A) 4
B) 1
C) 3
D) 2
Correct Answer: B
Solution :
Let n be the number of wrongly connected cells. Number of cells helping one another \[=(12-n)\] Total e.m.f. of such cells \[=(12-n)E\] Total e.m.f. of cells opposing = nE Resultant e.m.f. of battery \[=(12-n)E-nE\]\[=(12-2n)E\] Total resistance of cells = 12r (Q resistance remains same irrespective of connections of cells) With additional cells Total e.m.f. of cells when additional cells help battery = (12 ? 2n) E + 2E Total resistance = 12r + 2r = 14r \ \[\frac{(12-2n)E+2E}{14r}=3\] ......(i) Similarly when additional cells oppose the battery \[\frac{(12-2n)E-2E}{14r}=2\] ......(ii) Solving (i) and (ii), n = 1You need to login to perform this action.
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