| Date | May 2012 | Marks available | 8 | Reference code | 12M.2.SL.TZ2.7 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Calculate, Define, and Show that | Question number | 7 | Adapted from | N/A |
Question
Part 2 Electric potential difference and electric circuits
Ionized hydrogen atoms are accelerated from rest in the vacuum between two vertical parallel conducting plates. The potential difference between the plates is V. As a result of the acceleration each ion gains an energy of 1.9×10–18J.
Calculate the value of V.
The plates in (a) are replaced by a cell that has an emf of 12.0 V and internal resistance 5.00 Ω. A resistor of resistance R is connected in series with the cell. The energy transferred by the cell to an electron as it moves through the resistor is 1.44 ×10–18 J.
(i) Define resistance of a resistor.
(ii) Describe what is meant by internal resistance.
(iii) Show that the value of R is 15.0 Ω.
(iv) Calculate the total power supplied by the cell.
Markscheme
\(V = \frac{{1.9 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}}\);
=12V;
(i) ratio potential difference/voltage (across resistor) to current (in resistor) / \(\frac{V}{I}\)
with symbols defined;
(ii) some of the power/energy delivered by a cell is used/dissipated in driving the current though the cell itself;
the power loss can be equated to I2r where r represents the (internal) resistance of the cell;
To award [2] the resistance must be put into some context.
Award [1 max] for e.g. it is the resistance of the cell itself.
(iii) pd across \(R = \frac{{1.44 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}} = 9.00{\rm{V}}\);
pd across internal resistance=12.0-9.00(=3.00V);
current in circuit=\(\left( {\frac{{3.00}}{{5.00}} = } \right)0.600{\rm{A}}\);
\(R = \frac{{9.00}}{{0.600}}\);
(=15.0Ω)
(iv) 7.20 W;
Examiners report
However, this part was done well.
(i) Many have now learnt the definition of resistance that this syllabus requires. Some still continue however to provide (spurious) explanations of how resistance arises.
(ii) This was a description and many candidates were able to gain one point. But the second point for an analysis of the internal power dissipation of a cell was universally absent.
