A battery is formed by connecting 48 identical cells, each of emf 5 V and internal resistance 3 , in a mixed grouping. The cells are arranged in m parallel rows, each containing n cells connected in series. This battery is connected across an external resistance of 4 . If the values of m and n are chosen such that the power dissipated in the external resistance is maximum, the maximum power dissipated in the external

Step-by-step solution

For a mixed grouping of cells with m parallel rows and n cells in series per row, the equivalent emf and internal resistance are: _ eq = n r_ eq = nr m The power dissipated in the external resistance R is maximum when the external resistance is equal to the equivalent internal resistance of the battery: R = nr m Given R = 4 and r = 3 : 4 = n(3) m 3n = 4m n = 4m 3 The total number of cells is N = mn = 48 . Substituting n : m ( 4m 3 ) = 48 4m^2 = 144 m^2 = 36 m = 6 Thus, n = 4(6) 3 = 8 . The equivalent emf of the bat