Current Electricity Ques 111
111. 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 :
[2016]
(a) $5: 1$
(b) $5: 4$
(c) $3: 4$
(d) $3: 2$
Show Answer
Answer:
Correct Answer: 111.(d)
Solution:
- (d) When two cells are connected in series i.e., $(E_1+E_2)$ the balance point is at $50 cm$. And when two cells are connected in opposite direction i.e., $(E_1-E_2)$ the balance point is at 10 $cm$. 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} $
BETA
