Question

20 cell of internal resistance `0.5 Omega` and emf `1.5 V` are used to sent a current through an external resistance of (i) `500 Omega` (ii) `0.005 Omega` (iii) `2.5 Omega`. How would you arrange then to get the maximum current in each case ? Find the value of current in each case

Answer 1

Correct Answer - (i) `059A`(ii)` 50 A`(ii) 3.0A`
(i) As the value of external resistance `R` is very large as compared to internal resistance of the cells hence are shall get maximum current if cell are connected is series
Current `I = (n epsilon)/( R + nr) = (20 xx 15)/(500 + 20 xx 0.5) = 059A`
(ii) As the value of external resistance `R` is very small as compured to internal resistance of the cells hence are shall get maximum current if cell are connected is parallel
Current `I = (m epsilon)/( mR + r) = (20 xx 15)/(20 xx 005+ 0.5) = 50A`
(iii) As the value of external resistance `R` is very comparable with the internal resistance of the cells hence are shall get maximum current if cell are connected in mixed grouping
Let there ne `n` a raw and `m` row of cells when connected in mixed grouping .Then
`mn = 20 `....(i)
For maximum current
`R = (nr)/(m) or (n)/(m) = (R ) /(r) = (2.5)/(0.5) = 5 or n = 5 m`
From (i) `mxx 5 m = 20 m = 2 :. n = 5 xx 2 = 10`
Maximum current
`I = (m n epsilon )/(mR+ nr) = (20 xx 15)/(2 xx 25 + 10 xx 0.5) = 3.0 A`