Question

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:

• A. $5:1$
• B. $5:4$
• C. $3:4$
• D. $3:2$

Answer 1

Answer: (D)

$3:2$

If the emfs of the two cells are taken as $E_1$ and $E_2$ respectively.

So, $\frac{E_1+E_2}{E_1-E_2}=\frac{50}{10}$

or, $\frac{2E_1}{2E_2}=\frac{50+10}{50-10}$

or, $\frac{E_1}{E_2}=\frac{3}{2}$