It is desired to supply a current of 2A through a resistance of \[10\,\Omega .\] 20 cells are provided, each of them has 2V emf and internal resistance of \[0.5\,\Omega .\] Two students of class XIIth, Shikha and Shahana try their hand on the requirement. Shahana succeeds but Shikha not. |
(i) Justify the set up of Shahana. |
(ii) What might have gone wrong with Shikha, when she gets 12 A current in the load? |
(iii) What are the basic values shown by Shahana and Shikha in their work? |
Answer:
(i) Given, n = 20, E = 2 V, \[r=0.5\,\Omega \] and \[R=10\,\Omega \] If all cells are connected correctly in series to the load R, then \[I=\frac{nE}{R+nr}=\frac{20\times 2}{10+20\times 0.5}=\frac{40}{10+10}=\frac{40}{20}=2A\] It justifies the set up of Shahana. (ii) If one cell is wrongly connected in series circuit, then it will reduce the total emf of the circuit by the two times of its own emf. Let m cells are connected wrongly by Shikha, then we have \[{{I}_{1}}=\frac{(n-2m)E}{R+nr}\Rightarrow 1.2=\frac{(20-2\times m)2}{10+20\times 0.5}\] \[\Rightarrow (20-2m)=\frac{1.2\times (10+10)}{2}\Rightarrow 20-2m=12\] \[\therefore \] \[m=(20-12)/2=4\] It means, 4 cells are connected wrongly by Shikha. (iii) Shahana has good subject knowledge and she knows how to apply it in the practicals.
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