NEET | 2016Current Electricity and Electrical InstrumentHard

Question

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cellsare connected in series first to support one another and then in opposite direction. The balance points areobtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf’s is :

Options

A.

5 : 1

B.

5 : 4

C.

3 : 4

D.

3 : 2

Solution


E1+E2=λ50E1-E2=λ10
E1 + E2 = 5E1 - 5E2
6E2 = 4E1
32=E1E2

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