## How combination of cells are done?

Sometimes a combination of cells are used in a circuit to get the desired EMF which is not possible by a single cell. These cells are connected in different combinations as –

- Identical cells of same EMF connected in series.
- Different cells of different EMF connected in series.
- Identical cells of same EMF connected in parallel.
- Different cells of different EMF connected in parallel.

__Combination of identical cells in Series__

*In series combination of cells the negative terminal of one cell is connected with the positive terminal of the next cell and so on.*

Consider that ( n ) identical cells ( C_1 ), \ ( C_2 ), \ ........... \ ( C_n ) etc. each of EMF ( E ) and internal resistance ( r ) are connected in series to an external *resistance* ( R ) .

Since, cells are connected in series. So, the total EMF of combination of cells will be –

E_{eff} = n E

Also, all the internal resistances of the cells are in series.

- Therefore, effective internal resistance of the circuit will be –

r_{eff} = nr

This in turn is connected in series with the external resistance ( R ) .

- Therefore,
*equivalent resistance*of circuit will be –

R_{eq} = ( R + nr )

Therefore, current flowing through the *circuit* will be –

I = \left ( \frac {\text {Effective emf}}{\text {Equivalent resistance}} \right ) = \left ( \frac { nE }{ R + nr } \right ) …….. (1)

Thus, combination of cells in series should be used after considering about two points as mentioned below –

- External resistance is very large as compared to the internal resistance of cell.
- External resistance is too small as compared to the internal resistance of cell.

#### 1. When external Resistance is very large

When the external *resistance* in the circuit is very large such that ( R >> nr )

Then, \quad ( R + nr ) \cong R

Hence, equation (1) becomes –

I = \left ( \frac { nE }{ R } \right ) = n \left ( \frac { E }{ R } \right )

But, \left ( \frac { E }{ R } \right ) is the current supplied by one cell.

Therefore, it is useful to use a combination of cells connected in series to increase current in a circuit having very large external resistance as compared to internal resistance of cells. The total current flowing in the circuit becomes ( n ) times of the current supplied by one cell.

#### 2. When external Resistance is very small

When external *resistance* is very small such that ( R << n r )

Then, ( R + nr ) \cong n r

Hence, equation (1) becomes –

I = \left ( \frac { nE }{ nr } \right ) = \left ( \frac { E }{ r } \right ) which is equal to the current supplied by one cell.

Therefore, when a large number of cells are connected to a very small external resistance, then total current flowing in the circuit will be the same as that of current supplied by one cell.

Therefore, it is use less to use a combination of cells connected in series to increase the current in a circuit having very small external resistance as compared to the internal resistance of cells.

**Hence, combination of cells in series is of importance, only if the external resistance in the circuit is very large as compared to total internal resistance of a cells i.e. when ( R >> r ) **

__Combination of different cells in Series__

Consider two cells of EMF ( E_1 ), \ ( E_2 ) and internal resistances of ( r_1 ), \ ( r_2 ) respectively are connected in series.

Since, cells are connected in series so the total EMF of combination of cells will be –

E_{eff} = ( E_1 + E_2 )

Also, the *internal resistance* of the cells are in series. So, the total internal resistance is –

r_{ eq } = ( r_1 + r_2 ) .

Then, current flowing through the circuit will be –

I = \left ( \frac { E_{ eff }}{ r_{ eq }} \right ) = \left ( \frac { E_1 + E_2 }{ r_1 + r_2 } \right ) ……… (2)

__Combination of identical cells in Parallel__

*In a parallel combination of cells, negative terminals of all cells are connected together and in the same way all the positive terminals are connected together.*

A parallel combination of cells is shown in figure.

Consider ( n ) number of identical cells ( C_1 ), \ ( C_2 ), \ ........... \ ( C_n ) , each of EMF ( E ) and internal resistance ( r ) , are connected in parallel to an external resistance ( R ) .

Since, cells are connected in parallel so the total *EMF* of combination of cells will be –

E_{eff} = E

All the *internal resistances* of the cells are connected in parallel with each other. So, their effective internal resistance ( r_{eff} ) is obtained as –

\left ( \frac { 1 }{ r_{eff} } \right ) = \left [ \frac { 1 }{ r } + \frac { 1 }{ r } + \frac { 1 }{ r } + ........ \right ] = \left ( \frac { n }{ r } \right )

- Therefore, effective internal resistance is –

r_{eff} = \left ( \frac { r }{ n } \right )

Also, effective resistance ( r_{eff} ) and external resistance ( R ) are connected in series.

- Therefore, equivalent resistance of the circuit is –

R_{eq} = ( R + r_{eff} ) = \left ( R + \frac { r }{ n } \right )

Then, current flowing through the circuit will be –

I = \left ( \frac {\text {Effective emf}}{\text {Equivalent resistance}} \right ) = \left [ \frac { E }{ \left ( R + \frac { r }{ n } \right ) } \right ] = \left ( \frac { n E }{ nR + r } \right ) ………. (3)

Thus, combination of cells in parallel should be used after considering about two points as mentioned below –

- External resistance is very large as compared to the internal resistance of cell.
- External resistance is too small as compared to the internal resistance of cell.

#### 1. When external Resistance is very large

When external resistance is very large such that ( R >> r ) . Then –

( nR + r ) \cong nR

Hence, equation (3) becomes –

I = \left ( \frac { nE }{ nR + r } \right ) = \left ( \frac { nE }{ nR } \right ) = \left ( \frac { E }{ R } \right )

But, \left ( \frac { E }{ R } \right ) is the current supplied by one cell.

Therefore, it is use less to use a combination of cells connected in parallel to increase the current in a circuit having very large external resistance as compared to the internal resistance of cells.

#### 2. When external Resistance is very small

When external resistance is very small such that ( R << r )

Then ( nR + r ) \cong r

Hence, equation (3) becomes –

I = \left ( \frac { nE }{ nR + r } \right ) = \left ( \frac { nE }{ r } \right ) = n \left ( \frac { E }{ r } \right ) .

But, \quad \left ( \frac { E }{ r } \right ) is equal to the current supplied by one cell.

Therefore, it is useful to use a combination of cells connected in parallel to increase current in a circuit having small external resistance compared to internal resistance of cells. The total current flowing in the circuit becomes ( n ) times of the current supplied by one cell.

**Hence, combination of cells in parallel is of importance, only if the external resistance in the circuit is very small as compared to internal resistance of a cell i.e. when ( R << r )**

__Combination of different cells in Parallel__

Consider two cells of *EMF* ( E_1 ), \ ( E_2 ) and internal resistances ( r_1 ), \ ( r_2 ) respectively are connected in parallel.

If ( I_1 ) and ( I_2 ) are the currents supplied by individual cells then total current in the main circuit will be –

I = ( I_1 + I_2 ) ………. (a)

Since, cells are connected in parallel so the terminal potential difference of combination of cells will be the same. Let it is ( V ) . Then –

For the first cell \quad V = ( E_1 - I_1 r_1 )

Or, \quad I_1 = \left ( \frac {E_1 - V}{ r_1 } \right )

Similarly for the second cell \quad V = ( E_2 - I_2 r_2 )

Or, \quad I_2 = \left ( \frac {E_2 - V}{ r_2 } \right )

Putting the values of ( I_1 ) and ( I_2 ) in equation (a) we get –

I = \left [ \left ( \frac {E_1 - V}{ r_1} \right ) + \left ( \frac {E_2 - V}{ r_2} \right ) \right ]

Or, \quad I = \left ( \frac {E_1 r_2 + E_2 r_1}{ r_1 r_2} \right ) - V \left ( \frac { r_1 + r_2 }{ r_1 r_2} \right )

Therefore, \quad V = \left ( \frac {E_1 r_2 + E_2 r_1 }{ r_1 + r_2 } \right ) - I \left ( \frac { r_1 r_2 }{ r_1 + r_2} \right ) ………. (b)

Let, the parallel grouping of cells is replaced by a single cell of equivalent emf ( E_{ eq } ) and *internal resistance* ( r_{ eq } ) then –

V = ( E_{ eq } - I r_{ eq } ) ……… (c)

By comparing equations (b) and (c) we get –

E_{ eq } = \left ( \frac {E_1 r_2 + E_2 r_1 }{ r_1 + r_2 } \right )

And \quad r_{ eq } = \left ( \frac { r_1 r_2 }{ r_1 + r_2} \right )

Therefore, \left ( \frac { E_{ eq}}{ r_{ eq }} \right ) = \left ( \frac {E_1 r_2 + E_2 r_1 }{ r_1 r_2 } \right ) = \left ( \frac { E_1 }{ r_1 } + \frac { E_2 }{ r_2} \right )

Or, \quad E_{ eq} = \left [ \frac { E_1 }{ r_1 } + \frac { E_2 }{ r_2} \right ] r_{ eq } ………. (4)

See numerical problems based on this article.