A potentiometer is used as shown to compare the e.m.f.s of two cells. The balance points for cells X and Y are 0.70 m and 0.90 m respectively.

Question 41

A potentiometer is used as shown to compare the e.m.f.s of two cells.

The balance points for cells X and Y are 0.70 m and 0.90 m respectively.

If the e.m.f. of cell X is 1.1 V, what is the e.m.f. of cell Y?

A 0.69 V          B 0.86 V          C 0.99 V          D 1.4 V

Reference: Past Exam Paper – November 2007 Paper 1 Q34

Solution:

Answer: D.

At the balance length, the p.d. across the wire is equal to the e.m.f. of the cell.

p.d. ∝ L

For cell X: 1.1 V ∝ 0.70 m     (1)

For cell Y: emf ∝ 0.90 m        (2)

Divide (2) by (1),

emf / 1.1 = 0.90 / 0.70

emf = (0.90/0.70) × 1.1 = 1.4 V