How would you arrange 48 cells each of e.m.f \(2 \mathrm{~V}\) and internal resistance \(1.5 \Omega\) so as to pass maximum current through the external resistance of \(2 \Omega\) ? (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups

We have four possible arrangements of cells: (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups In each arrangement, we will calculate the equivalent e.m.f (E) and internal resistance (r) for each group of cells.

For cells connected in series, the equivalent E is the sum of the e.m.f of each cell, whereas the equivalent r is the sum of the internal resistances of all cells in the series. For cells connected in parallel, the sum of the reciprocals of E and r remains the same. (A) 2 cells in 24 groups: Here, 2 cells are connected in series and each group is connected in parallel. E = 2 + 2 = 4 V (E of cells connected in series) r = 1.5 + 1.5 = 3 Ω (r for cells connected in series) Equivalent E for the 24 groups connected in parallel = \(24 \times 4 = 96 \mathrm{~V}\) Equivalent r for the 24 groups connected in parallel = \(3/24 = 0.125 \mathrm{~ \Omega}\) (B) 4 cells in 12 groups: Here, 4 cells are connected in series and each group is connected in parallel. E = 2 + 2 + 2 + 2 = 8 V (E of cells connected in series) r = 1.5 + 1.5 + 1.5 + 1.5 = 6 Ω (r for cells connected in series) Equivalent E for the 12 groups connected in parallel = \(12 \times 8 = 96 \mathrm{~V}\) Equivalent r for the 12 groups connected in parallel = \(6/12 = 0.5 \mathrm{~ \Omega}\) (C) 8 cells in 6 groups: Here, 8 cells are connected in series and each group is connected in parallel. E = 8 * 2 = 16 V (E of cells connected in series) r = 8 * 1.5 = 12 Ω (r for cells connected in serie)s Equivalent E for the 6 groups groups connected in parallel = \(6 \times 16 = 96 \mathrm{~V}\) Equivalent r for the 6 groups connected in parallel = \(12/6 = 2 \mathrm{~ \Omega}\) (D) 3 cells in 16 groups: Here, 3 cells are connected in series and each group is connected in parallel. E = 2 + 2 + 2 = 6 V (E of cells connected in series) r = 1.5 + 1.5 + 1.5 = 4.5 Ω (r for cells connected in series) Equivalent E for the 16 groups connected in parallel = \(16 \times 6 = 96 \mathrm{~V}\) Equivalent r for the 16 groups connected in parallel = \(4.5/16 = 0.28125 \mathrm{~ \Omega}\)

We have the external resistance R = 2 Ω. We will use Ohm's law (I = E / (R + r)) to calculate the current for each arrangement. (A) I = \(96 / (2 + 0.125) = 96 / 2.125 = 45.18 \mathrm{~A}\) (B) I = \(96 / (2 + 0.5) = 96 / 2.5 = 38.4 \mathrm{~A}\) (C) I = \(96 / (2 + 2) = 96 / 4 = 24 \mathrm{~A}\) (D) I = \(96 / (2 + 0.28125) = 96 / 2.28125 = 42.09 \mathrm{~A}\)

Comparing the currents for each arrangement, we see that the maximum current is achieved in arrangement (A) with 45.18 A. Therefore, the best arrangement to pass the maximum current through the external resistor of 2 Ω is 2 cells in 24 groups.