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JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Critical Thinking

  • question_answer
    12 cells each having same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two cells which are in series. Current is 3 A when cells and battery aid each other and is 2 A when cells and battery oppose each other. The number of cells wrongly connected is

    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 = 1


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