A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells…

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of EMF's is :
  1. 5 : 1
  2. 5 : 4
  3. 3 : 4
  4. 3 : 2

Solution


as  EL


E 1 + E 2 E 1 E 2 = λ50 λ10
E1+E2=5E1-5E2
6E2=4E1
32=E1E2

Asked in: NEET 2016 (Phase 1)

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