SATHEE: Current Electricity - Result Question 94

Current Electricity - Result Question 94

99. A potentiometer wire is $100 c m$ 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 $c m$ and $10 c m$ from the positive end of the wire in the two cases. The ratio of emf’s is :

[2016]

(a) $5 : 1$

(b) $5 : 4$

(d) $3 : 2$

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Answer:

Correct Answer: 99. (d)

Solution:

(d) When two cells are connected in series i.e., $(E_{1} + E_{2})$ the balance point is at $50 c m$ . And when two cells are connected in opposite direction i.e., $(E_{1} - E_{2})$ the balance point is at 10 $c m$ . According to principle of potential

$\begin{aligned} \frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{50}{10} \\ \Rightarrow \frac{2 E_{1}}{2 E_{2}} = \frac{50 + 10}{50 - 10} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{3}{2} \end{aligned}$

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Current Electricity - Result Question 94

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