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Question 18M.2.HL.TZ2.b.ii

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Date May 2018 Marks available [Maximum mark: 2] Reference code 18M.2.HL.TZ2.b.ii
Level HL Paper 2 Time zone TZ2
Command term Determine Question number b.ii Adapted from N/A
b.ii.
[Maximum mark: 2]
18M.2.HL.TZ2.b.ii

The diagram shows a potential divider circuit used to measure the emf E of a cell X. Both cells have negligible internal resistance.

M18/4/PHYSI/SP2/ENG/TZ2/04

AB is a wire of uniform cross-section and length 1.0 m. The resistance of wire AB is 80 Ω. When the length of AC is 0.35 m the current in cell X is zero.

Determine E.

[2]

Markscheme

current leaving 12 V cell is 12 80 = 0.15 «A»

OR

E = 12 80  × 28

E = «0.15 × 28 =» 4.2 «V»

 

Award [2] for a bald correct answer

Allow a 1sf answer of 4 if it comes from a calculation.

Do not allow a bald answer of 4 «V»

Allow ECF from incorrect current

[2 marks]

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