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Current Electricity question

2016 · Q120
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Current Electricity question

2016 · Q120

NEETPhysicsCurrent ElectricityMCQ+4 / −1
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. A
    3 : 4
  2. B
    3 : 2
  3. C
    5 : 1
  4. D
    5 : 4
View written solutionFree

Correct answer: B

When two cells are connected in series i.e., E1 + E2 the balance point is at 50 cm. And when two cells are connected in opposite direction i.e., E1 – E2 the balance point is at 10 cm. According to principle of potential

E1+E2E1−E2=5010{{{E_1} + {E_2}} \over {{E_1} - {E_2}}} = {{50} \over {10}}E1​−E2​E1​+E2​​=1050​

⇒2E12E2=50+1050−10 \Rightarrow {{2{E_1}} \over {2{E_2}}} = {{50 + 10} \over {50 - 10}}⇒2E2​2E1​​=50−1050+10​

⇒E1E2=32 \Rightarrow {{{E_1}} \over {{E_2}}} = {3 \over 2}⇒E2​E1​​=23​

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