Cells Connected in Series, Parallel & Mixed Grouping — Class 12 Physics

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What Is Grouping of Cells in a Circuit?

In real circuits — whether in a torch, a car battery bank, or a lab experiment — a single cell rarely provides enough current or voltage. That is why cells are grouped together, either in series, in parallel, or in a combination of both (mixed grouping).

This topic is a high-weightage section of Class 12 Physics, Chapter 3 — Current Electricity, and appears regularly in JEE Main, JEE Advanced, and NEET question papers. According to the NTA syllabus, students must be able to derive equivalent EMF and internal resistance for all three configurations and apply them to solve circuit problems.

The core idea mirrors what you already know about resistors: just as resistors combine to give an equivalent resistance, a group of cells can be replaced by one equivalent cell. Understanding the condition for maximum current — especially in mixed grouping — is one of the most frequently tested concepts at the JEE level.

For step-by-step NCERT solutions on this chapter, refer to NCERT Solutions Class 12 Physics on eSaral.

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Cells Connected in Series Combination

Series Combination Formula and Derivation

Cells are said to be connected in series when the negative terminal of one cell is connected to the positive terminal of the next. The external resistance is connected between the free terminals of the first and last cells.

Setup: n identical cells, each with EMF = E and internal resistance = r.

Equivalent single cell:

• Equivalent EMF: E_eq = nE
• Equivalent internal resistance: r_eq = nr

Current through the circuit:

I = nE / (R + nr)

where R is the external load resistance.

Key Results for Series Combination

If cells are non-identical:

E_eq = ΣE_i     r_eq = Σr_i

I = ΣE_i / (R + Σr_i)

When Is Series Combination Useful?

Series combination is most effective when the external resistance R is large compared to the total internal resistance nr. Under this condition, the series arrangement genuinely multiplies the available voltage and drives significantly more current than a single cell could.

In JEE problems, always check whether cells are identical before applying the series formula. If one cell is reversed, subtract its EMF contribution but add its internal resistance — a common mistake that costs marks. Students in eSaral's JEE batch who practise this distinction consistently score full marks on grouping-of-cells MCQs.

Cells Connected in Parallel Combination

Parallel Combination Formula and Derivation

Cells are said to be connected in parallel when all positive terminals are joined at one node and all negative terminals at another. The external resistor connects these two nodes.

Setup: m identical cells, each with EMF = E and internal resistance = r.

Equivalent single cell:

• Equivalent EMF: E_eq = E (same as one cell)
• Equivalent internal resistance: r_eq = r/m

Current through the circuit:

I = E / (R + r/m)

Key Results for Parallel Combination

When Is Parallel Combination Useful?

Parallel combination is best suited when the external resistance R is small. The key advantage is that it reduces the effective internal resistance to r/m, allowing more current to be delivered to a low-resistance load without excessive internal losses.

This is exactly why batteries in power banks or UPS systems are often connected in parallel — they share the current load equally, extending both capacity and lifespan.

Mixed Grouping of Cells

In mixed grouping, n cells are connected in series per row, and m such rows are connected in parallel.

Total cells = p = n × m

Equivalent single cell:

• Equivalent EMF: E_eq = nE
• Equivalent internal resistance: r_eq = nr/m

Current through the circuit:

I = nE / (R + nr/m) = E / (R/n + r/m)

Power delivered to load:

P = E²R / (R/n + r/m)²

What Is the Condition for Maximum Current in Mixed Grouping?

Current (and power to load) is maximum when the denominator (R/n + r/m) is minimised.

R/n = r/m

R = nr/m

This means maximum current flows when external resistance equals the total internal resistance of the combination.

I_max = nE / 2R = mE / 2r

P_max = n²E² / 4R = m²E²R / 4r²

Mixed grouping is preferred in practical and exam scenarios because maximum current and maximum power transfer occur under the same condition (R = nr/m). In pure series or pure parallel configurations, these optima are separate. Faculty at eSaral — trained at IIT Bombay — specifically flag this point as a high-probability JEE Advanced reason-based question.

Series vs Parallel vs Mixed — Quick Comparison Table

This table is a quick-revision tool. Before any exam, reproduce it from memory — it covers almost every MCQ variation on this topic. You can also find related practice problems in the NCERT Solutions for Class 12 Physics chapter on Current Electricity.

Solved Examples

Example 1: Finding Internal Resistance in Parallel Combination

Question: Three identical cells, each of EMF 2V, are connected in parallel and then connected to a 5 Ω resistor. If the terminal voltage across the cells is 1.5 V, find the internal resistance of each cell.

Solution

For m = 3 cells in parallel:

I = E / (R + r/3)     and     V = IR

Terminal voltage:

V = ER / (R + r/3)

Rearranging:

r/3 = R(E − V)/V

Substituting values:

r/3 = 5(2 − 1.5)/1.5 = 5/3

r = 5 Ω

Answer: Internal resistance of each cell = 5 Ω

Example 2: Applying Kirchhoff's Laws to a Multi-Cell Circuit

Question: Determine currents I₁, I₂, and I₃ in the circuit (as described in the original figure with EMFs of 24V and 27V, and resistances of 2Ω, 6Ω, 4Ω).

Solution

Using Kirchhoff's Voltage Law (KVL) in loop abcda:

-2I₁ + 24 - 27 - 6I₂ = 0

2I₁ + 6I₂ = -3    ...(1)

Using KVL in loop cdfec:

-27 - 6I₂ + 4I₃ = 0

-6I₂ + 4I₃ = 27    ...(2)

Using Kirchhoff's Current Law (KCL) at junction c:

I₁ = I₂ + I₃    ...(3)

Substituting equation (3) into equation (2):

4I₁ - 10I₂ = 27    ...(4)

Solving equations (1) and (4):

I₁ = 3 A, I₂ = -1.5 A, I₃ = 4.5 A

The negative value of I₂ means its actual direction is opposite to the assumed direction in the figure.

For more practice problems on Kirchhoff's Laws and Current Electricity, explore the full NCERT Solutions Class 12 library on eSaral.

Like resistors, a circuit can also have a combination of cells connected in series, parallel or even both! And again like resistors, we can calculate current and voltages that can be replaced by an equivalent cell in a circuit which we will learn here. 

A circuit composed solely of components connected in series is known as a series circuit; likewise, one connected completely in parallel is known as a parallel circuit.

In a series circuit, the current that flows through each of the components is the same, and the voltage across the circuit is the sum of the individual voltage drops across each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component. [source]

Here we will see cells connected in series, parallel and mixed one by one along with examples.

Contents:

1. Cells connected in Series Combination
2. Cells connected in Parallel Combination
3. Cells connected in Mixed Grouping
4. Solved Examples

Cells connected in Series Combination

The cells are said to be connected in series if negative terminal first is connected to positive of second whose negative terminal is connected to positive of third cell. The external resistance is connected between free terminals of first and last cells.

[caption id="attachment_4615" align="aligncenter" width="255"] Cells connected in Series Combination[/caption]

Let 'n' identical cells each of emf E and internal resistance r be connected in series. The combination can be replaced by a single cell of emf nE and internal resistance nr. The current flowing through load ${\rm{I}} = {{{\rm{nE}}} \over {{\rm{R}} + {\rm{nr}}}}$

[caption id="attachment_4616" align="aligncenter" width="205"] Equivalent circuit for series combination[/caption]

1. 
   
   
   
   
   
   
   
2. 1. If ${\rm{nr}} < < {\rm{R}}$ then ${\rm{I}} = {{{\rm{nE}}} \over {\rm{R}}}$ If equivalent internal resistance nr is less than external resistance R then current in circuit is equal to n times circuit current due to single cell.
2. 1. If nr >> R then I = E/r. If equivalent internal resistance nr is greater than external resistance R then current in circuit is equal to short circuited current obtained from one cell.
2. 1. Maximum current can be drawn from series combination of cells if external resistance is very large as compared to equivalent internal resistance.
2. 1. If in series combination of n cells p cells are reversed than equivalent emf${E_{{\rm{eq}}}} = ({\rm{n}} - {\rm{p}}){\rm{E}} - {\rm{pE}} = ({\rm{n}} - 2{\rm{p}}){\rm{E}}\quad $ and $\quad {{\rm{r}}_{{\rm{eq}}}} = {\rm{nr}}\quad $
      
      So current ${\rm{I}} = {{({\rm{n}} - 2{\rm{p}}){\rm{E}}} \over {{\rm{nr}} + {\rm{R}}}}$
2. 1. If un-identical cells are connected in series then${E_{{\rm{eq}}}} = {{\rm{E}}_1} + {{\rm{E}}_2} + \ldots . = \Sigma {{\rm{E}}_{\rm{i}}}$ and ${r_{eq}} = {r_1} + {r_2} + \ldots . = \Sigma {r_i}$
      
      So current ${\rm{I}} = {{\Sigma {{\rm{E}}_{\rm{i}}}} \over {{\rm{R}} + \Sigma {{\rm{r}}_{\rm{i}}}}}$.

Cells connected in Parallel Combination

The cells are said to be connected in parallel if positive terminals of all the cells are connected together at one point and their negative terminals at another point. The external resistor is connected between these two points.

[caption id="attachment_4617" align="aligncenter" width="237"] Cells connected in Parallel Combination[/caption]

Let ‘m’ identical cells each of emf E and internal resistance r be connected in parallel. The combination can be replaced by a single cell of emf $E$ and internal resistance ${\rm{r}}/{\rm{m}}.$ The current ${\rm{I}} = {{\rm{E}} \over {{\rm{R}} + {\rm{r}}/{\rm{m}}}}$

[caption id="attachment_4618" align="aligncenter" width="218"] Equivalent circuit for parallel combination[/caption]

1. If ${\rm{r}}/{\rm{m}} < < {\rm{R}}$ then ${\rm{I}} = {{\rm{E}} \over {\rm{R}}}$ . If equivalent internal resistance ${\rm{r}}/{\rm{m}}$ is less than external resistance ${\rm{R}}$ then current in circuit is equal to current produced by a single cell.

1. If ${\rm{r}}/{\rm{m}} > > {\rm{R}}$ then ${\rm{I}} = {{{\rm{mE}}} \over {\rm{r}}}$. If equivalent internal resistance ${{\rm{r}} \over {\rm{m}}}$ is greater than external resistance then current in circuit is equal to ${\rm{m}}$ times the current produced by a short circuited cell.

1. Thus maximum current can be drawn from parallel combination of cells if external resistance is small as compared to net internal resistance of cells.

Mixed Grouping

Let there be $n$ identical cells in one row and $m$ rows of cells in parallel. The combination of cells can be replaced by a single cell of emf nE and internal resistance ${{{\rm{ nr }}} \over m}$ .

The current $I = {{nE} \over {R + {{nr} \over m}}} = {E \over {{R \over n} + {r \over m}}}$ where $\quad {\rm{n}} \times {\rm{m}} = {\rm{p}} = $ total number of cells.

[caption id="attachment_4619" align="aligncenter" width="279"] Cells connected in Mixed Grouping[/caption]

[caption id="attachment_4620" align="aligncenter" width="165"] Equivalent circuit for Mixed Grouping[/caption]

1. 
   
2. 1. The current in the circuit is maximum when ${{\rm{R}} \over {\rm{n}}} + {{\rm{r}} \over {\rm{m}}}$ is minimum
      
      So$\quad {{\rm{d}} \over {{\rm{dm}}}}\left( {{{\rm{R}} \over {\rm{n}}} + {{\rm{r}} \over {\rm{m}}}} \right) = 0\quad $
      
      or $\quad {{\rm{d}} \over {{\rm{dm}}}}\left( {{{{\rm{mR}}} \over {\rm{p}}} + {{\rm{r}} \over {\rm{m}}}} \right) = 0\quad $
      
      or $\quad {{\rm{R}} \over {\rm{p}}} - {{\rm{r}} \over {{{\rm{m}}^2}}} = 0$
      
      or $\quad {{\rm{R}} \over {\rm{p}}} = {{\rm{R}} \over {{\rm{mn}}}} = {{\rm{r}} \over {{{\rm{m}}^2}}}\quad $
      
      or $\quad {{\rm{R}} \over {\rm{n}}} = {{\rm{r}} \over {\rm{m}}}$
      
      In mixed grouping of cells current in circuit is maximum if ${{\rm{R}} \over {\rm{n}}} = {{\rm{r}} \over {\rm{m}}}\quad $ and $\quad {{\rm{I}}_{\max }} = {{{\rm{nE}}} \over {2{\rm{R}}}} = {{{\rm{mE}}} \over {2{\rm{r}}}}$
2. 1. In mixed grouping of cells power transferred to the load is maximum where external resistance R is equal to total internal resistance or $R = {{{\rm{ nr }}} \over m}\quad $ or $\quad {R \over n} = {r \over m}$. This shows power transfer is maximum when current is maximum.$P = {{{E^2}R} \over {{{\left( {{R \over n} + {r \over m}} \right)}^2}}}\quad $ and $\quad {P_{\max }} = {{{n^2}{E^2}} \over {4R}} = {{{m^2}{E^2}R} \over {4{r^2}}}$

1. In mixed combination of cells, current in circuit and power transferred to load become maximum under same condition. This is why it is preferred over series and parallel combination of cells.

Solved Examples

[esquestion]Three identical cells each of emf 2V and unknown internal resistance are connected in parallel. This combination is connected to a 5$\Omega $ resistor. If the terminal voltage across the cells is 1.5V. What is the internal resistance of each cell?

#sol#In parallel combination of three cells ${\rm{I}} = {{\rm{E}} \over {{\rm{R}} + {\rm{r}}/3}}$

Terminal potential difference $V = IR = {{ER} \over {R + r/3}}\quad $ or $\quad {r \over 3} = {{ER - RV} \over V}$

or $\quad r = 3\left( {{{ER - RV} \over V}} \right) = 3\left( {{{2 \times 5 - 5 \times 1.5} \over {1.5}}} \right) = 5\Omega $

[/esquestion]

[esquestion]Determine currents ${{\rm{I}}_1},{{\rm{I}}_2}$ and ${{\rm{I}}_3}?$

#sol#Using II law in loop abcda $ - 2{{\rm{I}}_1} + 24 - 27 - 6{{\rm{I}}_2} = 0\quad $ or $\quad 2{{\rm{I}}_1} + 6{{\rm{I}}_2} = - 3$ $ \ldots ..(1)$

Using II law in loop cdfec we get $\quad - 27 - 6{{\rm{I}}_2} + 4{{\rm{I}}_3} = 0\quad $ or $\quad - 6{{\rm{I}}_2} + 4{{\rm{I}}_3} = 27\quad \ldots \ldots (2)$

From junction rule at c we get $\quad {{\rm{I}}_1} = {{\rm{I}}_2} + {{\rm{I}}_3}$ $ \ldots ..(3)$

From 2 and 3 we get $\quad - 6{{\rm{I}}_2} + 4\left( {{{\rm{I}}_1} - {{\rm{I}}_2}} \right) = 27\quad $ or $\quad 4{{\rm{I}}_1} - 10{{\rm{I}}_2} = 27\quad \ldots \ldots $ (4)

Solving 1 and 4 we get ${{\rm{I}}_1} = 3{\rm{A}},\quad {{\rm{I}}_2} = - 1.5{\rm{A}}\quad $ so ${{\rm{I}}_3} = {{\rm{I}}_1} - {{\rm{I}}_2} = 4.5{\rm{A}}$

Negative ${{\rm{I}}_2}$ means the direction should have been opposite to shown in figure.

[/esquestion]     Also Read:

• Types and Effects of Electric Current
• Ohm’s Law and Resistance
• Combination of Resistances
• EMF and Internal Resistances of a Cell
• Cells Connected in Series, parallel and Mixed
• Kirchhoff’s Circuit Law
• Electric Currents in Conductors
• Wheatstone Bridge
• Post office Box
• Wheatstone Meter Bridge
• Moving Coil galvanometer
• Ammeter and Voltmeter
• Potentiometer Working Principle