Questions Q1. * A circuit is set up as shown in the diagram. The battery has negligible internal resistance.

Size: px

Start display at page:

Download "Questions Q1. * A circuit is set up as shown in the diagram. The battery has negligible internal resistance."

Adrian Johns
5 years ago
Views:

1 Questions Q1. * A circuit is set up as shown in the diagram. The battery has negligible internal resistance. (a) Calculate the potential difference across the 40 Ω resistor. (2) Potential difference =... (b) A thermistor is connected in parallel with the 40 Ω resistor as shown. The thermistor is initially at a temperature of 100 C and its resistance is 20 Ω. As the thermistor cools down, its resistance increases.

2 Explain what happens to the current through the battery as the temperature of the thermistor decreases. (3) (Total for Question = 5 marks) Q2. The following circuit is used to monitor the temperature in a greenhouse. The battery has no internal resistance. (a) The graph shows how the resistance of the thermistor varies with temperature.

3 (i) Use the graph to find the resistance of the thermistor at 20 C. (1) Resistance =... (ii) Calculate the reading on the voltmeter when the thermistor is at 20 C. (3) Reading on the voltmeter =... (b) Explain what will happen to the reading on the voltmeter as the temperature of the greenhouse decreases. (2) (Total for Question = 6 marks) Q3. (a) A kettle is rated at 1 kw, 220 V. Calculate the working resistance of the kettle. (2) Resistance =... (b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

4 Calculate how much energy has been supplied. (2) Energy =... (c) Different countries supply mains electricity at different voltages. Many hotels now offer a choice of voltage supplies as shown in the photograph. (i) By mistake, the kettle is connected to the 110 V supply. Assuming that the working resistance of the kettle does not change, calculate the time it would take for the same amount of water to reach boiling point. (3) Time =... (ii) Explain what might happen if a kettle designed to operate at 110 V is connected to a 220 V supply. (2)

5 (Total for Question = 9 marks) Q4. Select one answer from A to D and put a cross in the box ( ) The battery in the circuit has negligible internal resistance and an e.m.f. of 12 V. The potential difference across the parallel combination is A 0V B 4V C 6V D 8V (Total for Question = 1 mark) Q5. The diagram shows a circuit set up by a student.

6 (a) Both voltmeters have a resistance of 10 MΩ. The reading on V1 is 6 V and the reading on V2 is zero. Explain these readings. (2) (b) The student replaces the 10 Ω resistor of unknown resistance R. The reading on V1 is now 4 V. Calculate the value of R (3) R =... (Total for Question = 5 marks)

7 Q6. The diagram shows a uniform wire XY across which a potential difference V0 is applied. Which of the following correctly shows the output potential difference across XZ? (Total for Question = 1 mark) Q7. A combination of resistors is connected to a 12 V supply of negligible internal resistance. The potential difference between points A and B is A 4V B 6V C 8V D 12 V

8 (Total for question = 1 marks) Q8. The diagram shows a potential divider circuit that contains a negative temperature coefficient thermistor. The temperature of the room containing the circuit increases. Select the row of the table that correctly shows the changes in readings on the meters. (Total for question = 1 mark) Examiner's Report Q1. (a) Another calculation which was generally well answered. Candidates need to be suspicious of answers that give a p.d. greater than the e.m.f. of the battery! A lot of candidates chose to do this calculation by finding the current in the circuit and then the p.d. across the 40 Ω resistor. A small but significant number of candidates found the current using just the 80 Ω resistor with the

9 9V. This shows that some candidates have a very poor grasp of even the most simple of circuits. (b) Answers generally lacked sufficient detail. Only a minority of candidates seemed able to methodically work their way through the argument, including all the relevant steps such as the essential one that the e.m.f./p.d. had to be constant for a larger total resistance to result in a smaller current. Candidates needed to think about what happened to the parallel combination first and after that, the effect on the total resistance. All too often these two stages were missed and there were references to resistance increasing which were too vague. Results Plus: Examiner Comments The first sentence is a repeat of what is given in the question. The answer effectively begins half way along the second line. There is a correct reference to the p.d. staying constant but there is insufficient detail as to how the candidate has come to the conclusion that the total resistance will increase. Again, there are three marks so three different points must be made. Q2. This was a relatively straightforward question, using potential dividers and thermistors. Nearly all candidates were able to read the value of resistance of the graph. Although, only the most able candidates are able to cope with the concept of a potential divider, this question can be correctly answered by finding the total resistance, then the current and finally the potential difference across the 1 kilo-ohm resistor.

10 Unfortunately, most candidates who chose to follow this route did not find the total resistance and did a current calculation using one of the resistances, instead of the total resistance. A significant number of candidates calculated a p.d. greater than 6 V and did not think that this was wrong. For the last part, it was necessary for candidates to identify that, it was the resistance of the thermistor that increased.

11 Results Plus: Examiner Comments This candidate found a current using just the thermistor's resistance. In the last part, there is no mention of the thermistor.

13 Results Plus: Examiner Comments An example where the reading on the voltmeter is greater than 6 V. The last part shows that the candidate has little understanding about potential dividers. Results Plus: Examiner Tip Check if your answer is sensible. The p.d. across one component in a circuit can't be greater than the p.d. of the supply. Q3. (a) -(b) Both parts were generally well answered, with the majority of candidates scoring full marks. Many candidates chose to answer (b) using E =VIt rather than the more simple E =Pt. (c) There were many methods seen in answer to this question, with quite a few candidates scoring three marks for the calculation. The common wrong answer was three minutes, because candidates did not realise that reducing the p.d. would also reduce the current. The question was about energy/power and so candidates need to think in terms of a power equation. The last part of the question again required candidates to refer to an equation, preferably E = V2/R but V=IR was accepted. Despite the question saying that resistance stayed the same, many candidates tried to answer this in terms of increased resistance.

14 Results Plus: Examiner Comments A very complex way to get to the answer but it is correct. This also scored two marks for the last part because there is a reference to an equation. Q4. No Examiner's Report available for this question

15 Q5. (a) About a quarter of answers gained at least one mark, with half of those getting the second mark as well. Those who approached this by carrying out circuit calculations often found it quite straightforward. For the remainder, a number of incomplete approaches and misconceptions were seen, with little understanding of the use of voltmeters with very high resistance. Some candidates simply said that if one reading is 6 V the other must be 0 V to give a total of 6 V, but this information was all in the question. Candidates frequently considered the parallel section of the circuit in isolation and suggested that the voltmeter had no current because all the current flowed through the 10 W resistor instead, and did not realise that there was essentially no current in the whole circuit. There were a number of references to voltage flowing and also to current flowing as far as V1 and then stopping. Results Plus: Examiner Comments This candidate has mentioned that V1 is high, but has not made it clear that it is very high in comparison to the parallel combination which contains an equally high resistance. The statement that potential is 'getting absorbed' is not an adequate description. Results Plus: Examiner Tip

16 When comparing things it is generally necessary to make a contextual reference to each of them in the response. Results Plus: Examiner Comments This response fails to recognise that the potential difference across parallel components is equal and is attempting to suggest the reading on V2 is zero because its current is zero due to the current going through the resistor instead. It is comparing the resistor and its voltmeter and ignoring V1. Candidates are expected to know that a voltmeter has a very high resistance specifically to ensure it draws very little current. (b) This was slightly less well answered than part (a), with about a quarter of answers gaining at least one mark, and a quarter of those also getting the second and third marks. The most common way to get a single mark was finding the resistance of the parallel section of the circuit, although most candidates thought this was the final answer required, ignoring V2, or perhaps thinking it had been removed. Some candidates got the final answer after assuming the current in the circuit divided equally between the parallel branches, although this is not always the case.

17 Results Plus: Examiner Comments This is a typical response in which the candidate has just found the resistance of the parallel section of the circuit, as if the resistor and V2 had both been removed and replaced by the new resistance R. Results Plus: Examiner Tip Be careful to correctly identify the context. In this case, altering the original circuit diagram or drawing a new one would have clarified the situation. Results Plus: Examiner Comments This candidate has also only calculated the new resistance of the parallel section, but by the alternative method of applying the ratio of the potential differences. Q6.

18 Q7. No Examiner's Report available for this question Q8. No Examiner's Report available for this question Mark Scheme Q1. Q2.

19 Q3.

20 Q4. Q5.

21 Q6. Q7. Q8. Powered by TCPDF (

Calculate the total resistance of this combination. (3)

1 The circuit shows a combination of three resistors. 22 Ω 47 Ω 620 Ω Calculate the total resistance of this combination. Total resistance = (Total for Question = 3 marks) 2 (a) Sketch a graph to show