User interface language: English | Español

Date May 2018 Marks available 2 Reference code 18M.2.SL.TZ2.4
Level Standard level Paper Paper 2 Time zone 2
Command term State Question number 4 Adapted from N/A

Question

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.

State what is meant by the emf of a cell.

[2]
a.

Show that the resistance of the wire AC is 28 Ω.

[2]
b.i.

Determine E.

[2]
b.ii.

Markscheme

the work done per unit charge

in moving charge from one terminal of a cell to the other / all the way round the circuit

 

Award [1] for “energy per unit charge provided by the cell”/“power per unit current”

Award [1] for “potential difference across the terminals of the cell when no current is flowing” 

Do not accept “potential difference across terminals of cell”

[2 marks]

a.

the resistance is proportional to length / see 0.35 AND 1«.00»

so it equals 0.35 × 80

«= 28 Ω»

[2 marks]

b.i.

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]

b.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.

Syllabus sections

Core » Topic 5: Electricity and magnetism » 5.3 – Electric cells
Show 30 related questions
Core » Topic 5: Electricity and magnetism
Core

View options