Questions Q1. Select one answer from A to D and put a cross in the box ( ) Which one of the following graphs correctly shows the relationship between potential difference (V) and current (I) for a filament lamp? A B C D (Total for Question = 1 mark) Q2. Select one answer from A to D and put a cross in the box ( ) The graph shows the relationship between potential difference V and current I for a fixed 20 Ω resistor and a filament lamp.
The resistor and lamp are placed in series with a 9 V battery of negligible internal resistance. The current in the circuit is A 0.1 A B 0.2 A C 0.3 A D 0.4 A (Total for Question = 1 mark) Q3. (a) State Ohm's law. (2) (b) Using the axes below sketch graphs to show how resistance varies with potential difference for a fixed resistor and a 1.5 V filament lamp. (3)
(c) The filament of a lamp is made of metal. Explain why the lamp does not demonstrate Ohm's law. (2) (Total for Question = 7 marks) Q4. Select one answer from A to D and put a cross in the box ( ) The graph shows how the current I varies with potential difference V for an electrical component. Which row of the table gives the resistance of the component at V2 and describes how the resistance changes from V1 to V2?
(Total for Question = 1 mark) Q5. (a) The diagram shows the circuit used to investigate how the current varies with potential difference for an electrical component P. The circuit contains an ammeter and a voltmeter. (i) On the diagram, label the ammeter A and the voltmeter V. (1) (ii) The position of the contact of the potential divider is moved so that the reading on the voltmeter becomes zero. Label this position Z. (1) (b) The graph shows how the current I varies with potential difference V for two electrical components P and Q.
(i) State the value of the current for which the resistance of P is the same as the resistance of Q and determine this value of resistance. (3) Current =... Resistance =... *(ii) Component Q is a filament lamp. Explain the shape of its graph. (3)
(c) A potential divider consisting of component P and a resistor R is connected to a 12 V supply. The lamp Q and a voltmeter are connected to the circuit as shown. The supply has a negligible internal resistance. The reading on the voltmeter is 4.0 V. (i) Use the graph in part (b) to determine the current in the resistor R. (2) Current =... (ii) Calculate the resistance of the resistor R. (2) Resistance =... (iii) The lamp Q is removed. Explain, without further calculation, how the voltmeter reading would change. (3)
(Total for Question = 15 marks) Q6. Select one answer from A to D and put a cross in the box ( ) The graphs show the variation of potential difference V with the current I for three components. The three components are a metal wire at constant temperature, a filament lamp and a diode. Which row of the table correctly identifies these graphs? (Total for Question = 1 mark) Q7. Which of the following current potential difference (I V) graphs shows the correct behaviour for a filament bulb?
A B C D (Total for Question = 1 mark) Q8. Explain why an ammeter must be placed in series to measure current through a component must have a very low resistance. (3) (Total for Question = 3 marks) Q9.
The graph shows the current potential difference characteristic for an electrical component. (a) State the name of the component. (1) (b) State the resistance of the component when the potential difference is - 0.7 V. (1) (c) Calculate the resistance of the component when the potential difference is + 0.7 V. (2) Resistance =... (d) State a practical use for this component. (1) (Total for question = 5 marks)
Q10. A student investigates how the current through a filament light bulb varies with the potential difference across it. (a) Draw a diagram of a circuit the student could use to obtain suitable measurements for a range of potential difference from 0 V to 12 V. (3) (b) The student's results are shown on the graph. The student decides to draw a tangent to the curve at a potential difference of 6 V and use the gradient of the tangent to determine the resistance of the bulb. (i) Explain why this is not a correct method to determine the resistance. (2) (ii) Calculate the resistance of the bulb when the potential difference across it is 6 V. (2)
Resistance =... *(c) Describe and explain the change in the resistance of the bulb as the potential difference across it is increased. (4) (Total for question = 11 marks) Q11. Which of the following current-potential difference (I-V) graphs correctly shows the behaviour of a diode?
A B C D (Total for question = 1 mark) Q12. A student is taking measurements in order to determine the resistance of a component in a circuit. He connects a voltmeter in parallel with the component and an ammeter in series with the component. Explain why the voltmeter should have a very high resistance. (2)
(Total for question = 2 marks) Q13. (a) Sketch a graph to show how current varies with potential difference for a filament lamp. (2) (b) The temperature of a filament lamp increases as the current through it increases. Explain this in terms of the structure of a metal. (3) (Total for question = 5 marks) Q14.
The graph shows how potential difference V varies with current I for a circuit component. Which of the following could be the circuit component? A copper wire B filament lamp C fixed resistor D thermistor (Total for question = 1 mark) Q1. No Examiner's Report available for this question Q2. No Examiner's Report available for this question Q3. (a) A significantly large number of candidates could not state Ohm's Law. Although mathematically, resistance is defined by the equation R=V/I, this equation is not a statement of Ohm's Law. Apart from the candidates who merely quoted the equation, a large number of candidates tried to use the equation and stated that resistance was directly proportional to potential difference.
This is a typical answer which shows that the student does not know Ohm's Law but the answer scored one mark for the temperature constant comment. Results Plus: Examiner Tip Make sure that students have a list of definitions and Laws to learn. (b) The majority of candidates misread this question and assumed that current -potential difference graphs were being asked for. For the minority who did read the question and drew the correct graph for the fixed resistor, hardly any realised that for the filament lamp, an initial non zero value of resistance was needed. This is the correct resistance graph for the filament but the resistance of the filament lamp has a value when the potential difference is zero. Results Plus: Examiner Tip Candidates need to read the question carefully (c) This was generally well answered, with many candidates scoring both marks. Some candidates
answered along the lines of 'it didn't demonstrate Ohm's Law because current was not proportional to potential difference'. While this is a correct statement, it isn't an explanation and so did not gain any marks. Marks were also lost when candidates merely referred to the temperature or the resistance changing rather than in this case, them both increasing. Q4. The question was about resistance changes for a component and working out resistance at a point and where overall resistance was increasing or decreasing. The common wrong answer was C, so candidates knew the resistance at a point but not how it was varying. Q5. (a) (a)(i) Nearly every candidate labelled the voltmeter and ammeter the correct way round. (a)(ii) Over half the candidates were unable to label Z correctly. The likelyhood for a candidate, for seeing 0 volts on the voltmenter, would be to place Z at the top or bottom end of the resistor but on most incorrect responses, Z was placed randomly around the circuit. The concept of potential dividers continually confuses candidates from year to year and needs to be addressed, whether it be through additional practical work or simpler examples given initially when it is taught so that candidates can appreciate the fundamental idea that the voltage across such a device can change, depending on how much of the device is in the part of the circuit that the volmeter is connected across.
(i) and (ii) both correct. (ii) Although the position of Z is not exactly on the resistor, as the reading on the voltmeter would also be 0 volts, this postition was allowed for the mark. A typical incorrect answer.
(b) (b)(i) Virtually all candidates could read the correct value from the graph. Resistance was usually found correctly. Some candidates did try to find the gradient of the graph, scoring only 1 mark for use of the formula V=IR. (b)(ii) Candidates found this question more challenging, many trying to describe the shape of the graph rather than explaining its shape. Some candidates did think that the graph was ohmic at small currents and a mark was given for this. Most candidates could identify that the temperature was increasing but not identify that it was the current causing the increase and not the voltage as commonly seen. Most candidates managed to get a mark for the idea that the resistance is increasing. At this point only the most able candidates were able to get the last marking point about the rate of increase of the current decreasing. Many attempted to describe the decreasing gradient but were too vague, just mentioning a smaller increase in current as the voltage increases. This problem with describing rates was repeated in question 16aii as to try to describe non linear graphs without a mention of rate is much more difficult. It is expected that candidates will be taught to comment on trends/pattern seen with data obtained in the practical aspect of the course (6PH03, section d, statement A5)and an understanding of the terms linear, proportional, directly proportional, rate is expected from candidates.
(b)(i) 3 marks. (b)(ii) No marks as candidate has just attempted to describe the shape of the graph and not explain it.
(b)(i) 3 marks. A range of features presented, but the answer could have been improved with reference to 'nearshore', 'background' etc - which would provide more structure to the answer. Results Plus: Examiner Tip In this case you need to see what has increased initially (the voltage) and then what has increased as a result of this increase (the current) and then describe the effect that the increased current has on the material eg heating effect, resistance increases and then relate this back to the graph eg the rate of increse of current now decreases.
(b)(i) All 3 marks. (b)(ii) The candidate has implied that temperature increases due to the increasing voltage but correctly identified that the resistance increases. 1 mark. They have attempted to describe the decreasing rate of increase of current but
have not quite described the idea that for the same voltage increase, at higher voltages, a lower current is produced. (b)(i) 3 marks. (b)(ii) 3 marks, increase in current at a lower rate correctly described. (c) (i-ii)
(c)(i)this question was not well answered and, along with (c(iii)) demonstrates a poor understanding of Kirchoff's laws. Most candidates failed to realise that they had to read two currents from the graph, those who did find 0. 3A and 0.5 A did usually remember to add them together. Candidates were also seen to read the current from the graph at 8V and getting an answer of 0.6A, not realising that the graph is not for component R but for P and Q. (c)(ii) Candidates managed to score better in this question as the incorrect current found in part (c)(i) was allowed as an error carried forward. The most difficult part for the candidates was to realise that, as they were finding the resistance of R, they had to use the potential difference across R of 8V and not 4V. Once the values to be used were identified, candidates could use V=IR successfully but often did not score any marks for this question as they did not have the correct voltage.
(c)(i) Incorrect voltage of 8V used to read from the graph. No marks. (c)(ii) Correct use of V=IR with the incorrect current from part (i) and the correct voltage of 8V. 2 marks. Results Plus: Examiner Tip As there is not a graph for the pd across R remember that the current through R is the sum of the individual currents through P and Q which can be found from the graph.
(i) Only 1 current read from the graph. both currents are needed for the first mark. (ii) 4 V is the voltage across P and Q and not R so no marks.
(c) (iii) Over half the candidates gained a mark for identifying that the reading increases. The explanations that accompanied this statement were mostly incorrect with many candidates describing how the current only has to go through 1 component so will increase so there will be an increase in the voltage. Any attempts to describe the effect in terms of current did not score. Good answers correctly described how the resistance of p is now greater than when it was in parallel, voltage increases across P and hence voltmeter reading increases. Scores no marks. Scores 3 marks. The candidate has clearly discussed and understood the change of resistance of the section the voltmeter is connected across and hence the increase in voltage and voltmeter reading. Q6. 65% of candidates answered this question correctly.
Q7. Candidates not choosing B usually selected D, suggesting that they are familiar with the general form of the graph for a non-ohmic conductor and that they know a filament bulb is one. They need to concentrate on a way of deciding which shape matches a positive temperature coefficient and which a negative temperature coefficient. Q8. The majority of candidates scored at least one mark, the most commonly awarded being the mark for a comment about current being the same in series or splitting up for a parallel arrangement. They often suggested that the current would change if the resistance changed but failed to be specific about an increase in resistance for the circuit or that the current would decrease. Q9. (a) About two-thirds of the entry identified a diode or LED. Other suggestions included thermistor, LDR, lamp, copper and resistor. Although the diode is made from semiconductor, this is not sufficient to name the component. A typical incorrect answer. Along with the most common answer, thermistor, this does have a non-linear characteristic.
Results Plus: Examiner Tip Make sure you are familiar with the I-V characteristics of all the components named in the specification for either orientation of axes, as well as how to explain them. (b) The most common answer by far, was zero, with infinity seen only about one time in four. Even candidates who had written R = 0.7 V / 0 A often stated zero as the answer. A few calculated the resistance for +0.7 V. Most were happy with this answer and did not consider the physical implication. They would have benefited from looking at the graph and considering that it shows zero current for all negative potential differences, meaning that the resistance is very large, not very small. An example of a calculation with an incorrect answer. Candidates sometimes have trouble multiplying by zero as well. Another way to arrive at this incorrect result. This candidate may be aware that anything multiplied by zero is zero because 'undefined' is given as an answer. Results Plus: Examiner Tip It is always worth checking whether the Physics is consistent with a numerical answer. Zero is one answer here, but that would mean current was unhindered, not reduced to nothing. (c) This calculation presented few difficulties, with the vast majority completing it successfully. A few tried to use the gradient of the graph and some misread the scale, but unit errors were rare in the answers.
The calculation is inverted here. Because the candidate has gone straight to this incorrect step, no credit can be given for use of the relevant equation. If R = V/I had been written down first, the candidate might well have substituted correctly. Results Plus: Examiner Tip It is always worth taking a moment to write out the relevant equation, rather than doing too much in your head at once. This candidate has attempted to use the gradient of the graph to find the resistance. 0 marks Results Plus: Examiner Tip Resistance is defined as potential difference divided by current. It is not the gradient of a graph of potential difference against current. It only takes that value when the ratio is the same for every value on the graph - i.e. when resistance is constant and the graph is a straight line through the origin.
This gets a mark for 'use of' by substituting values of pd and current, but there is a power of ten error in the potential difference, so the final answer is wrong. (d) In the majority of cases, a device that may contain a diode. Very often, a fire alarm was mentioned without stating the actual use of the diode in the device. Simple answers that gained credit just referred to the diode allowing current to flow in only one direction. A typical answer naming an electronic device. This is not a specific use of the diode and it does not explain its function, if any. An example of a correct answer.
Q10. a The majority got at least two marks for the circuit, but only two fifths got three marks. This is a very commonly used circuit and they should have had quite a bit of experience in using it, so it is somewhat surprising. The arrangements for achieving a variable power supply were often quite unorthodox wiring arrangements, but, when included, they generally worked. Meters were usually connected correctly, but sometimes the bulb was missing or the ammeter connected before a parallel section so that it measured the wrong current. The meters were sometimes used to determine the resistance of the wrong component. Standard symbols were not always used. bi Only about a fifth gained any credit here, usually for answers relating to the graph not being a
straight line. A common erroneous suggestion was that it was incorrect because the gradient is I /V, so its inverse should be used. bii The great majority completed this successfully, although some lost a mark because their graph reading was a bit inaccurate. Although the question stated that use of the gradient was incorrect, a number of students still used precisely that method. c This is a standard explanation at this level and GCSE, so it was disappointing that, while the majority got at least one mark, only about half extended that with a second mark. The most common marks were the first two on the mark scheme increased resistance and increased temperature. As to the rest, candidates generally had an idea of the explanation, but poor expression and imprecise use of technical vocabulary again cost marks. In particular, descriptions relating to the amplitude of oscillation of lattice ions and frequency of collisions of electrons with lattice ions often lacked the comparative element with respect to the prior situation; they just described collisions and oscillations without indicating an increase. Others were vague about what was vibrating or which particles were colliding with which. Often the increase in resistance was caused by the electrons vibrating more or because electrons had more collisions with electrons. Sometimes it was because it was more difficult for electrons to get through or it was due to increased friction. Q11. 75% of candidates gave the correct response. Q12. With only about half of the candidates getting a single mark for mentioning a very small current through the voltmeter, this question demonstrated a generally poor understanding of current and potential difference applied to circuits. Candidates who suggested the current through the voltmeter should be small usually failed to appreciate the significance of this. A common answer was that the current should pass through the component rather than the voltmeter, as if there was a fixed current in the circuit. In the context of their answers, these candidates did not appreciate that the potential difference across the voltmeter would be the same as that across the component and that the current through the component would therefore be unaffected. In this question's context, determining resistance, it is the ratio of potential difference to current that matters and not the particular values anyway. The key point was that the current through the ammeter should be the same as the current
through the component. As the question mentioned the voltmeter and not the ammeter, many discussed the effect on potential difference, quite a few mentioning 'voltage through the component'. A few candidates discussed total resistance but could rarely express their answers clearly enough to gain credit, usually saying little more than that 1/R would be very small. Some candidates failed to score a mark through lack of detail, such as stating that the ammeter reading would not be accurate without saying why or just saying the voltmeter current should be low rather than very small. This response is awarded 1 mark for describing the possible effect on the resistance of the parallel section of the circuit, but it does not say how this would affect the measurement of the resistance of the component, i.e. by causing a greater current in the ammeter than the current in the component. Results Plus: Examiner Tip You need to be able to describe the reasons for the resistance of meters in terms of current and potential difference in series and parallel circuits.
This candidate achieves one mark for saying (twice) that the current in the voltmeter is negligible. In an explanation this needs to be linked to the final outcome, which is a correct determination of the resistance of the component. The response says that all current passes through the component, but the key missing effect of the negligible voltmeter current here is that the current measured by the ammeter is the same as the current in the component. Q13. (a) A large number failed to label the axes, so they could not get the mark for the correct curve. A large number who committed themselves to labelled axes drew the curve the wrong way. Quite a few with labelled axes and the correct curve did not take sufficient care with their drawing and added a 'hook' showing decreasing current at the end. Most candidates gained a mark for symmetry, but only a third scored full marks.
This sketch has the general trend for the change in current and p.d., but goes too far in showing a decrease in current. The line is not drawn for negative values. The curve doesn't match the axes, as this sketch indicates decreasing resistance at higher currents. Results Plus: Examiner Tip You should be able to sketch and interpret I-V graphs for components with either orientation of axes.
A correct version. (b) The majority of candidates applied the correct model but answered the wrong question here, commonly writing in detail about the effect of temperature on resistance and often concluding with a statement that current is reduced, directly contradicting the question. These candidates were still able to gain credit for establishing a link between energy dissipation and collisions of electrons with lattice ions. Most candidates who started with reference to I = navq realised that only v could increase and made a better start, although a few thought n would increase. Candidates giving the correct explanation sometimes failed to score the second mark through a lack of detail, such as describing increased collisions but not saying what was colliding with what, or failing to refer to an increase, for example just stating that there are collisions of electrons with lattice ions. Some candidates only referred to collisions between electrons. This candidate is not answering the question asked. The candidate is required to explain the increase in temperature but instead starts with an increase in temperature and describes its effect. Even there it should refer to increasing lattice ion vibration and not just lattice ion vibration. Results Plus: Examiner Tip It is valuable to revise using past paper questions, but read questions carefully to be sure they
aren't just similar at first glance to previous questions. The increasing rate for flow of electrons is identified, but there is not sufficient detail in the rest of the answer. Interactions with the lattice ions need to be included.
This is a good answer overall which scored two marks. A reference to the metal ions vibrating more is required for the final mark because it is describing an increase in temperature. Q14. The incorrect answer of choice was B, which is the correct shape with axes reversed. Mark Scheme Q1.
Q2. Q3. Q4.
Q5.
Q6. Q7. Q8. Q9.
Q10.
Q11. Q12.
Q13. Q14.
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