It is desired to supply a current of `2A` through a resistance of `10Omega` As many as `20`calls are pravided each of `e.m.f.2V` and resistance `0.5Omega` Two friends Rajeev and sanjeev try their hand on the problem Rajeev success but sanjeev falls
Read the above passage and answer the following question:
(i) Justify the set tip up of Rajeev?
(ii) What neight have gone with sanjeev when he gets be gets `1.4A` current in the external load?
(iii) What are the basic value shown by Rajeev and sanjeev in their work ?

Explanation

Here`n = 20: epsilon = 2V: r = 0.5Omega : R = 10 Omega`
If all the `20` cells are currectly connected in series to the exerted load of resistance `R` resistance `R` then corrent in load is
`I = (n epsilon )/(R + nr) = (20 xx 2)/(10 + 20 xx 0.5) = 2A`
`R` mean Rajeev set up is correct
(ii) If one cell is wrogly connected in series arrengement of cells , it then reduce the total `e.m.f.` of the set up by an amount to two times the `e.m.f` cell Let `m` cell be connected wrongly by sanjeev in series of total `n` cell , then be get the current `1.4A` in the external of resistance `R( = 10 Omega )` .Thereforte
`I = 14 = ((n - 2m)epsilon)/((R +nr)) = ((20 - 2m) xx 2)/(10+20 xx 0.5)`
On solving we get `m = 3` it means three cell are connected wrong by sanjeev ltb rgt (iii) Rajeev has proper knowledge and careful handing of the apparation .That is why be got the required result from his setup sanjeev 's knowladge is incomplate and his handing is careless That is why he could not get the required Hence to get proper result one has to be careful and should plane all loopholes