Question 18M.2.HL.TZ2.4
| Date | May 2018 | Marks available | [Maximum mark: 8] | Reference code | 18M.2.HL.TZ2.4 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Comment, Determine, Show that, State | Question number | 4 | Adapted from | N/A |
The diagram shows a potential divider circuit used to measure the emf E of a cell X. Both cells have negligible internal resistance.
State what is meant by the emf of a cell.
[2]
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]
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.
Show that the resistance of the wire AC is 28 Ω.
[2]
the resistance is proportional to length / see 0.35 AND 1«.00»
so it equals 0.35 × 80
«= 28 Ω»
[2 marks]
Determine E.
[2]
current leaving 12 V cell is = 0.15 «A»
OR
E = × 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]
Cell X is replaced by a second cell of identical emf E but with internal resistance 2.0 Ω. Comment on the length of AC for which the current in the second cell is zero.
[2]
since the current in the cell is still zero there is no potential drop across the internal resistance
and so the length would be the same
OWTTE
[2 marks]