Q#41 (Past Exam Paper – November 2007 Paper 1 Q34)
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
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
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