8 cells each of internal resistance
Question :- 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,
(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,
(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 
The emf each row = nE, here E = 1.5V
∴ Current in external circuit,
As mnRr = constant, so I will be maximum if,
Hence for maximum current,
So now
From equation (1),
And
n = 2 × 2 = 4
Thus, 4 cells in series in a row and 2 such rows of cells in parallel.
Max. current,
