12 cells, each of emf 1.5 V and internal resistance `0.5Omega` are arranged in m rows each containing n cells connected series as shown in figure. Calculate the values of n and m for which this combination would send maximum current through an external resistance of 1.5 `Omega` .

For the circuit shown here, calculate the potential difference between points B and C.

Explanation

Suppose m calls are arranged in m rows each containing n cells connected in series, then
`m xx = 12`
Total internal resistance = Total extermal resistance
`(nr)/(m) = R`
`(n xx 0.5)/(m ) = 1.5 `
n =3 m
On Solving Eqs (i) and (ii)
m =2 and n =6
Therefore 12 cells should be connect in two rows .

(i) Applying kirchhoff first law at junction D we get
`I = I_(1) +I_(2)`
Applying kirchhoff.s second law to loop ACBA
`2I + I + 2I_(1) =2 -1 `
`3I + 2 I_(1) =1 `
Applying kirchhoff.s second law to loop CDBC
`3I_(2) +I_(2)-2I_(1) =3-1 `
`4I_(2) - 2I_(1) =2`
On solving Eqs (i) (ii) and (iii) we get
`I_(1)= (1)/(13) A , I_(2) = (6)/(13)` A .
And `I = (5)/(13)` A
Potential difference between the points B and C
`V_(1) = I_(1) xx 2 = (2)/(13)` V