| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.SL.TZ2.4 |
| Level | Standard level | Paper | Paper 2 | Time zone | 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.
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.
Show that the resistance of the wire AC is 28 Ω.
Determine E.
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]
the resistance is proportional to length / see 0.35 AND 1«.00»
so it equals 0.35 × 80
«= 28 Ω»
[2 marks]
current leaving 12 V cell is \(\frac{{12}}{{80}}\) = 0.15 «A»
OR
E = \(\frac{{12}}{{80}}\) × 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]
