Grouping of Cells

Grouping of Identical Cells :

(i) Cells in Series

(ii) Cells in Parallel

(iii) Mixed Grouping

Cells in Series:

When ‘n’ identical cells each of EMF ‘E’ and internal resistance ‘r’ are connected in series to an external resistance ‘R’, then

total emf of the combination = n E

effective internal resistance = n r

Total Resistance = R + nr

Current through external resistance

$\large i = \frac{n E}{R + n r}$

If R << n r then $\large i = \frac{E}{r}$ = current from one cell

If R >> n r then  $\large i = \frac{nE}{R}$

Wrongly Connected Cells :

(a)By mistake if ‘m’ cells out of ‘n’ cells are wrongly connected to the external resistance ‘R’

(b)total emf of the combination = (n – 2m)E

(c)total internal resistance = n r

(d)total resistance = R + n r

(e)current through the circuit $\large i = \frac{(n-2m)E}{R+nr}$

Cells in Parallel:

When ‘n’ identical cells each of EMF ‘E’ and internal resistance ‘r’ are connected in parallel to an external resistance ‘R’, then

Total emf of the combination = E

effective internal resistance = r/n

total resistance in the circuit $\large = R + \frac{r}{n}$

Current through the external resistance $\large i = \frac{E}{R + \frac{r}{n}} = \frac{nE}{nR + r}$

If R >> r/n, then i = E/R = current from one cell.

If R << r/n, then i = nE/r

Mixed Grouping:


If n identical cell’s are connected in a row and such m rows are connected in parallel as then

Equivalent emf of the combination Eeq = nE

Equivalent internal resistance of the combination $\large r_{eq} = \frac{nr}{m}$

Main current flowing through the load $\large i = \frac{nE}{R +\frac{nr}{m} } = \frac{m n E}{mR + nr}$

Condition for maximum power $\large R = \frac{nr}{m}$

The current will be maximum if , External Resistance = Total Internal Resistance of the cells

Problem: Three cells of emf 2.0 V , 1.8 V and 1.5 v are connected in series . There internal resistances are 0.05 Ω , 0.7 Ω and 1 Ω respectively . If this battery is connected to an external resistance of 4 Ω , calculate (i) the total current flowing in the circuit . (ii) the potential difference across the terminal of the cell of emf 1.5 V while in use .

Solution: (i) Total emf of the circuit = 2.0 + 1.8 + 1.5 = 5.3 V

Total Resistance of the circuit = 4 + 0.05 + 0.7 + 1 = 5.75 Ω

Current in the circuit , $\displaystyle I = \frac{Total \; emf}{Total \; Resistance}$

$\displaystyle I = \frac{5.3}{5.75} = 0.9 A $

(ii) Potential difference across the terminal of the cell of emf 1.5 V

V = e – I r

V = 1.5 – 0.9 × 1 = 0.6 V

Problem : A battery of emf E and internal resistance r gives a current of 0.5 A with an external resistance of 12 Ω and a current of 0.25 A with an external resistance of 25 Ω . Calculate (i) internal resistance of the cell (ii) emf of the cell

Solution : Let R = External resistance in the series of cell ;

r = Internal resistance of the cell ;

Current in the circuit , $\displaystyle I = \frac{E}{R + r}$

In first case : $\displaystyle 0.5 = \frac{E}{12 + r}$ …(i)

In second case : $\displaystyle 0.25 = \frac{E}{25 + r}$ …(ii)

By solving equation (i) & (ii) we get ,

r = 1 Ω ; E = 6.5 V

 

 Also Read :

→ Electric Current , Drift velocity & Deduction of Ohm’s Law
→ Combination of Resistors
→Concept of EMF & Terminal P.D
→Kirchhoff ‘ s Laws & Solved Examples
→Wheatstone Bridge Principle & Metre Bridge
→Ammeter and Voltmeter
→Potentiometer & its Applications
→Sensitivity of Potentiometer
→Charging of capacitors
→Discharging of Capacitors
→ Heating effect of currents

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