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Mixed grouping of cells | Mixed grouping of identical cells

Mixed grouping of cells | Mixed grouping of identical cells :- There are three types of grouping of identical cells :-

Series Grouping of cells
Parallel Grouping of cells
Mixed Grouping of cells

(1) Series Grouping of cells

(Mixed grouping of cells | Mixed grouping of identical cells)

In series grouping of identical cells, several identical cells are connected end-to-end in an electrical circuit where the positive terminal of one cell is connected to the negative terminal of the next cell, and so on. This setup is typically used to increase the total voltage output of the battery pack.

Consider ‘n’ cells each of emf ε and internal resistance r are connected in series to an external resistor R as shown in figure below :

Equivalent emf of n cells in series,

$\displaystyle {{\varepsilon }_{eq}}=\varepsilon +\varepsilon +.....upto\text{ }n\text{ }terms$

$\displaystyle \Rightarrow {{\varepsilon }_{eq}}=n\varepsilon$

Equivalent internal resistance of n cells in series,

$\displaystyle {{r}_{eq}}=r+r+.....upto\text{ }n\text{ }terms$

$\displaystyle \Rightarrow {{r}_{eq}}=nr$

Total resistance of the circuit = (nr + R)

Current flowing in the circuit,

$\displaystyle I=\frac{n\varepsilon }{nr+R}$ …..(1)

Special cases :-

Case (i). If R << nr, then R can be neglected in comparison to nr. From equation (1) we get,

$\displaystyle I=\frac{n\varepsilon }{nr}=\frac{\varepsilon }{r}$

Thus same current flows through the external resistor as the current due to a single cell.

Case (ii). If R >> nr, then nr can be neglected in comparison to R. From equation (1) we get,

$\displaystyle I=n\left( \frac{\varepsilon }{R} \right)$

Thus, the current in the external resistor is n times the current due to a single cell.

Hence, we conclude that

Series combination of cells is advantageous when R >> nr, then $\displaystyle I=n\left( \frac{\varepsilon }{R} \right)$
Series combination of cells is not advantageous when R << nr, then $\displaystyle I=\frac{\varepsilon }{r}$ .

In short using series combination of identical cells, the current drawn is maximum if the value of external resistor is very high as compared to the total internal resistance of the cells.

(2) Parallel Grouping of cells

(Mixed grouping of cells | Mixed grouping of identical cells)

Parallel grouping of cells involves connecting the positive terminals of multiple cells together and the negative terminals together to get more current. Consider ‘m’ cells each of emf ε and internal resistance r are connected in parallel and this combination is connected to an external resistor R as shown in figure below :

In parallel grouping of cells the equivalent emf is same as that of a single cell, hence

Equivalent emf of m cells in parallel = emf of a single cell = ε

Equivalent internal resistance of m cells in parallel,

$\displaystyle \frac{1}{{{r}_{p}}}=\frac{1}{r}+\frac{1}{r}+.....upto\text{ }m\text{ }terms$

$\displaystyle {{r}_{p}}=\frac{r}{m}$

As R and r_p are in series so,

Total resistance of the circuit = R + r/m

∴ Current in the external resistance R is given by,

$\displaystyle I=\frac{\varepsilon }{R+r/m}=\frac{m\varepsilon }{mR+r}$ …..(2)

Special cases :-

Case (i). If R << r/m, then R can be neglected in comparison to r/m. From equation (2) we get,

$\displaystyle I=m\left( \frac{\varepsilon }{r} \right)$

Thus, the current in the external resistor is m times the current due to a single cell.

Case (ii). If R >> r/m, then r/m can be neglected in comparison to R. From equation (2) we get,

$\displaystyle I=\frac{\varepsilon }{R}\,$

Thus same current flows through the external resistor as the current due to a single cell.

Hence, we conclude that

Parallel combination of cells is advantageous when R << r/m (or r >> mR), then $\displaystyle I=m\left( \frac{\varepsilon }{r} \right)$
Parallel combination of cells is not advantageous when R >> nr (or r << mR), then $\displaystyle I=\frac{\varepsilon }{R}\,$ .

In short using parallel combination of identical cells, the current drawn is maximum if the value of external resistor is very low as compared to the total internal resistance of the cells.

(3) Mixed Grouping of cells

(Mixed grouping of cells | Mixed grouping of identical cells)

If some cells are connected in series and to form a row and a number of such rows of cells are connected in parallel, then this type of grouping of cells is called mixed grouping. This approach allows for customization of the voltage to get maximum power.

Let n identical cells each of emf ε and internal resistance r in a row are connected in series and m such rows are connected in parallel. This arrangement is connected to an external resistor R as shown in figure below :-

Total emf of each row (of whole combination) = nε

Total internal resistance of each row = nr

As m identical rows of cells are connected in parallel, therefore,

Total internal resistance (r_p) of all the cells is given by,

$\displaystyle \frac{1}{{{r}_{p}}}=\frac{1}{nr}+\frac{1}{nr}+.....upto\text{ }m\text{ }terms$

$\displaystyle {{r}_{p}}=\frac{nr}{m}$

Total resistance of the circuit = R + nr/m

The current in the external resistor R is given by,

$\displaystyle I=\frac{n\varepsilon }{R+nr/m}=\frac{mn\varepsilon }{mR+nr}$

Note :-

(1). The current in the mixed grouping of cells will be maximum when (mR + nr) is minimum, i.e.,

$\displaystyle \left[ {{\left( \sqrt{mR} \right)}^{2}}+{{\left( \sqrt{nr} \right)}^{2}}-2\sqrt{mnRr} \right]+2\sqrt{mnRr}=\text{minimum}$

$\displaystyle \Rightarrow {{\left( \sqrt{mR}-\sqrt{nr} \right)}^{2}}+2\sqrt{mnRr}=\text{minimum}$

Here $\displaystyle 2\sqrt{mnRr}$ cannot be zero, so I will be maximum when :-

$\displaystyle {{\left( \sqrt{mR}-\sqrt{nr} \right)}^{2}}=0$

$\displaystyle \sqrt{mR}-\sqrt{nr}=0$

$\displaystyle \Rightarrow mR=nr$

$\displaystyle \Rightarrow R=\frac{nr}{R}$

Hence current drawn by mixed grouping of cells will be maximum if the value of external R is equal to the total internal resistance of all the cells.

(2). When one cell is wrongly connected in series combination of n identical cells, each emf ε, then total emf is reduced by 2ε.

so the effective emf = (nε – 2ε)

Here,

$\displaystyle {{V}_{B}}-\varepsilon +\left[ \varepsilon +\varepsilon +.......(n-1)\text{ }terms \right]={{V}_{A}}$

$\displaystyle \Rightarrow \left( {{V}_{A}}-{{V}_{B}} \right)=\left( n\varepsilon -2\varepsilon \right)$

(3). When n cells each of internal resistance r are connected wrongly in series, then total internal resistance of the combination = nr. This is because there is no effect of order of cells on total resistance in series combination.

(4). The series combination of cells is used to get more voltage, parallel combination is used to get more current and mixed grouping of cells is used to get more power.

Example.

8 cells each of internal resistance 0.5 Ω and emf 1.5V are used to send a current through an external resistor of (a) 200 Ω (b) 0.002 Ω (c ) 1.0 Ω. How would you arrange them to get the maximum current in each case? Find the value of current in each case.

Solution :

Here, total number of cells = 8,

r = 0.5 Ω and ε = 1.5 V

(a). When external resistor R = 200 Ω, then R >> r, so for maximum current, the cells are to be connected in series in the circuit.

Total internal resistance of 8 cells = 8r

Current in circuit,

$\displaystyle I=\frac{8\varepsilon }{R+8r}=\frac{8\times 1.5}{200+8\times 0.5}=\frac{12}{204}=0.059A$

(b) When R = 0.002 Ω, then R << r, so for maximum current, the cells are to be connected in parallel in the circuit.

Total internal resistance of 8 cells = r / 8

Total resistance of circuit = R + r / 8 = 0.002 + 0.5 / 8 = 0.0645 Ω

Effective emf of all cells = emf of each cell = 1.5 V

Current in circuit,

$\displaystyle I=\frac{1.5}{0.0645}=23.26A$

(c ) When R = 1.0 Ω, then R is comparable to r. For maximum current, the cells are to be connected in mixed grouping.

Let there be m rows of cells in parallel with n cells in series, in each row. Then,

Total number of cells, mn = 8 …..(1)

Total internal resistance of combination = nr/m

∴ Total resistance in the circuit $\displaystyle =\left( R+\frac{nr}{m} \right)$ , here R = 1Ω

The emf each row = nE, here E = 1.5V

∴ Current in external circuit,

$\displaystyle I=\frac{nE}{R+\frac{nr}{m}}=\frac{mnE}{mR+nr}=\frac{mnE}{{{\left( \sqrt{mR}-\sqrt{nr} \right)}^{2}}+2\sqrt{mnRr}}$

As mnRr = constant, so I will be maximum if,

$\displaystyle {{\left( \sqrt{mR}-\sqrt{nr} \right)}^{2}}=0$

$\displaystyle \Rightarrow mR=nr$

Hence for maximum current,

$\displaystyle \Rightarrow R=\frac{nr}{m}$

So now

$\displaystyle 1.0=\frac{0.5n}{m}\Rightarrow n=2m$

From equation (1),

$\displaystyle m\times 2m=8\Rightarrow m=2$

And

n = 2 × 2 = 4

Thus, 4 cells in series in a row and 2 such rows of cells in parallel.

Max. current,

$\displaystyle I=\frac{n\varepsilon }{R+\frac{nr}{m}}=\frac{mn\varepsilon }{mR+nr}=\frac{8\times 1.5}{2\times 1+4\times 0.5}=3A$

Mixed grouping of cells | Mixed grouping of identical cells

(1) Series Grouping of cells

(Mixed grouping of cells | Mixed grouping of identical cells)

(2) Parallel Grouping of cells

(Mixed grouping of cells | Mixed grouping of identical cells)

(3) Mixed Grouping of cells

(Mixed grouping of cells | Mixed grouping of identical cells)

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