circuit current of one cell

The parallel combination is useful when the external resistance is small compared with the equivalent internal resistance. Its advantage is the lower internal resistance, and not a higher EMF.

Non-Identical Cells in Parallel

For cells of different EMFs and internal resistances,

$$\frac{1}{r_{eq}} = \sum \frac{1}{r_i}, \qquad E_{eq} = \frac{\sum E_i / r_i}{\sum 1 / r_i}$$

Mixed Grouping of Cells

In mixed grouping, $n$ identical cells are connected in series in each row, and $m$ such rows are connected in parallel. The total number of cells is $p = nm$.

Mixed grouping of cells: m rows of n cells in series, with the rows connected in parallel

Each row has EMF $nE$ and internal resistance $nr$. The $m$ rows are in parallel, so the group acts as one cell of EMF $nE$ and internal resistance $nr/m$.

Equivalent circuit for mixed grouping: one cell of EMF nE and internal resistance nr/m

$$E_{eq} = nE, \qquad r_{eq} = \frac{nr}{m}$$ $$I = \frac{nE}{R + nr/m} = \frac{E}{\dfrac{R}{n} + \dfrac{r}{m}}$$

Condition for Maximum Current in Mixed Grouping

For a fixed total number of cells $p = nm$ and a fixed $R$, the current is maximum when $\dfrac{R}{n} + \dfrac{r}{m}$ is minimum. Put $n = p/m$:

$$\frac{d}{dm}\left(\frac{mR}{p} + \frac{r}{m}\right) = 0 \;\Rightarrow\; \frac{R}{p} - \frac{r}{m^2} = 0 \;\Rightarrow\; \frac{R}{n} = \frac{r}{m}$$

The same result follows from the AM-GM inequality, since $\dfrac{R}{n} + \dfrac{r}{m} \geq 2\sqrt{\dfrac{Rr}{nm}}$, with equality when $\dfrac{R}{n} = \dfrac{r}{m}$.

$$R = \frac{nr}{m}, \qquad I_{max} = \frac{nE}{2R} = \frac{mE}{2r}$$

So the current is maximum when the external resistance equals the total internal resistance of the group. The number of cells per row and the number of rows are then

$$n = \sqrt{\frac{pR}{r}}, \qquad m = \frac{p}{n}$$

Power Delivered to the Load

$$P = I^2R = \frac{E^2R}{\left(\dfrac{R}{n} + \dfrac{r}{m}\right)^2}$$

For a given arrangement, the power in $R$ is maximum when $R = \dfrac{nr}{m}$:

$$P_{max} = \frac{n^2E^2}{4R} = \frac{m^2E^2R}{4r^2} = \frac{nmE^2}{4r}$$

For a fixed number of cells, the same condition $R = nr/m$ gives the maximum current and the maximum power.

Series vs Parallel vs Mixed Grouping: Comparison

Parameter

Series ($n$ cells)

Parallel ($m$ cells)

Mixed ($n \times m$ cells)

Equivalent EMF

$nE$

$E$

$nE$

Equivalent internal resistance

$nr$

$r/m$

$nr/m$

Current

$\dfrac{nE}{R + nr}$

$\dfrac{E}{R + r/m}$

$\dfrac{nE}{R + nr/m}$

Works best when

$R$ is large ($R \gg nr$)

$R$ is small ($R \ll r/m$)

$R = nr/m$

Approximate current in that case

$nE/R$

$mE/r$

$nE/2R$

Typical use

Torch, remote control (cells connected in series)

Sharing the load, more capacity

Battery packs

Reproduce this table from memory before the exam. It covers most of the MCQ variations on cells connected in series, parallel and mixed grouping.

Solved Examples

Example 1: Cells Connected in Series With One Cell Reversed

Six identical cells, each of EMF $2\ \text{V}$ and internal resistance $0.5\ \Omega$, are connected in series, with one cell connected in the opposite direction. Find the current through an external resistance of $4\ \Omega$.

Solution: Here $n = 6$ and $p = 1$.

$$E_{eq} = (n - 2p)E = (6 - 2)\times 2 = 8\ \text{V}, \qquad r_{eq} = nr = 6 \times 0.5 = 3\ \Omega$$ $$I = \frac{8}{4 + 3} = \frac{8}{7} \approx 1.14\ \text{A}$$

Example 2: Internal Resistance in a Parallel Combination

Three identical cells, each of EMF $2\ \text{V}$ and unknown internal resistance, are connected in parallel. The combination is connected to a $5\ \Omega$ resistor, and the terminal voltage across the cells is $1.5\ \text{V}$. Find the internal resistance of each cell.

Solution: For $m = 3$ cells in parallel, $I = \dfrac{E}{R + r/3}$, and the terminal voltage is

$$V = IR = \frac{ER}{R + r/3} \quad \Rightarrow \quad \frac{r}{3} = \frac{R(E - V)}{V}$$ $$\frac{r}{3} = \frac{5(2 - 1.5)}{1.5} = \frac{5}{3} \quad \Rightarrow \quad r = 5\ \Omega$$

Example 3: Best Arrangement in Mixed Grouping

Twenty identical cells, each of EMF $1.5\ \text{V}$ and internal resistance $0.5\ \Omega$, are connected to an external resistance of $2.5\ \Omega$. Find the arrangement for maximum current, and the current.

Solution: For maximum current, $n = \sqrt{\dfrac{pR}{r}} = \sqrt{\dfrac{20 \times 2.5}{0.5}} = 10$, so $m = \dfrac{20}{10} = 2$.

$$I_{max} = \frac{nE}{2R} = \frac{10 \times 1.5}{2 \times 2.5} = 3\ \text{A}$$

The table checks this against the other arrangements of 20 cells:

$n$ (cells per row)

$m$ (rows)

Current

20

1

2.40 A

10

2

3.00 A

5

4

2.40 A

4

5

2.07 A

2

10

1.15 A

1

20

0.59 A

Example 4: Non-Identical Cells in Parallel

A cell of EMF $2\ \text{V}$ and internal resistance $1\ \Omega$ is connected in parallel with a cell of EMF $4\ \text{V}$ and internal resistance $2\ \Omega$. Find the equivalent EMF and internal resistance.

Solution:

$$\frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{2} = 1.5 \quad \Rightarrow \quad r_{eq} = \frac{2}{3}\ \Omega \approx 0.67\ \Omega$$ $$E_{eq} = \frac{2/1 + 4/2}{1/1 + 1/2} = \frac{4}{1.5} \approx 2.67\ \text{V}$$

Practice Questions on Cells Connected in Series, Parallel and Mixed Grouping

  1. Four cells connected in series, each of EMF $1.5\ \text{V}$ and internal resistance $0.5\ \Omega$, are joined to a $5\ \Omega$ resistor. Find the current.
  2. Three cells, each of EMF $2\ \text{V}$ and internal resistance $1\ \Omega$, are connected in parallel to a $1.5\ \Omega$ resistor. Find the current.
  3. Thirty-six cells, each of EMF $2\ \text{V}$ and internal resistance $1\ \Omega$, are used with an external resistance of $4\ \Omega$. Find the arrangement for maximum current, and the maximum current.

Answers: 1. $I = \dfrac{6}{5 + 2} \approx 0.86\ \text{A}$. 2. $I = \dfrac{2}{1.5 + 1/3} \approx 1.09\ \text{A}$. 3. $n = 12$, $m = 3$, and $I_{max} = \dfrac{12 \times 2}{2 \times 4} = 3\ \text{A}$.

Common Mistakes With Cells Connected in Series, Parallel and Mixed Grouping

These mistakes cost marks in tests on cells connected in series and in parallel.

Mistake

Correct approach

Adding the EMFs of cells connected in parallel

The EMF stays $E$ in parallel, and only the internal resistance changes to $r/m$

Using $r/m$ for cells connected in series

In series, the internal resistance is $nr$

Forgetting that a reversed cell still adds internal resistance

$E_{eq} = (n - 2p)E$, but $r_{eq} = nr$

Mixing up $n$ and $m$ in mixed grouping

$n$ is the number of cells in each row (in series), and $m$ is the number of rows (in parallel)

Using $R = nr/m$ for any $R$

It is the condition for the best arrangement of a fixed number of cells, or for maximum power in a given arrangement

All of these are part of the Current Electricity notes for Class 12, JEE and NEET.

Frequently Asked Questions

What are cells connected in series, and what is the equivalent EMF?

Cells are connected in series when the negative terminal of one cell is joined to the positive terminal of the next. For $n$ identical cells, each of EMF $E$ and internal resistance $r$, the equivalent EMF is $nE$ and the equivalent internal resistance is $nr$. The current through an external resistance $R$ is $I = \dfrac{nE}{R + nr}$.

What is the formula for cells connected in series with one cell reversed?

If $p$ cells out of $n$ cells connected in series are reversed, the equivalent EMF is $(n - 2p)E$, and the equivalent internal resistance is still $nr$, because every cell has internal resistance whichever way it is connected. The current is $I = \dfrac{(n - 2p)E}{R + nr}$.

Why does parallel combination not increase the total EMF?

In a parallel combination, all positive terminals are at one potential and all negative terminals are at another, so the potential difference stays equal to that of one cell, $E$. What changes is the internal resistance, which falls to $r/m$. This lets the group supply a larger current to a small external resistance.

What is the equivalent internal resistance of cells connected in parallel?

For $m$ identical cells, each of internal resistance $r$, connected in parallel, the equivalent internal resistance is $r/m$, and the equivalent EMF is $E$. For non-identical cells, $\dfrac{1}{r_{eq}} = \sum \dfrac{1}{r_i}$ and $E_{eq} = \dfrac{\sum E_i/r_i}{\sum 1/r_i}$.

What is the condition for maximum current in mixed grouping?

For a fixed total number of cells $p = nm$ and a fixed external resistance $R$, the current is maximum when $\dfrac{R}{n} = \dfrac{r}{m}$, that is, when $R = \dfrac{nr}{m}$. The external resistance then equals the total internal resistance of the group. The maximum current is $I_{max} = \dfrac{nE}{2R} = \dfrac{mE}{2r}$.

How do you find the number of rows and cells per row for maximum current?

Use $n = \sqrt{\dfrac{pR}{r}}$ and $m = \dfrac{p}{n}$, where $p$ is the total number of cells, $R$ is the external resistance and $r$ is the internal resistance of one cell. For example, 20 cells with $r = 0.5\ \Omega$ and $R = 2.5\ \Omega$ give $n = 10$ and $m = 2$.

How do you decide between series, parallel and mixed grouping of cells?

Use cells connected in series when the external resistance is large compared with $nr$, because the higher EMF drives more current. Use the parallel combination when the external resistance is small compared with $r/m$. When the load is in between, use mixed grouping with $R = nr/m$. eSaral's Current Electricity notes explain this with worked examples.

Is the grouping of cells in the JEE and NEET syllabus?

The JEE Main syllabus lists the combination of cells in series and in parallel under Current Electricity, and NEET Physics also has the Current Electricity chapter. Check the latest official syllabus for the exact topics. eSaral's NCERT Solutions for Class 12 Physics cover the chapter questions.

What is the difference between cells connected in series and in parallel?

In series, the EMFs add ($nE$) and the internal resistances add ($nr$), and the same current flows through every cell. In parallel, the EMF stays $E$ and the internal resistance becomes $r/m$, and the current divides among the cells. Series suits large loads, and parallel suits small loads.

What is the maximum power delivered to the load in mixed grouping?

For a given arrangement, the power in the external resistance is maximum when $R = nr/m$, and it is $P_{max} = \dfrac{n^2E^2}{4R} = \dfrac{nmE^2}{4r}$. At this condition, half of the total power is lost inside the cells.

Why are batteries and power banks made with cells in series-parallel?

Series strings raise the voltage, and parallel strings raise the capacity and the current the pack can supply. Many battery packs therefore use a mixed arrangement, such as several cells in series and several such strings in parallel.

Where can I practise problems on cells connected in series, parallel and mixed grouping?

Start with the solved examples and the practice questions on this page. Then solve the chapter questions with eSaral's NCERT Solutions for Class 12 Physics, and the JEE Main questions on the Current Electricity chapter.

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eSaral Academic and Editorial Team

Team eSaral is the collective author profile for educational content created by eSaral’s teachers and academic contributors. The team draws on expertise from IIT graduates, doctors, experienced educators and subject specialists to develop resources for JEE, NEET and school students. Our articles aim to explain concepts clearly and help students study with confidence.

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