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**9. Combinations of cells**

A cell is used to maintain current in an electric circuit. We cannot obtain a strong current from a single cell. Hence need arises to combine two or more cells to obtain a strong current. Cells can be combined in three possible ways:

**(A) **In series, **(B) **In parallel, and **(C) **In mixed grouping.

**(A) Cells in Series**

In this combination, cells are so connected that -ve terminal of each cell is connected with the +ve terminal of next and so on. Suppose n cells are connected in this way. Let e.m.f and internal resistance of each cell are E and r respectively.

Net e.m.f of the cells = nE. Total internal resistance = nr. Hence total resistance of the circuit = nr + R.

If total current in the circuit is I, then ...(1)

**Case (i) : **If nr < < R, then i.e., if total internal resistance of the cells is far less than external resistance, then current obtained from the cells is approximately equal to n times the current obtained from a single cell. Hence cells, whose total internal resistance is less than external resistance, just be joined in series to obtain strong current.

**Case (ii) : **If nr >> R, then i.e., if total internal resistance of the cells is much greater than the external resistance, then current obtained from the combination of n cells is nearly the same as obtained from a single cell. Hence there is no use of joining such cells in series.

**(B) Cells in Parallel**

**(I) When E.M.F's and internal resistance of all the cells are equal : **In this combination, positive terminals of all the cells are connected at one point and negative terminals at other point. Figure shown such cells connected in parallel across some external resistance R. Let e.m.f and internal resistance of each cell are E and r respectively.

Because all the cells are connected in parallel between two points, hence e.m.f of battery = E.

Total internal resistance of the combination of n cells = r/n

Because external resistance R is connected in series with internal resistance, hence total resistance of the circuit = (r/n) R

If current in external resistance is I, then

=

**Case (I) : **If r << R, the i.e., if internal resistance of the cells is much less than external resistance, then total current obtained from combination is nearly equal to current given by one cells only. Hence there is no use of joining cells of low internal resistance in parallel.

**Case (II) : **If r >> R, then i.e., if the internal resistance of the cells is much higher than the external resistance, then total current is nearly equal to n times the current given by one cell. Hence cells of high internal resistance must be joined in parallel to get a strong current.

**(II) When emf's and internal resistance of all the cells connected in parallel are different.**

In this case, total current in external resistance is obtained with the help of Kirchhoff's laws. Figure shows three cells of e.m.f E_{1}, E_{2} and E_{3} and internal resistances r_{1}, r_{2} and r_{3} connected in parallel across some external resistance R. Suppose currents given by three cells are i_{1}, i_{2} and i_{3}. Hence according to Kirchhoff's first law, total current I in external resistance R, is given by

I = i_{1} i_{2} i_{3} ...(1)

Applying Kirchhoff's 2nd law to closed mesh ABEF we get

IR i_{1}r_{1} = E_{1} or i_{1} = ...(2)

Similarly, for closed meshes ABDG and ABCH, we get

....(3)

and ....(4)

Substituting eq. (2), (3) and (4) in eq. (1), we have

or or

If n cells are joined in parallel, then and E_{eq}., r_{eq}.

**(C) Cells in Mixed Grouping**

In this combination, a certain number of cells are joined in series in various rows, and all such rows are then connected in parallel with each other.

Suppose n cells, each of e.m.f E and internal resistance r, are connected in series in every row and m such rows are connected in parallel across some external resistance R, as shown in figure.

Total number of cells in the combination = mn. As e.m.f. of each row = nE and all the rows are connected in parallel, hence net e.m.f of battery = nE.

Internal resistance of each row = nr. As m such rows are connected in parallel, hence total internal resistance of battery =

Hence total resistance of the circuit =

If the current in external resistance is I, then

It is clear from above equation that I will be maximum when

This will be possible when the quantity [Ã–nr - Ã–mR]^{2} is minimum. Because this quantity is in square, it can not be negative, hence its minimum value will be equal to zero, i.e.,

mR = nr or R =

**i.e., In mixed grouping of cells, current in external resistance will be maximum when total internal resistance of battery is equal to external resistance.**

** Because power consumed in the external resistance or load = I ^{2}R, hence when current in load is maximum, consumed power in it is also maximum, Hence consumed power in the load will also be maximum when R = .**

**Ex.30 Find the current in the loop.**

**Sol. **The given circuit can be simplified as

= â‡’

**Ex.31 Find the emf and internal resistance of a single battery which is equivalent to a combination of three batteries as shown in figure.**

**Sol. **

Battery (B) and (C) are in parallel combination with opposite polarity. So, their equivalent

r_{BC }= 1 W

Now,

r_{ABC} = 2Î© **Ans.**

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