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 50cm and 10cm from the positive end of the wire in the two cases. The ratio of emf's is:

Question

Toppr Admin · Accepted Answer

If the emfs of the two cells are taken as E1 and E2 respectively-So- E1-E2E1-x2212-E2-5010or- 2E12E2-50-1050-x2212-10or- E1E2-32

If the emfs of the two cells are taken as E1 and E2 respectively.So, E1+E2E1−E2=5010or, 2E12E2=50+1050−10or, E1E2=32

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, $\frac{2 E_{1}}{2 E_{2}} = \frac{50 + 10}{50 - 10}$

or, $\frac{E_{1}}{E_{2}} = \frac{3}{2}$