Batteries in Series and Parallel 

Batteries in series are connected end-to-end in such a way that the high potential terminal of one battery connects to the lower potential terminal of the given battery. The total voltage of the series combination is the sum of the voltages of the individual batteries, while the capacity (amp-hour rating) remains the same as that of a single battery.Batteries in parallel are connected with all positive terminals linked together and all negative terminals linked together. This arrangement maintains the same voltage as a single battery but increases the total capacity (amp-hour rating) as the capacities of the individual batteries add up.

1.0Basic Definitions Associated with Cells and Batteries

Cell: A cell is a device that provides the necessary potential difference to maintain a continuous flow of current in an electric circuit. It consists of two electrodes, typically rods or plates, which are immersed in a chemical solution known as the electrolyte. 

2.0Symbol of a Cell

1. EMF(Electromotive Force)(E):The voltage across the terminals of a cell when it is not delivering any current is called EMF of the cell. The EMF of a cell is the energy supplied by the cell per unit charge as it moves through the entire circuit, including the cell itself.

EMF depends on:

• Nature of Electrolyte
• Metal of Electrodes

EMF does not depends on:

• Area of plates
• Distance between the electrodes
• Quantity of Electrolyte
• Size of Cell

2. Internal Resistance of a Cell(r):Resistance offered by the electrolyte of the cell when an electric current flows through it is known as internal resistance.

Dependency of internal resistance r is :

• Distance between two electrodes increases $\Rightarrow r\text{ increases }$
• Area dipped in electrolyte increases $\Rightarrow r\text{ decreases }$
• Concentration of electrolyte increases $\Rightarrow r\text{ increases }$
• Temperature increases $\Rightarrow r\text{ decreases }$

3. Terminal Potential Difference(V)

• When current is drawn through a cell or current is supplied to it then the potential difference across its terminals is called terminal voltage.

• When I is current is drawn from the cell then terminal voltage V is less than its EMF,

V = E – Ir

(a) During Discharging of a Cell:

The current flows inside the cell from the cathode to the anode.

Current,

$I=\frac{E}{R+r}$

E=I R+I r=V+I r

V = E – Ir

• When the cell is in use, the potential difference across it is lower than its electromotive force (EMF).
• As the current drawn from the cell rises, the terminal potential difference decreases. When a substantial current is drawn from a cell, its terminal potential difference decreases significantly.

(b) When Cell is getting Charged:

Current inside the cell is from anode to cathode.

Current $I=\frac{V-E}{r}$

$\Rightarrow V=E+Ir$         

During charging the terminal potential difference is greater than the emf of the cell.

(c) When cell is in the open circuit:

In open circuit $R=\infty$

$\therefore I=\frac{E}{R+r}=0\Rightarrow V=E$

In an open circuit, terminal potential difference is equal to EMF and is the maximum potential difference which a cell can provide.

(d) When cell is short circuited:

In short Circuit R=0 

$I=\frac{E}{R+r}=\frac{E}{r}\text{ and }V=IR=0$

Note: In short circuit, current from the cell is maximum and terminal potential difference is zero.

3.0Combination of Cells in Series     

• When cells are connected in series the total EMF of the series combination is equal to the sum of the EMF of the individual cells and the internal resistance of the cells will also be in series.
• A series combination of cells is used when high EMF is required.

$E_{\text{eq. }}=E_1+E_2+E_3+\dots \dots ..E_n$

$r_{eq.}=r_1+r_2+r_3+\dots \dots .\dots \dots .r_n$

$R_{eq}=r_1+r_2+r_3+\dots \dots \dots ..r_n+R$

Current $i=\frac{E_{eq.}}{r_{eq.}+R}$

If all n cells are identical then

$i=\frac{nE}{nr+R}$

• If $nr\gg R,i=\frac{E}{r}$= Current from any cell when short circuited
• If n r<<R, $i=\frac{nE}{R}=n$ ✕ Current from any one cell,when connected with the external Resistance.

When Cells supports each other

$i=\frac{E_1+E_2}{R+\left(r_1+r_2\right)}$

T.P.D of First Cell =$E_1-i{r}_1$

T.P.D of Second Cell =$E_2-i{r}_2$

When Cells are Connected with Opposite Polarity

Let $E_1>E_2$

$i=\frac{E_1-E_2}{R+\left(r_1+r_2\right)}$

T.P.D of first cell =$E_1-i{r}_1$

T.P.D of second cell =$E_2+i{r}_2$

4.0Combination of Cells in Parallel

• In a Parallel combination of cells, the positive  terminal of one cell is connected to the positive terminal of another cell and the negative terminal of one cell is linked to the negative terminal of the other cell.
• Parallel combination of cells is used when high current is required in the circuit.

$E_{\text{eq. }}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}+\dots \dots }{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\dots .}$

$\frac{1}{r_{eq.}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\dots \dots$

Terminal Potential Difference(V)$=E_{\text{eq. }}-I{r}_{\text{eq. }}.$

When cell support each other

$E_{\text{eq. }}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$

$r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$

When cells are connected with opposite polarity

Let $E_1>E_2$

$E_{eq}=\frac{E_1{r}_2-E_2{r}_1}{r_1+r_2}$

$r_{eq}=\frac{r_1\cdot r_2}{r_1+r_2}$

Note: If n identical cells are connected in parallel and then :-

$E_{\text{eq. }}=n\frac{\frac{E}{r}}{\frac{n}{r}}$

$E_{\text{eq. }}=E$

$r_{eq.}=\frac{r}{n}$

Current in the circuit $i=\frac{E}{R+\frac{r}{n}}=\frac{nE}{nR+r}$

• If $r\ll nR,i=\frac{E}{R}$ = Current from any one cell when connected with external resistance.
• If $r\gg >nR,i=\frac{nE}{r}=n$ ✕ Current from any cell when short-circuited.

5.0Circuit Analysis Involving Combination of Cells

In series ‘n’ identical cells are connected and there are ‘m’ such branches in the circuit then,

• Total number of cells in the circuit =n m
• Internal Resistance of the cells connected in a row =n r
• Total EMF of the cells connected in a row =$E_{\text{net }}=nE$
• Since there are m such rows,
• Total internal resistance of the circuit $r_{net}=\frac{nr}{m}$
• Total EMF of the circuit=Total EMF of the cells connected in a row =$E_{\text{net }}=nE$
• Current in the circuit $I=\frac{E_{net}}{R+r_{net}}=\frac{nE}{R+\frac{nr}{m}}$

Note: Current I in the circuit is maximum when external resistance in the circuit is equal to the total internal resistance of the cells.

Special Case

If in a battery of n identical cells, m cells are wrongly connected.

• Let m be the number of wrongly connected cells.
• Number of cells helping one another =(n-m)
• Total EMF of such cells=(n-m) E
• Total EMF of cells opposing =m E
• Resulting EMF of battery =(n-m) E-m E=(n-2 m) E
• Total internal resistance of cells =n r 

(∴ Resistance remains same irrespective of way of connection of cells.)

• $E_{net}=(n-m)E$ (Condition n>2 m)
• $I_{net}=\frac{(n-2m)E}{R+nr}$

Points to Remember

• At the time of charging a cell, when current is being supplied to it, the terminal voltage is greater than the EMF(E)

V=E+I r    so V>E

•  Series combination is useful when internal resistance of the cell is less than external resistance.
• Parallel combination is useful when internal resistance of the cell is greater than external resistance.
• Internal Resistance of an ideal cell =0 
• If external resistance is zero, then current delivered by the battery is maximum.

6.0Sample Questions on Batteries in Series and Parallel

Q-1.When a resistance of 4 is connected to a cell then a current 2A flows through it. If the cell is connected to a resistance of 9 then current is decreased by 50%.Find the current through the cell if its terminals are directly connected by a connecting wire.

Solution:

$i=\frac{E}{R+r}=2A$

$i=\frac{E}{4+r}$ ……….(1)

$i^{\prime }=\frac{i}{2}=\frac{E}{9+r}$ …….(2)

by $\frac{(1)}{(2)}$

$\frac{2}{1}=\frac{E}{4+r}\times \frac{9+r}{E}$

$8+2r=9+r$ $\Rightarrow r=1\Omega$

By using equation (1)

$E=10\mathrm{V}$

Current through cell during short circuit $=\frac{E}{r}=\frac{10}{1}=10\mathrm{A}$

Q-2.Find out the reading of Voltmeter?

Solution:

Current Flowing in the Circuit, $i=\frac{E}{R}=\frac{10-4}{20+10}=\frac{1}{5}A$

$\Rightarrow V_B-10\times \frac{1}{5}+10=V_A$

$\Rightarrow V_A-V_B=8\mathrm{V}$

Q-3.Find out the reading of Ammeter?

Solution:

$E_{eq.}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}=\frac{2\times 2+2\times 2}{2+2}=2V$

$R_{net}=2+1=3\Omega$

$\Rightarrow I=\frac{2}{3}A$

Q-4. If $V_B-V_A$=4 V, Find the value of x?

Solution:

$V_B-V_A=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$

$\Rightarrow 4=\frac{5x+20}{10+x}$

$\Rightarrow 4(10+x)=5x+20$

$\Rightarrow 40+x=20+5x$

$\Rightarrow x=20\Omega$

Q-5. When a cell is connected to a resistor whose resistance is equal to the cell's internal resistance, the potential difference across the cell's terminals is measured to be 3 volts. What is the EMF of the cell?

Solution:

$I=\frac{E}{R+r}$

$I=\frac{E}{r+r}=\frac{E}{2r}$ ….(1)          (∴R=r)

Potential Drop, V=I r

$I=\frac{V}{r}$ ………..(2)

From equation (1) and (2),we get

$\frac{E}{2r}=\frac{V}{r}\Rightarrow E=2V$

$E=2V=2\times 3=6\mathrm{V}$