AS Electricity Part II

309 minutes

309 marks

Q1. Columns A and B show some of the results from an experiment in which the current Ithrough a component X was measured for various values of the potential difference V applied across it.

column A column B column C column D

potential difference V / V current I/ mA (V – 0.55) / V In (I / mA)

0.70 12.5

0.75 17.0

0.80 22.0

0.85 29.0

0.90 39.0

0.95 51.5

(a) Draw a diagram of a circuit which could have been used to obtain these results.

(2)

(b) (i) Calculate the resistance of X when the potential difference is 0.70 V.

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(ii) By considering one other value of potential difference, explain whether or not X is an ohmic conductor.

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(c) It is suggested that for potential differences greater than 0.55 V, the current voltage relationship for X is of the form.

I = A ek(V–0.55)

where A and k are constants.

(i) Complete column C and column D in the table above

(ii) Plot a graph of 1n (I /mA ) on the y-axis against (V – 0.55) on the x-axis.

(Allow one sheet of graph paper)

(iii) Use your graph to determine the constants k and A.

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(iv) On the basis of your graph, discuss the validity of the above relationship.

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(Total 15 marks)

Q2. (a) On the axes in Figure 1 draw I –V characteristics for two components, A and B, both of which obey Ohm’s law. Component B has a lower resistance than component A. Label your characteristics clearly as A and B.

Figure 1(2)

(b) On the axes in Figure 2 draw the I – V characteristic for a silicon semiconductor diode, giving any relevant voltage values.

Figure 2(3)

(c) Figure 3 shows the I – V characteristic of a filament lamp. Explain the shape of this characteristic.

Figure 3

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(Total 9 marks)

Q3. (a) A metal wire of length 1.4 m has a uniform cross-sectional area = 7.8 × 10–7 m2. Calculate the resistance, R, of the wire.resistivity of the metal = 1.7 × 10–8 Ωm

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(b) The wire is now stretched to twice its original length by a process that keeps its volume constant. If the resistivity of the metal of the wire remains constant, show that the resistance increases to 4R.

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(Total 4 marks)

Q4. (a) The circuit shown in Figure 1 may be used to determine the internal resistance of a battery. An oscilloscope is connected across the battery as shown. Figure 2 represents the screen of the oscilloscope.

Figure 1 Figure 2

The time base of the oscilloscope is switched off throughout the experiment. Initially the switches S1 and S2 are both open. Under these conditions the spot on the oscilloscope screen is at A.

(i) Switch S1 is now closed, with S2 remaining open. The spot moves to B. State what the deflection AB represents.

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(ii) Switch S1 is kept closed and S2 is also closed. The spot moves to C. State what the deflection AC represents.

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(iii) The vertical sensitivity of the oscilloscope is 0.50 V div–1. Calculate the current through the 14 Ω resistor with both switches closed.

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(iv) Hence, calculate the internal resistance of the battery.

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(b) The oscilloscope is now connected to an alternating voltage source of rms value 3.5 V.

(i) Calculate the peak value of the alternating voltage.

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(ii) Draw on Figure 3 what you would expect to see on the oscilloscope screen, if the time base is still switched off and the voltage sensitivity is altered to 2.0 V div–1.

Figure 3(3)

(Total 9 marks)

Q5. (a) In the circuit shown in Figure 1, the battery has an emf of 12 V and negligible internal resistance.

PQ is a potential divider, S being the position of the sliding contact. In the position shown, the resistance between P and S is 180 Ω and the resistance between S and Q is 60 Ω.

Figure 1

(i) Calculate the current, I, in the circuit, assuming that there is no current through the voltmeter V.

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(ii) What property of the voltmeter allows us to assume that no current flows through it?

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(iii) What is the reading on the voltmeter?

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(b) The circuit in Figure 1 is modified as shown in Figure 2, by exchanging the voltmeter for a load R, whose resistance is about the same as the resistance of section SQ of the potential divider.

Figure 2

Explain, without calculation, why the current through the battery increases in value from that in part (a).

You may be awarded marks for the quality of written communication in your answer.

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(Total 6 marks)

Q6. A battery of emf 24 V and negligible intemal resistance is connected to a resistor network as shown in the circuit diagram in the diagram below.

(a) Show that the resistance of the single equivalent resistor that could replace the four resistors between the points A and B is 50 Ω.

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(b) If R1 is 50 Ω, calculate

(i) the current in R1,

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(ii) the current in the 60 Ω resistor.

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(Total 8 marks)

Q7. (a) A student wishes to measure the resistance of a fixed length of uniform constantan wire. The apparatus available includes a battery, a switch, a milliammeter and a voltmeter.

You may be awarded marks for the quality of written communication in your answer.

(i) Draw a circuit diagram using the apparatus listed above. Include in your diagram an extra piece of apparatus which will enable a range of measurements to be made.

(ii) State how the student should make the necessary measurements, ensuring that a range of readings is recorded.

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(iii) Describe how the results would be used to determine an accurate value for the resistance of the wire.

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(b) A heating element for an electric fire consists of a single strand of nichrome wire wound around an insulator. The heater is required to produce 1.2 kW when connected to the 230 Vrms ac mains.

(i) Calculate the working resistance of the nichrome wire.

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(ii) Calculate the length of nichrome wire required to make the element.

cross-sectional area of the wire = 9.4 × 10–8 m2

resistivity of nichrome = 1.1 × 10–6 Ω m

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(Total 13 marks)

Q8. (a) X and Y are two lamps. X is rated at 12 V, 24 W and Y at 6.0 V, 18 W. Calculate the current through each lamp when it operates at its rated voltage.

X: .................................................................................................................

Y: ..................................................................................................................(2)

(b) The two lamps are connected in the circuit shown. The battery has an emf of 27 V and negligible internal resistance. The resistors R1 and R2 are chosen so that the lamps are operating at their rated voltage.

(i) What is the reading on the voltmeter?

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(ii) Calculate the resistance of R2.

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(iii) Calculate the current through R1.

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(iv) Calculate the voltage across R1.

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(v) Calculate the resistance of R1.

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(Total 9 marks)

Q9. In the circuit shown, the battery has an emf of 12 V and an internal resistance of 2.0 Ω. The resistors A and B each have resistance of 30 Ω.

Calculate

(i) the total current in the circuit,

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(ii) the voltage between the points P and Q,

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(iii) the power dissipated in resistor A,

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(iv) the energy dissipated by resistor A in 20 s.

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Q10. Four resistors, each having resistance of 50 Ω, are connected to form a square. A resistance meter measured the resistance between different corners of the square. Determine the resistance the meter records when connected between the following corners.

(a) Between A and C, as in Figure 1.

Figure 1

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(b) Between A and B, as in Figure 2.

Figure 2

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(Total 5 marks)

Q11. In the circuit shown the battery has emf and internal resistance r.

(a) (i) State what is meant by the emf of a battery.

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(ii) When the switch S is open, the voltmeter, which has infinite resistance, reads 8.0 V. When the switch is closed, the voltmeter reads 6.0 V.Determine the current in the circuit when the switch is closed.

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(iii) Show that r = 0.80 Ω.

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(b) The switch S remains closed. Calculate

(i) the power dissipated in the 2.4 Ω resistor,

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(ii) the total power dissipated in the circuit,

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(iii) the energy wasted in the battery in 2 minutes.

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(4)(Total 8 marks)

Q12. An oscilloscope is connected to a sinusoidal ac source as shown in Figure 1.The frequency and the voltage output of the ac source can be varied.

Figure 1

At a certain frequency the ac signal has an rms output of 7.1 V. Figure 2 shows the trace obtained on the screen of the oscilloscope when one horizontal division corresponded to a time of 5.0 ms.

Figure 2

(a) Calculate, for the signal shown in Figure 2,

(i) the peak voltage,

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(ii) the frequency.

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(b) The voltage output and frequency of the signal are now changed so that the peak voltage is 80 V and the frequency is 200 Hz.

State which two controls on the oscilloscope have to be altered so that four full cycles again appear on the screen but the peak to peak distance occupies the full screen.

Determine the values at which these two controls have to be set.

control 1: …...................................................................................................

value of the setting: ......................................................................................

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control 2: …...................................................................................................

value of setting: ............................................................................................

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(Total 8 marks)

Q13. The graph shows the I - V characteristics for two conductors, the tungsten filament of a lamp and a length of constantan wire.

(a) State, with a reason, which conductor obeys Ohm’s law across the full voltage range.

conductor: ....................................................................................................

reason: .........................................................................................................(2)

(b) (i) Calculate the resistance of the tungsten filament when V = 1 V and V = 10 V.

V = 1 V: ...............................................................................................

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V = 10 V: .............................................................................................

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(ii) Explain why the values of resistance, calculated in part (b)(i), differ from each other.

You may be awarded marks for the quality of written communication in your answer.

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(c) Use the graph to determine the resistivity of constantan, given that the wire is 0.80 m long with a uniform cross-sectional area of 6.8 × 10–8 m2.

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(d) A student is required to obtain the I – V characteristic for a filament lamp using a datalogger, so that the data can be fed into a computer to give a visual display of the characteristic.

Draw a labelled circuit diagram for such an experiment.(An account of the experiment is not required).

(3)

(Total 12 marks)

Q14. (a) In the circuit in Figure 1, the battery, of emf 15 V and the negligible internal resistance, is connected in series with two lamps and a resistor. The three components

each have a resistance of 12 Ω.

Figure 1

(i) What is the voltage across each lamp?

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(ii) Calculate the current through the lamps.

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(b) The two lamps are now disconnected and reconnected in parallel as shown in Figure 2.

Figure 2

(i) Show that the current supplied by the battery is 0.83 A.

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(ii) Hence show that the current in each lamp is the same as the current in the lamps in the circuit in Figure 1.

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(c) How does the brightness of the lamps in the circuit in Figure 1 compare with the brightness of the lamps in the circuit in Figure 2?

Explain your answer.

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(Total 8 marks)

Q15. A student carried out an experiment to investigate how the resistance of a coil made from iron wire varied as its temperature changed. In the experiment, a constant potential difference of 2.00 V was maintained across the coil as the student measured the current through it at various temperatures.

(a) Draw a circuit diagram to show how such measurements could be made using a 3.0 V battery and other necessary apparatus.

(3)

(b) The results are shown in the table.

temperature θ / °C p.d. across the coil / V current / mA resistance R / Ω

10 2.00 790

20 2.00 740

35 2.00 680

50 2.00 620

65 2.00 590

80 2.00 550

The resistance, R, of a metal wire varies with temperature according to the equation

R = R0 (1 + αθ)

where R0 is the resistance of the wire at 0 °C, θ is the temperature of the wire in °C, and αis a constant which depends on the material from which the wire is made.

(i) Complete the table to show the value of the resistance, R, of the iron coil at each temperature.

(ii) Use the data to plot a graph to confirm the equation.

(One sheet of graph paper should be provided.)(6)

(c) (i) Use your graph to determine the value of the resistance of the coil at 0 °C (R0).

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(ii) Use your graph to determine the value of the constant α and state its unit.

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(d) Explain why the resistance of a metal wire changes when its temperature is increased.

You may be awarded marks for the quality of written communication in your answer.

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(Total 16 marks)

Q16. (a) Give an expression for the resistivity of a material in the form of a uniform wire.Define all the symbols used.

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(b) A thin film of carbon may be used in some electronic systems. Typical dimensions of such a film are shown in Figure 1.

Figure 1

(i) Calculate the resistance of the carbon film to a current I passing through it as shown in Figure 1.

resistivity of carbon = 4.0 × 10–5 Ω m

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(ii) Without recalculating the resistance of the carbon film, explain how you would expect the resistance to change if the current flowed as in Figure 2. You should consider the numerical ratio or factor by which each dimension affecting the resistance has changed.

You may be awarded marks for the quality of written communication in your answer.

Figure 2

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(Total 6 marks)

Q17. In the circuit shown, a battery of emf and internal resistance r is connected to a variable resistor R. The current I and the voltage V are read by the ammeter and voltmeter respectively.

(a) The emf is related to V, I and r by the equation

= V + Ir

Rearrange the equation to give V in terms of , I and r.

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(b) In an experiment, the value of R is altered so that a series of values of V and the corresponding values of I are obtained. Using the axes, sketch the graph you would expect to obtain as R is changed.

(2)

(c) State how the values of and r may be obtained from the graph.

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(Total 5 marks)

Q18. (a) Figure 1 shows two possible arrangements of connecting three resistors, each resistor having a resistance of 40 Ω.

Figure 1

Calculate the equivalent resistance in each case.

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(b) The designer of a heating element for the rear window of a car decides to connect six separate heating elements together as shown in Figure 2. Each element has a resistance of 6.0 Ω and the unit is connected to a 12 V dc supply having zero internal resistance.

Figure 2

(i) Calculate the current in each single element.

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(ii) With the aid of a similar calculation give a reason why the heater would not be as effective if all six were connected in series.

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(Total 8 marks)

Q19. A battery is connected across a uniform conductor. The current in the conductor is 40 mA.

(i) Calculate the total charge that flows past a point in the conductor in 3 minutes.

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(ii) Using data from the Data Sheet calculate the number of electron charge carriers passing the same point in the conductor in this time.

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(iii) If 8.6 J of energy are transferred to the conductor in this time, calculate the potential difference across the conductor.

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(iv) Calculate the resistance of the conductor.

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Q20. A student investigates the variation of electric potential with distance along a strip of conducting paper of length l and of uniform thickness. The strip tapers uniformly from a width 4w at the broad end to 2w at the narrow end, as shown in Figure 1. A constant pd is applied across the two ends of the strip, with the narrow end at positive potential, Vl, and the broad end at zero potential. The student aims to produce a graph of pd against distance x, measured from the broad end of the strip.

Figure 1

(a) Draw a labelled circuit diagram which would be suitable for the investigation.

(2)

(b) The student obtained some preliminary measurements which are shown below.

pd, V/V 0 2.1 4.5 7.2

Distance, x/m 0 0.100 0.200 0.300

By reference to the physical principles involved, explain why the increase of V with xis greater than a linear increase.

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(c) The potential, V, at a distance x from the broad end is given by

V = k – 1.44Vl ln (2l – x),

where Vl is the potential at the narrow end, and k is a constant.

(i) The student’s results are given below. Complete the table.l = 0.400 m

distance x/m potential V/V (2l – x)/m ln (2l – x)

0.100 2.1 0.700 – 0.357

0.200 4.5

0.270 6.4

0.330 8.3

0.360 9.3

0.380 10.1

(ii) Plot a graph of V against ln (2l – x) and explain whether or not it confirms the equation.

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(iii) Use the graph to calculate Vl.

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(Total 16 marks)

Q21. A sinusoidal alternating voltage source of frequency 500 Hz is connected to a resistor of resistance 2.0 kΩ and an oscilloscope, as shown in Figure 1.

Figure 1

(a) The rms current through the resistor is 5.3 mA. Calculate the peak voltage across the resistor.

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(b) The settings on the oscilloscope are

timebase: 250 µs per division, voltage sensitivity: 5.0 V per division.

Draw on the grid, which represents the screen of the oscilloscope, the trace that would be seen.

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(4)(Total 6 marks)

Q22. (a) For a conductor in the form of a wire of uniform cross-sectional area, give an equation which relates its resistance to the resistivity of the material of the conductor. Define the symbols used in the equation.

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(2)

(b) (i) An electrical heating element, made from uniform nichrome wire, is required to dissipate 500 W when connected to the 230 V mains supply. The cross-sectional area of the wire is 8.0 × 10–8 m2. Calculate the length of nichrome wire required.

resistivity of nichrome = 1.1 × 10–6 Ω m

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(ii) Two heating elements, each rated at 230 V, 500 W are connected to the 230 mains supply

(A) in series, (B) in parallel.

Explain why only one of the circuits will provide an output of 1 kW.

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(Total 8 marks)

Q23. The graph shows how the resistance, RR, of a metal resistor and the resistance, RTh, of a thermistor change with temperature.

(a) Give the values of the resistance RR and RTh at a temperature of 200 °C.

RR ..............................................................RTh

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(b) The resistor and thermistor are connected in series to a 12V battery of negligible internal resistance, as shown in Figure 1.

Figure 1

(i) Calculate the voltage across the terminals AB when both the resistor and thermistor are at 200 °C.

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(ii) Assuming that the temperature of the resistor always equals the temperature of the thermistor, deduce the temperature when the voltage across the resistor equals the voltage across the thermistor.

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(c) A lamp and a switch are now connected across the terminals AB, as shown in Figure 2. The temperature of the thermistor does not change from that obtained in part (b)(ii).

Figure 2

(i) The lamp is rated at 2.0 W at a voltage of 6.0 V. Calculate the resistance of the lamp at this rating.

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(ii) The switch S is now closed. Explain, without calculation, why the voltage across the thermistor will fall from the value in part (b)(ii).

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(Total 9 marks)

Q24. (a) The graph in Figure 1 shows the I – V characteristic for a semiconductor diode.

Figure 1

In order to produce this characteristic in the laboratory, a student is given suitable apparatus, including a data logger. The output of the data logger is connected to a computer to give a visual display.

(i) Draw a labelled circuit diagram of the apparatus that would be used to obtain the characteristic from O to A in Figure 1.

(ii) Describe how the apparatus would be used.

You may be awarded marks for the quality of written communication in your answer.

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(iii) What change would have to be made to the circuit in order to obtain the

characteristic from O to B in Figure 1?

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(b) On the axes in Figure 2 draw the characteristic for a filament lamp.

Figure 2

Explain why the I V characteristic has the shape you have drawn.

You may be awarded marks for the quality of written communication in your answer.

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(Total 11 marks)

Q25. In the circuit shown in Figure 1, the battery, of emf 6.0V, has negligible internal resistance.

Figure 1

(a) Calculate the current through the ammeter when the switch S is

(i) open,

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(ii) closed.

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(b) The switch S is now replaced with a voltmeter of infinite resistance.Determine the reading on the voltmeter.

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(Total 5 marks)

Q26. The graph shows how the current through a thermistor varies with the potential difference across it.

(a) Draw the circuit of an experimental arrangement which could be used to collect the data necessary to produce this graph. On your circuit diagram label clearly a component which would enable the current to be changed continuously across the range.

(4)

(b) (i) Using information obtained from the graph, calculate the resistances of the thermistor when the current is 0.10 mA and also when the current is 0.60 mA.

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(ii) Using the results of part (b) (i) deduce how the resistance of the thermistor changes as its temperature increases.

You may be awarded marks for the quality of written communication in your answer.

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(Total 9 marks)

Q27. (a) In the circuit shown in Figure 1, the battery has an emf of 6.0 V. With the switch closed and the lamp lit, the reading on the voltmeter is 5.4 V.

Figure 1

Explain without calculation, why the voltmeter reading is less than the emf of the battery.

You may be awarded marks for the quality of written communication in your answer.

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(b) A torch is powered by two identical cells each having an emf of 1.5 V and an internal resistance r. The cells are connected in series. The torch bulb is rated at 1.6 W and the voltage across it is 2.5 V.

(i) Draw the circuit described.

(ii) Calculate the internal resistance of each cell.

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(c) In the circuit in Figure 2 the cell has emf and internal resistance r. The voltage V across the cell is read on the voltmeter which has infinite resistance, and the current I through the variable resistor R is read on the ammeter.

Figure 2

By altering the value of the variable resistor R, a set of values of V and I is obtained. These values, when plotted, give the graph shown in Figure 3.

Figure 3

Show how the values of and r may be obtained from this graph. Explain your method.

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(Total 11 marks)

Q28. (a) A student is given three resistors of resistance 3.0 Ω, 4.0 Ω and 6.0 Ω respectively.

(i) Draw the arrangement, using all three resistors, which will give the largest resistance.

(ii) Calculate the resistance of the arrangement you have drawn.

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(iii) Draw the arrangement, using all three resistors, which will give the smallest resistance.

(iv) Calculate the resistance of the arrangement you have drawn.

.............................................................................................................

.............................................................................................................

.............................................................................................................(5)

(b) The three resistors are now connected to a battery of emf 12 V and negligible internal resistance, as shown in Figure 1.

Figure 1

(i) Calculate the total resistance in the circuit.

.............................................................................................................

.............................................................................................................

(ii) Calculate the voltage across the 6.0 Ω resistor.

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.............................................................................................................

.............................................................................................................(4)

(Total 9 marks)

Q29. (a) A set of decorative lights consists of a string of lamps. Each lamp is rated at 5.0 V, 0.40 W and is connected in series to a 230 V supply.

Calculate

(i) the number of lamps in the set, so that each lamp operates at the correct rating,

.............................................................................................................

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(ii) the current in the circuit,

.............................................................................................................

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(iii) the resistance of each lamp,

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(iv) the total electrical energy transferred by the set of lights in 2 hours.

.............................................................................................................

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.............................................................................................................(5)

(b) When assembled at the factory, one set of lights inadvertently contains 10 lamps too many. All are connected in series. Assume that the resistance of each lamp is the same as that calculated in part (a) (iii).

(i) Calculate the current in this set of lights when connected to a 230 V supply.

.............................................................................................................

.............................................................................................................

.............................................................................................................

(ii) How would the brightness of each lamp in this set compare with the brightness of each lamp in the correct set?

.............................................................................................................(3)

(Total 8 marks)

Q30. A battery of emf and internal resistance r is connected in series to a variable resistor R and an ammeter of negligible resistance. A voltmeter is connected across R, as shown in the figure below.

(a) (i) State what is meant by the emf of the battery.

.............................................................................................................

.............................................................................................................

(ii) The reading on the voltmeter is less than the emf. Explain why this is so.

.............................................................................................................

.............................................................................................................(2)

(b) A student wishes to measure and r. Using the circuit shown in the figure above the value of R is decreased in steps and at each step the readings V and I on the voltmeter and ammeter respectively are recorded. These are shown in the table.

reading on voltmeter/V reading on ammeter/A

8.3 0.07

6.8 0.17

4.6 0.33

2.9 0.44

0.3 0.63

(i) Give an expression relating V, I, and r.

.............................................................................................................

(ii) Draw a graph of V (on the y-axis) against I (on the x-axis) on graph paper.

(Allow one sheet of graph paper)

(iii) Determine the values of and r from the graph, explaining your method.

: ………...........................................................................................

............................................................................................................

r: .........................................................................................................

.............................................................................................................(8)

(Total 10 marks)

Q31. (a) A student wishes to measure the resistivity of the material of a uniform resistance wire. The available apparatus includes a battery, a switch, a variable resistor, an ammeter and a voltmeter.

(i) Draw a circuit diagram which incorporates some or all of this apparatus and which enables the student to determine the resistivity of the material.

(ii) State the measurements which must be made to ensure that a reliable value of the resistivity is obtained.

.............................................................................................................

.............................................................................................................

.............................................................................................................

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(iii) Explain how a value of the resistivity would be obtained from the measurements.

.............................................................................................................

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.............................................................................................................

.............................................................................................................

.............................................................................................................(10)

(b) A wire made from tin with cross-sectional area 7.8 × 10–9 m2, has a pd of 2.0 V across it. Calculate the minimum length of wire needed so that the current through it does not exceed 4.0 A.

resistivity of tin = 1.1 × 10–7 Ω m

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................(2)

(Total 12 marks)

Q32. In the circuit shown in the figure below, the battery, of negligible internal resistance, is connected to two resistors which form a potential divider.

(a) (i) Calculate the current through the ammeter.

.............................................................................................................

(ii) A 20 Ω resistor is now connected between X and Y. State and explain, without further calculation, whether the current through the ammeter will increase or decrease.You may be awarded marks for the quality of written communication in your answer.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................(4)

(b) The 20 Ω resistor is now removed and replaced with a voltmeter. Stating the assumption made, show that the reading on the voltmeter is 4.9 V.

......................................................................................................................

......................................................................................................................

......................................................................................................................(2)

(c) The voltmeter is now removed and the terminals X and Y joined together with a wire. Calculate the reading on the ammeter.

......................................................................................................................

......................................................................................................................

......................................................................................................................(2)

(Total 8 marks)

Q33. (a) A steady current of 0.25 A passes through a torch bulb for 6 minutes.Calculate the charge which flows through the bulb in this time.

......................................................................................................................

......................................................................................................................

......................................................................................................................(2)

(b) The torch bulb is now connected to a battery of negligible internal resistance. The battery supplies a steady current of 0.25 A for 20 hours. In this time the energy transferred in the bulb is 9.0 × 104 J. Calculate

(i) the potential difference across the bulb,

.............................................................................................................

.............................................................................................................

.............................................................................................................

(ii) the power of the bulb.

.............................................................................................................

.............................................................................................................

.............................................................................................................(3)

(Total 5 marks)

Q34. The circuit in Figure 1 shows a sinusoidal ac source connected to two resistors, R1 and R2, which form a potential divider. Oscilloscope 1 is connected across the source and oscilloscope 2 is connected across R2.

Figure 1

(a) Figure 2 shows the trace obtained on the screen of oscilloscope 1. The time base of the oscilloscope is set at 10 m/s per division and the voltage sensitivity at 15 V per division.

Figure 2

For the ac source, calculate

(i) the frequency,

.............................................................................................................

.............................................................................................................

(ii) the rms voltage.

.............................................................................................................

.............................................................................................................

.............................................................................................................(4)

(b) The resistors have the following values: R1 = 450 Ω and R2 = 90 Ω.Calculate

(i) the rms current in the circuit,

.............................................................................................................

(ii) the rms voltage across R2.

.............................................................................................................(2)

(c) Oscilloscope 2 is used to check the calculated value of the voltage across R2. The screen of oscilloscope 2 is identical to that of oscilloscope 1 and both are set to the same time base. Oscilloscope 2 has the following range for voltage sensitivity: 1 V per div., 5 V per div., 10 V per div. and 15 V per div.State which voltage sensitivity would give the most suitable trace. Explain the reasons for your choice.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................(3)

(Total 9 marks)

Q35. The battery of an electric car consists of 30 cells, connected in series, to supply current to the motor, as shown in the figure below.

(a) Assume that the internal resistance of each cell is negligible and that the pd across each cell is 6.0 V.

(i) State the pd across the motor.

.............................................................................................................

(ii) The battery provides 7.2 kW to the motor when the car is running. Calculate the current in the circuit.

.............................................................................................................

.............................................................................................................

.............................................................................................................

(iii) The battery can deliver this current for two hours. Calculate how much charge the battery delivers in this time.

.............................................................................................................

.............................................................................................................

.............................................................................................................

(iv) Calculate the energy delivered to the motor in the two hour period.

.............................................................................................................

.............................................................................................................

.............................................................................................................(7)

(b) In practice, each cell has a small but finite internal resistance. Explain, without calculation, the effect of this resistance on

• the current in the circuit, and

• the time for which the battery can deliver the current in part (a)(ii).

You may assume that the motor behaves as a constant resistance.

You may be awarded additional marks to those shown in brackets for the quality of written communication in your answer.

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................

......................................................................................................................(3)

(Total 10 marks)

M1. (a) circuit diagram to show:ammeter in series, voltmeter in parallel (1)variable source (e.g. battery + rheostat or potential divider) (1)

2

(b) (i) RX = = 56 Ω (1)

(ii) RX = (e.g.) = 23 (Ω) (1)

RX depends on current (or voltage) non-ohmic3

(c) (i) col C col D0.15 2.530.20 2.830.25 3.090.30 3.370.35 3.660.40 3.94

four pairs of values correct (1)all six pairs correct and col D to no more than 4 s.f (1)

(ii) axes labelled (1)suitable scales chosen (1)at least five points plotted correctly (1)acceptable straight line (1)

(iii) k = gradient (1)

gradient = = 5.7 (V–1) (1)

intercept on y-axis =ln A (1)

(intercept = 1.68 gives) A = e1.68 = 5.4 (mA) (1)

unit for k or A correct (1)

(iv) the points define a straight line (1)valid over given range (1)

max 10[15]

M2. (a)

straight line in both quadrants,through origin for A and B (1)greater gradient for B (1)2

(b)

characteristic to show:positive current increasing slowly and then rapidly (1)at ≈ 0.6V (1)negative current either zero or just < zero (1)

3

(c) as voltage increases, current increases (1)current heats filament (1)therefore resistance increases (1)correct argument to explain curvature(1)mirror image in negative quadrant (1)

max 4[9]

M3. (a) R = (1)

= 0.031 Ω (1) (0.0305 Ω)2

(b) constant volume gives l1A1 = l2A2

[or l2 = 2l1 and A2 = A1/2] (1)

R = =4R (1)

[or calculation with l2 = 2.8 (m) and A2 = 3.9 (m2) (1)]

gives R = 0.124 Ω (1)]2

[4]

M4. (a) (i) the emf (of the battery) (1)

(ii) the voltage across the battery when current flows[or terminal voltage or pd supplied to the circuit]

(iii) V = (3 × 0.5) = 1.5 (V) (1)current = (1.5/14) = 0.11 A (1) (0.107 A)

(iv) ( = V + Ir and emf = 3.5 × 0. 5 = 1.75 (V) gives)

1.75 = 1 .5 + 0.1 07r (1)

r = 2.3 Ω

[or use of = I (R + r) with I = 0. 107 gives r = 2.4 Ω

and I = 0. 11 gives r = 1.9 Ω]

(allow C.E. for value of I from (iii))6

(b) (i) peak value = 3.5√2 = 4.9 V (1)

(ii) oscilloscope screen to showvertical line of height 2.5 divisions above central axis (1)and below central axis (1)

3[9]

M5. (a) (i) total resistance = 180 + 60 = 240 (Ω) (1)(V = IR gives) 12 = I 240 and I = 0.05 A (1)

(ii) very large or infinite resistance (1)

(iii) V = 0.05 × 60 = 3.0V (1)

[or statement that V = of 12 V]

[or use of potentiometer equation

Vout = Vin = 3.0 V]

(allow C.E. for value of I from (a)(i))4

(b) parallel resistance gives lower equivalent resistance[or resistance of lower section of potentiometer reduced] (1)total resistance in circuit reduced (1)current through battery increases since V constant (1)

max 2QWC 2

[6]

M6. (a) first pair in parallel (1)

= = gives R’ = 20 (Ω) (1)

second pair in parallel gives R” = 30(Ω) (1)

resistance between A and B = 20 + 30 (1) (= 50 Ω)

(allow C.E. for values of R’ and R")4

(b) (i) total resistance = 50 + 50 = 100 Ω (1)(V = IR gives) 24 = I 100 and I = 0.24 A (1)

(ii) current in 60 Ω = I (1)= 0.080 (A) (1)[or alternative method](allow C.E. for value of I from (b)(i))

4[8]

M7. (a) (i) battery, milliammeter, and wire in series (1) (1)voltmeter across the wire (1)variable resistor/potential divider in series (1)

(ii) alter variable resistor (1)to obtain a series of values of I and V (1)QWC 1

(iii) plot a graph of V against I (1)gradient = R (1)QWC 1

[or calculate R = V/I for each reading and take mean]8

(b) (i) (P = gives) 1200 =

R = 44.1Ω (1)

(ii) R = (1)

l = (1)

= 3.8 m (1)

(allow C.E. for value of R in (i))

5[13]

M8. (a) (i) for X: (P = VI gives) 24 = 12I and I = 2 A (1)for Y 18 = 6I and I = 3 A (1)

2

(b) (i) 12 V (1)

(ii) voltage across R2 (= 12 – 6) = 6 (V) (1)I = 3 (A) (1)(V = IR gives) 6 = 3R2 and R2 = 2Ω (1) (allow C.E. for I and V from (a) and (b)(i))

[or V = I(Ry + R2) (1) 12 = 3(2 + R2) (1) R2 = 2Ω (1)]

(iii) current = 2 (A) + 3 (A) = 5 A (1)(allow C.E. for values of the currents)

(iv) 27 (V)– 12 (V) = 15 V across R1 (1)

(v) for R1, 15 = 5 R1 and R1 = 3Ω (1)(allow C.E. for values of I and V from (iii) and (iv)7

[9]

M9. (i) (V = IR gives) 12 = (30 + 30 + 2)I (1)

I = = 0.19 A (1) (0.194 A)

(ii) VPQ = 12 – (0.19 × 2) (1) = 11.6 V (1)

(allow C.E. for incorrect I in (i))

[or VPQ = 0.19 × 60 = 11.6 V] (I = 0.194 A gives 11.6 V)

[or VPQ = 12 × = 11.6 V

(iii) (PA = I2R gives) PA = (0.19)2 × 30 = 1.08 (1) W (1)

[or PA = ]

(allow C.E. for incorrect I in (i) or incorrect V in (ii))

(iv) (E = PAt gives) E = 1.08 × 20 (1)

= 21.6 J (1)

(allow C.E. for incorrect PA in (iii))[8]

M10. (a) between A and C: (each) series resistance = 100Ω (1)

(parallel resistors give) + = gives RAC = 50Ω (1)2

(allow C.E. for incorrect series resistance)

(b) between A and B: series resistance = 150Ω (1)

parallel = (1)

(allow C.E. for series resistance)

RAB = 37.5Ω (1) (38Ω)3

[5]

M11. (a) (i) energy changed to electrical energy per unit charge/coulombpassing through[or electrical energy produced per coulomb or unit charge][or pd when no current passes through/or open circuit] (1)

(ii) I = = 2.5 A (1)

(iii) (use of = I(R + r) gives) = V + Ir and 8 = 6 + Ir (1)

substitution gives 8 – 6 = 2.5r (1) (and r = 0.8 Ω)4

(b) (i) (use of P = I2R gives) PR = 2.52 × 2.4 = 15 W

[or P = VI gives P = 6 × 2.5 = 15 W] (1)

(allow C.E. for value of I from (a))

(ii) PT = 15 + (2.52 × 0.8) (1)

= 20 (W) (1)

(allow C.E. for values of PR and I)

(iii) E = 5 × 2 × 60 = 600 J (1)

(allow C.E. for value of P from (i) and PT from (ii))4

[8]

M12. (a) (i) (use of Vrms = gives) V0 = 7.1√2 = 10 V (1)

(ii) T = 10 (ms) (1)

(use of f = gives) f = = 100 Hz (1)3

(b) control 1: time base (1) (or time period)

(use of T = gives) T = = 5 × 10–3 (s) (1)

setting = 2.5 ms (div–1) (1)

control 2: voltage sensitivity or Y-plate setting (or Y-gain) (1)setting = 20 V (div–1) (1)

5[8]

M13. (a) constantan (wire) (1)I V[or straight line through origin][or constant gradient][or V and I increasing proportionally] (1)

2

(b) (i) (for V = 1 V, use of V = IR gives)

= 1.5 Ω (1) (1.47 ± 0.02Ω)

(for V = 10 V) R = = 4.4 or 4.5 (Ω) (1)

(4.46 ± 0.02 (Ω))

(ii) as current increases (1)greater heating effect (or temperature of filament increases) (1)R increases with temperature (1)

max 4QWC 2

(c) (1)

(R = 1/gradient, or use of values gives) R = = 5.5 (Ω) (1) (5.45 (Ω))

= 4.6 × 10–7 Ωm (1) (4.63 × 10–7 Ωm)

(allow C.E. for value of R)3

(d)

Variable supply or rheostat, (in series with filament) or pot. divider (1)filament and current sensor in series (1)voltage sensor across filament (1)connections to datalogger (1)

max 3[12]

M14. (a) (i) 5 V (1)

(ii) RT = 36 (Ω)(use of V = IR gives) 15 = I × 36 and I = 0.42 A (1)

3

(b) (i) equivalent resistance of the two lamps (1)

RT = 6 + 12 = 18 (Ω) and 15 = I × 18 (1) (to give I = 0.83 A)

(ii) current divides equally between lamps (to give I = 0.42 A)(or equivalent statement) (1)

3

(c) same brightness (1)(because) same current (1)

2[8]

M15. (a) circuit to show: ammeter in series with coil (1) voltmeter across coil (or across pot.div. output) (1) potential divider or variable resistor correctly drawn (1)

3

(b) (i) values of R/Ω (all correct to 3 or 4 sig.fig.) 2.53(2)

2.70(3)2.94(1)3.23(3.226 to 4 sig.fig.)3.39(0)3.64(3.636 to 4 sig.fig.) (1)

(ii) suitable graph to give straight line (e.g. R vs θ) (1)axes labelled (1)5 points plotted correctly (1)

appropriate scales (1)best fit line (drawn) (1)

6

(c) (i) R0 (= y intercept) = 2.38 Ω (± 0.2 Ω) (1)

(ii) gradient = αR0 (1)

= = 1.58 × 10–2 (Ω °C–1) (± 0.08 × 10–2(Ω °C–1) (1)

α = = 6.6(4) × 10–3 (± 0.3(4) × 10–3) (1) °C–1(or K–1) (1)

(allow C.E. for values of R0 from (i) and gradient values)5

(d) resistance caused by scattering/collisions ofcurrent/unattached/free/conduction electrons with vibratingions/atoms (1)increase in temperature makes the (+) ions/atoms vibrate more (1)unattached/free/conduction electrons scatter/collide more[or progress of electrons made more difficult] (1)

max 2QWC 1

[16]

M16. (a) ρ = (1)

R = resistance (of wire), A = cross-sectional area,l = length (of wire) (1)

2

(b) (i) R = (1)

= 75 Ω (1)

(ii) length has decreased causing resistance to decrease (1)area increased, causing resistance to decrease (1)each changed by factor of 1.5 × 103 (1)

4QWC 1

[6]

M17. (a) V = –Ir + (1)1

(b) straight line (within 1st quadrant) (1)negative gradient (1)

2

(c) : intercept on voltage axis (1)r: gradient (1)

2[5]

M18. (a) (i) (1)

R = = 13 Ω (1) (13.3)

(ii) two resistors in parallel give 20 (Ω) (1)R = 20 + 40 = 60 (Ω) (1)max 3

(b) (i) three resistors in parallel give (= 2 (Ω))

and total resistance = 4 (Ω) (1)

total current = = 3 (A) (1)

(allow C.E. for value of total resistance)

current in each element 1.0 A (1)

(allow C.E. for value of total current)

[or 6 V across each setresistance of each set = 2 Ω, gives current througheach set = 3 (A)current in each element = 1.0 A][or 6 V across each set/resistor,resistance of one resistor = 6 Ω,gives current in each element = 1.0 A]

(ii) six resistors in series gives R = 36 (Ω) and I = = 0.3 (A) (1)

heating effect (I2R) much reduced [or less power] (1)5

[8]

M19. (i) I = [or Q = It] (1)

Q = 40 × 10–3 × 3 × 60 = 7.2 C (1)

(ii) number of electrons = = 4.5 × 1019 (1)

(allow C.E. for value of Q from (i)

(iii) V = (1)

=

(allow C.E. for value of Q from (i)

(iv) (use of V = IR gives) R = (1)

(allow C.E. for value of V from (iii))[6]

M20. (a) circuit diagram to show:wide end of conducting strip to – of battery, narrow end to + (1)voltmeter between wide end and probe (1)

2

(b) resistance gradient increases as x increases (1)because strip becomes narrower (as x increases) (1)current constant throughout strip (1)voltage gradient = current × resistance gradient, sovoltage gradient increases as x increases (1)

4

(c) (i) (2l – x) ln (2l – x)(0.700) (–0.357)0.60(0) –0.5110.53(0) –0.6350.47(0) –0.7550.44(0) –0.8210.42(0) –0.868

1st column correct to 2 s.f. (1)2nd column correct to 3.s.f. (1) (1)(only 4 values correct, (1))

(ii) suitable scales (1)axes labelled and units included (1)

5 points correctly plotted (1)acceptable straight line (1)straight line confirms equation because equation is of formy = mx + c with negative gradient (1)

(iii) gradient = (–) = (–) 15.4 (V) (1)1.44 V1 = 15.4 gives V1 = 11 V (1) (10.7 ± 0.2 V) [alternative: V = V1 when x = l and ln (2l – x) (= ln 0.4) = 0.92 (1)at ln (2l – x) = 0.92, graph gives V1 = 11 V (1) (10.7 ± 0.2 V)]

10[16]

M21. (a) (V = IR gives) Vrms = (5.3 × 10–3 × 2 × 103) = 10.6 (V) (1)V0 = Vrms √2 = 10.6√2 = 15 V (1) (14.99 V)[or calculate I0 (= 7.5 mA) and then V0]

2

(b) (use of T = gives) T = = 2 × 10–3 = 2 0(ms) (1)

trace to show:correct wave shape (sinusoidal) (1)correct amplitude (3 divisions) (1)correct period (8 divisions) (1)

4[6]

M22. (a) R = (1)

ρ is resistivity, l is the length of the wire, A is the cross-sectional area (1)2

(b) (i) P = (1)

R = = 106(Ω)(1) (105.8 Ω)

l = = = 7.7 m (1) (7.69 m)

(allow C.E. for incorrect value of R )

(ii) in series, voltage across each < 230 V or pd shared (1)

power (= V2/R) is less than 500 W in each (1)

in parallel, voltage across each = 230 V (1)

correct rating, conclusion (1)

[or, in series, high resistance or combined resistance (1)

low current (1)

in parallel, resistance is lower, higher current (1)

more power, justified (1)]max 6

[8]

M23. (a) at 200 °C : RR = 130 ± 1 (Ω), RTh = 18 ± 1Ω (1)1

(b) (i) VAB = Vin (1)

= 12 × (1)

= 1.5 V (1) (1.46 V)

(allow C.E. for values from (a))

(ii) Rth = RR occurs at 50 °C (1) 4

(c) (i) (use of P = gives) Rb = = 18 Ω (1)

[or use of P = VI and calculate I]

(ii) (S open, RTh ≈ 90 Ω)bulb and thermistor in parallel (1)gives lower resistance than thermistor on its own (1)total resistance in circuit decreases (1)current increases VR > 6 V (1)(hence Vth < 6 V i.e. decreases)[or use of potentiometer equation, or ratio of resistances andshare of pd]

max 4[9]

M24. (a) (i) circuit details:diode (correct symbol) connected in correct direction,plus ammeter in series with diode (1)variable battery, variable resistor or potentiometer (1)voltmeter across diode (1)ammeter and voltmeter connected to data logger (1)

(ii) description: current and voltage recorded on data logger (1)change voltage/current in (small) steps by changingvariable resistor/potentiometer/battery (1)

QWC 1

(iii) reverse the diode or battery (1)max 6

(b) correct characteristic in positive and negative quadrant (1) (1)current heats the filament[or discussion of collisions with molecules] (1)hence resistance increases (1)rate of increase in I decreases [or suitable explanation] (1)same effect when current reversed (1)

max5QWC 1

[11]

M25. (a) (i) (total) resistance = (20 + 60) (Ω) (1)

(V = IR gives) I = = 0 075 A (1)

(ii) with S closed, (effective) resistance = 20 (Ω) (1)

I = =0.3 A (1)max 3

(b) use of same current as in part (i) (1)voltmeter reading = 0.075 × 60 = 4.5 V (1)

[or use potentiometer equation 6 × = 4.5 V](allow C.E. for value of I from (a)(i)

2[5]

M26. (a) circuit diagram to show:(milli)ammeter in series with thermistor, which must have correct symbol (1)voltmeter across thermistor (1)variable resistor/variable power supply/potentiometer in series withthermistor (1)variable component labelled (1)

4

(b) (i) resistance (when I = 0.1 mA) = (44 ± 0.1(ma)/0.1)= 44 ± 1 × 103 (Ω) (1)resistance (when I = 0.6 mA) = (17/0.6) = 28 × 103 Ω (1)

(ii) resistance decreases with increasing current (1) (from (b) (i))as current increases thermistor heats up (1)

resistance decreases as temperature increases (1)5

QWC 1[9]

M27. (a) battery has internal resistance (1)current passes through (this resistance) (1)work done/voltage lost, which reduces the value of the emf (1)

3QWC 1

(b) (i) circuit diagram to show:two cells in series (1)two resistors, each labelled r (1)

(ii) (use of P = IV gives) 1.6 = 2.5 I (1) (I = 0.64 (A))(use of = V + Ir gives) 3.0 = 2.5 +0.64 × 2r (1) (1)0.5 = 1.28r and r = 0.39 Ω (1)[or Rbulb =2.52/1.6 = 3.9 (Ω) and 2.5 = 3.9 × I gives I = 0.64 (A)‘lost volts’ = (3 – 2.5) = 0.5 (V) i.e. 0.25 (V) per cell0.25 = 0.64r and r = 0.39 Ω]

5

(c) = V + Ir gives V = –Ir + (equation of straight line) (1)intercept on y-axis gives (1)gradient gives (–)r (1)

3[11]

M28. (a) (i) three resistors in series (1)

(ii) R = 3.0 + 4.0 + 6.0 = 13 Ω (1)

(iii) three resistors in parallel (1)

(iv) (1)

R = 1.3 Ω (1)5

(b) (i) two resistors in parallel give and R’ = 2.0 (Ω) (1)

total resistance = (2 + 4) = 6.0 Ω (1)4

(ii) divide the emf in the ratio of 2 : 4 (1)to give 4.0 V (1)[or any suitable method]

[9]

M29. (a) (i) no of bulbs = = 46 (1)

(ii) (use of P = VI gives) I = = 0.080 A (1)

(iii) resistance of each bulb = = 63 Ω (62.5 Ω)

(allow C.E. for number of bulbs and value of I)

[or R = 62.5 Ω

or = 62.5 Ω]5

(iv) energy consumed by the set = 0.4 × 46 × (2 × 60 × 60) (1)= 132 kJ (1)(allow C.E. for number of bulbs from (i))

(b) (i) no of bulbs = 56, gives total resistance = 62.5 × 56 (Ω) (= 3500) (1)

I = = 0.066 A (1) (0.0657 A)

(use of 63 Ω gives 0.065 A)

(allow C.E. for no. of bulbs in (a) (i) and R in (a) (iii))

(ii) bulbs would shine less bright (1)3

[8]

M30. (a) (i) electrical energy produced (in the battery) per unit charge (1)

[or potential/voltage across terminals when there is no current]

(ii) there is a current (through the battery) (1)

voltage ‘lost’ across the internal resistance (1)Max 2

(b) (i) = V + Ir (1)

(ii) labelled scales (1)correct plotting (1)best straight line (1)

: intercept on y axis (1) = 9.2 (± 0.1) V (1)

r : (−) gradient = = 14.2 Ω (1) (range 14.0 to 14.3)8

[10]

M31. (a) (i) circuit diagram to show:wire, ammeter, battery, (variable resistor) and switch in series (1)[or potentiometer with ammeter in correct position]voltmeter across the wire (1)

(ii) (method: constant length of wire)measure length (of wire) (1)measure diameter (of wire) (1)measure voltage (across) and current (through wire) (1)

vary resistor to obtain different voltage and current (1)alternative[(method: variable length of wire)measure length (each time) (1)measure diameter (1)(for full length of wire) measure voltage and current (1)voltmeter to shorter lengths, measure voltage (and current) (1)]

(iii) (use of) ρ (to calculate ρ) (1) (for either method)calculate A from (πr2) (1) (for either method)(method: constant length of wire)

determine for (one) length (1)repeat readings (for same length and) take mean of ρ or R (1)[or plot graph of V vs I to give mean R (1)

or gradien = (1)]alternative[(method: variable length of wire)

determine for each length (1)calculate ρ for each length and take mean (1)[or graph of R vs l (1) with correct gradient (1)]

10

(b) (use of gives) (1)

l = 0.035 m (1)2

[12]

M32. (a) (i) (use of V = I R gives) 12 = I × 270 and I = 44.(4) mA (1)

(ii) two resistors in parallel give resistance less than 110 Ω(1)( ) total resistance decreases (1)current increases (1)

4

(b) V = 44.4 × 10−3 × 110 (1) (= 4.9 V)

[or ]

(assumption) no current flows through voltmeter (1)

[or voltmeter has very large or infinite resistance]2

(c) total resistance (in circuit) = 160 (Ω) (1)

12 = I × 160 and I = 75 mA (1)2

[8]

M33. (a) (1)

∆Q = 0.25 × 6 × 60 = 90 C (1)2

(b) (i) (1) [or E = VIt]

(1)

(ii) (use of gives) (1)

[or P = IV gives P = 0.25 × 5 = 1.2(5) W](allow C.E. in alternative method for value of V from (i))

3[5]

M34. (a) (i) T = 40(ms) (1)

Hz (1)(allow C.E. for value of T)

(ii) peak voltage (= 3 × 15) = 45 (V) (1)

rms voltage =32 V (1)(31.8 V)

4

(b) (i) Irms = = 59mA (1)

(58.9mA)(use of 32 V gives 59(.2) mA)(allow C.E. for value of Vrms from (a))

(ii) Vrms = 59 × 10–3 × 90 = 5.3(1) V (1)

(allow C.E. for value of Irms from (i)) [or V2 =V1 ]2

(c) Vpeak = 5.31× =7.5(1) (V) (1)best choice: 5 V per division (1)

(allow C.E. for incorrect Vrms and for suitable reason)

reason: others would give too large or too small a trace (1)3

[9]

M35. (a) (i) 180V (1)

(ii) (use of P = VI gives) (1)= 40A (1)(allow C.E. for value of v from (i))

(iii) (use of Q = It gives) Q = 40 × 2 × 60 × 60 (1)= 2.9×105C (1)(2.88 × 105C)

(allow C.E. for value of I from (ii))

(iv) E = QV (1)(or Pt or VIt)=2.9 × 105 × 180 = 52 (.2) × 106J (1)

(use of Q = 2.88 × 105 gives 5.18 × 106 J)(allow C.E. for values of Q and V)

7

(b) (i) pd across each cell/battery is reduced (1)[or because of resistance in each cell]current in circuit is reduced (1)

(total) charge circulated by battery remains the same (1)

[or valid energy reasons]time for which (reduced) current flows is increased (1)

3[10]

##

Although most candidates in part (a) drew an adequate circuit, usually with a variable resistor to change the pd, some failed to draw the voltmeter in parallel with the component X.

Part (b) worked well, with most candidates calculating the resistance correctly in both parts (i) and (ii). However, several of these candidates were not aware of the property of an ohmic conductor and so failed to gain full credit.

In part (c)(i) the majority of candidates gained both marks although some calculated

instead of (V – 0.55). In part (ii) many candidates gained at least three ofthe possible four marks through correct labelling, plotting and drawing the line. In part (iii) most candidates gained credit for calculating the gradient of the line and for relating the gradient correctly to the equation, but few candidates were able to calculate the constant A from the graph; they were usually unaware that the intercept on the y axis was ln A. Units for k and A were usually incorrect or absent.

E2. In part (a) most candidates drew the required straight line for component A, although a few did not bother to draw it in the negative quadrant, thereby losing a mark. The difficulty arose in deciding whether the straight line for B should have a larger or smaller gradient than line A. Overall about 50 % of the candidates had it correct.

The characteristic for the silicon semiconductor in part (b) was, on the whole, not well drawn. Faults were: showing a steady increase in current, with the characteristic at about 45° without showing a rapid rise; showing the current increasing directly from (0,0) without a slow increase section and showing the negative current at a considerable distance below the axis. It was good however to note that when a slow increase in current was shown, that the majority of candidates knew that the upcurving occurred at around 0.6 V.

The explanation of the filament lamp characteristic in part (c) was not done well on the whole. Many candidates did not indicate that the current in the filament increased the temperature of the filament, but it was known that resistance increased with temperature. A large number of candidates, having arrived at this point in the explanation then stated that therefore the current was less. It must be stressed that the current does not decrease, but that the rate of rise of current decreases. Many candidates opted for the gradient argument and were awarded marks for a correct explanation. Few candidates made reference to the fact that a current in the opposite direction produced the same effect, thus creating a mirror image of the characteristic

shown in the positive quadrant.

E3. Part (a) was straightforward and merely involved inserting data into an equation, usually recollected correctly. The answer here was 0.0305 but many candidates reduced this to 0.03 Ω, which incurred a significant figure error.

Part (b) proved to be a simple exercise to most candidates, but many struggled with the mathematics and the algebra if they tried to rearrange the equation. The majority of candidates simply inserted new values in the equation and showed that the new value of resistance was four times the previous value.

E4. In part (a), although many candidates gave the emf as the answer to (i), they struggled with part (ii) and failed to put into words what they obviously knew. Very few referred to the voltage across the battery when a current flowed, but rather used a vague description of the voltage across the circuit. Terminal pd was a phrase which occurred infrequently but which the examiners were pleased to see. The calculation in part (iii) was straightforward but many candidates incurred a significant figure penalty here by giving the answer as 0.1 A, i.e. not reducing correctly from the answer of 0.107 A. Part (iv) realised many correct answers and it was pleasant to see that candidates were using correctly the equation involving the emf and the internal resistance.

Part (b) involved a use of the oscilloscope which has not been tested before, and it was encouraging to find a large number of correct answers. The initial calculation was carried out correctly by the large majority of candidates and they also drew the vertical line on the screen correctly. The common error in this section was simply showing two points, without the line joining them. This did incur a penalty.

E5. The calculation in part (a)(i) was reasonably well done, although a significant number of candidates used only the resistance of the upper part of the potential divider to calculate the current and ignored the lower section. The property of the voltmeter was generally well known, although a significant number of candidates assumed that the resistance was zero or very low. Part (iii) usually realised correct results, even by those candidates who had the first part incorrect; they achieved this by using the potentiometer equation.

Although there was only a maximum of two marks for part (b) it was disappointing to find that many candidates gained only one or were not awarded any marks. Explaining what happens in a circuit in terms of Ohm’s law, without the accompanying calculation is a technique that needs to be developed. No doubt, many of the candidates who failed on the explanation would have gained full marks if a calculation had been required. Many candidates who argued correctly that the total resistance in the circuit decreased, then made the simple statement that the current would therefore increase, without justifying this in terms of a constant voltage.

E6. Part (a) was the calculation of the equivalent resistance of a network of resistors consisting of resistors connected in series and in parallel. The majority of candidates gained full marks on this section and were not troubled by the calculation. However, it is worth pointing out that since the final answer of 50 Ω was given in the question, then in order to gain full marks it was necessary to show that the two equivalent series resistors were being added together.

Part (b) did not prove to be as easy; the problem in (i) was that many candidates gave the total resistance as 50 Ω rather than 100 Ω. No consequential error for calculating the current was allowed and frequently no marks were awarded for this section. It was possible in part (b)(ii) to gain the two marks even if the answer to (i) was incorrect, but very few candidates managed to gain these marks. The usual error was giving the current in the circuit as 24/20, i.e. ignoring the second batch of parallel resistors. Again, many candidates, having calculated the total current correctly, assumed that 2/3 would pass through the 60 Ω resistor, not realising that the greater the resistor, the lower the current for a given voltage.

E7. This question gave candidates the opportunity to describe an experiment and also gain the ‘quality of written communication’ marks. Most of the descriptions were satisfactory, but examiners were disappointed by the large number of candidates failing to explain clearly that measurements of voltage and current were repeated for different settings of the variable resistor. Examiners had to read between the lines to see if the candidates were in fact saying this. It is difficult to imaging a description of a simpler experiment being asked and it was sad to find that many candidates did not gain high marks because of their inability to express themselves clearly.

The circuit was usually drawn correctly, using either a variable resistor or a potentiometer as the extra piece of apparatus. The potentiometer caused some consequential problems however, because the ammeter would then be inserted in the part of the circuit leading into the potentiometer and not in series with the wire.

Many candidates obviously did not read the question thoroughly since switches were missing and the actual wire whose resistance was required, was frequently omitted. Several candidates used a variable source instead of a variable resistor. This was not accepted. The other common error which occurred on far too many occasions was misreading the question completely and embarking on an experiment to determine the resistance of a variable length of wire and hence the resistivity of the material. In such answers, marks were given for the circuit, but none for the description.

The majority of candidates knew how to obtain the resistance from the observed values using either a graphical method or a calculation of R = V/I and then taking the mean. Several candidates took the mean of all the voltage measurements and the mean of all the current measurements and thus obtained one value of the resistance. This was considered to be poor practice as a method of determining the value of resistance.

The calculation in part (b) was usually correct and many candidates gained full marks. One error which occurred frequently and which incurred a penalty, was not treating the rms voltage as the

working voltage, but converting it to peak voltage. The other error was incorrect use of the equation relating resistivity and resistance.

E8. This question involved the analysis of a relatively difficult circuit, which included two lamps and two resistors. The question however, was so structured that the majority of candidates were able to work through and gain full marks. Others, unfortunately, although making a reasonable attempt, failed to gain many marks. In part (a), the majority of candidates calculated the correct value of the currents passing through each lamp.

In part (b), obtaining the correct answers to parts (i) and (ii) depended to a large extent on realising that the reading on the voltmeter equalled the voltage across lamp X. Many candidates missed this point, but were still able to gain some marks. In part (ii) the error that was committed regularly was determining the resistance of lamp Y instead of the resistance of resistor R2. But at least, more candidates realised that the same current passed through lamp Y and R2. Answers to parts (iii) and (iv) used the answer to part (a) as a starting point, but many candidates failed to realise that the current through R1 was the sum of the current through the two lamps. Considerable guesswork took over at this stage and although most of it was wrong, candidates could still get a mark for part (v) by using the answers obtained to parts(iii) and (iv).

E9. This question worked well and many candidates gained full marks. The majority of the other candidates only failed to gain maximum marks because of a unit error or significant figure error. Disappointingly, many answers were expressed as a fraction. It should be noted that this practice is not acceptable and the first answer expressed as a fraction was treated as a significant figure error.

In part (i) the error which occurred most frequently was ignoring the internal resistance of the battery. The correct answer was 0.19 A, to two significant figures, but many candidates rounded this down to 0.2 A, which apart from incurring a penalty, also, when carried forward to part (ii), gave a voltage across the resistors of 12 V. This implied that there was no voltage developed across the internal resistance of the battery. Although many candidates produced such an answer no one noted that such a situation was not possible. Many answers to part (ii), when carrying forward an incorrect value of the current from part (i), gave an answer well in excess of 12 V. Again this did nor seem to worry the candidates.

In part (iii) many candidates made the error of calculating the power dissipated in the total external resistance instead of in resistor A alone. The unit of power was usually correct as was the unit of energy in part (iv). Many candidates arrived at the correct answer in part (iv). Consequential errors were carried forward throughout the whole question. This gave many candidates the chance to gain some marks even if their initial calculation and subsequent answer was incorrect.

E10. In this example of calculating equivalent resistance, the same resistor network was used twice, the equivalent resistance being calculated between different terminals. The majority of candidates had no difficulty with the calculations, but it was worrying to find many answers where the candidates had attempted a solution, not by calculation, but with phrases such as “electricity takes the path of least resistance and therefore the effective resistance (in part (b)) is 50 Ω.”

It was surprising to find that a significant number of candidates obtained the correct result in part (b) but failed on part (a), since part (b) was deemed to be the most difficult of the two. Considerable arithmetical difficulty was encountered by many candidates with the reciprocal of the resistance when calculating the resistance of parallel resistors.

E11. Candidates found this question very accessible and many gained full marks. In part (a) the meaning of emf seems to be reasonably well understood with most candidates opting for the voltage when no current passed through the circuit. Others defined it correctly in terms of energy per unit coulomb. There were, unfortunately, many candidates who, apparently, had not encountered the definition of emf and merely quoted electromagnetic force, or even tried to define it in terms of a force in the circuit. The calculation of the current in part (ii) was well done and in part (iii) correct substitution of values into the equation = V + Ir gave r = 0.80 Ω.

Part (b) was involved with calculation of power and energy and although the majority of candidates obtained the correct answer for the power dissipated in the 2.4 Ω resistor, fewer had the correct answer for the total power dissipated in the circuit and a disappointing number had the correct value for the energy wasted the battery. The usual answer to the last part was to give the energy in the complete circuit. Whether this was due to inaccurate reading of the question or due to lack of understanding could not be decided.

E12. High marks were usually obtained in this question. Very few candidates failed to gain full marks in part (a) although it was sad to see some giving the peak voltage as Vrms/√2. In part (b) a large number of candidates were obviously familiar with the correct terminology of the oscilloscope controls, but other terms were also accepted. The calculations of the new settings were generally well performed.

E13. The large majority of candidates gained the two available marks in part (a) without much difficulty and supported their choice of conductor with the correct argument.

Some care was required in part (b) in reading the graph accurately and the fact that the majority of candidates simply put down 0.7 A at a voltage of 1.0 V, showed that they gave the graph no more than a cursory glance. It should also be pointed out that drawing vertical and horizontal lines in thick pencil or in ink on the graph very often obliterates the required point and accurate values can then not be read. The explanation in part (b)(ii) of why the resistance increased was

satisfactorily done in terms of the increased temperature of the filament, although for a complete answer it should be pointed out that it is the current that heats up the filament. Very often there was no direct reference to the temperature of the filament, but instead a general phrase such as “...leads to an increase in temperature...” or “ heat is created...”.

The calculation on resistivity in part (c) was more often than not carried out successfully. The errors which did occur was taking the resistance as the gradient, and not as the inverse of the gradient and omitting or using wrong units.

Some circuit diagrams in part (d) were very well drawn and the majority of candidates obtained full marks. The usual omission was a variable resistor or variable supply to alter the voltage. If a potentiometer is used then it is important to show the ammeter in series with the filament and not in series with the power supply. Some candidates lost a mark by connecting the voltage and current sensor directly to a computer rather than through the data logger.

E14. The question involved straightforward calculations on voltage, resistance and current. In part (a)(i) it was hoped that candidates would have spotted the correct voltage across each lamp by inspection. Surprisingly, even those who managed to get the wrong answer in part (i) nevertheless ignored their answer and proceeded from first principles to obtain the correct answer to part (ii).

Part (b) involved the same circuit components as in part (a) but connected differently. The majority of candidates showed that the current from the battery was the value given in the question. Using this value they then proceeded to argue or calculate the current in each lamp. Those candidates who merely halved the current value obtained in part (i) without any reasoning did not gain the mark.

Although the question told the candidates that the current through each lamp was the same in both circuits it was disappointing to find in part (c) how many candidates tried to argue that the brightness of the bulbs in the 2nd circuit would be different to that in the first, the main thrust of their argument being that the voltage across each bulb was different and therefore that the brightness would be different.

E15. Circuit diagrams presented in part (a) were of variable quality and although many candidates scored full marks, a significant number of diagrams did not include a variable resistor or a potential divider. The thermistor symbol was often given as the symbol for a variable resistor.

The table in part (b) was usually completed correctly although some candidates failed to give the values of R to an appropriate number of significant figures or rounded off incorrectly. In part (ii), most candidates correctly plotted a suitably labelled graph of resistance on the y-axis against temperature on the x-axis. However, a significant number of candidates chose an unsuitable scale for the resistance axis and some candidates were careless in drawing the best-fit line.

The determination of R0 from the intercept on the resistance axis was done successfully in part (c). Many correct solutions were also seen in part (ii), although candidates who plotted

temperature on the y-scale and resistance on the x-scale were usually unable to make progress at this point. A few candidates attempted to use the equation and data from the table instead of using the graph and therefore were unable to gain full credit. A significant number of candidates who measured the gradient correctly, equated the gradient to a or else calculated α correctly, but gave the wrong unit.

In part (d), although most candidates were aware that increasing the temperature caused increased vibration of the metal ions, and thus made the progress of electrons through the wire more difficult, few correctly explained the cause of resistance. They usually failed to mention that the conduction electrons passing along the wire repeatedly collided with the metal ions. Weaker candidates considered the motion of particles without indicating whether the particles were free electrons, metal atoms or ions.

E16. This question was the one question that consistently gained full or high marks. The expression in part (a) was invariably correct and it was pleasing to see that candidates usually referred to the cross-sectional area and not just the area.

In part (b), although the majority of candidates gained the correct answer of 75 Ω for the resistance, others failed with the calculation although the correct data had been inserted into the equation. The biggest error was failure to give the cross-sectional area correctly and a large number of candidates gave the volume of the carbon film instead. Such an error was treated as a Physics error and both marks were consequently lost. The explanation in part (ii) was, on the whole, done very well, with the majority of candidates arguing correctly that the resistance decreased because the cross-sectional area increased and the length decreased. Comparatively few attempted to estimate the factor by which each of these dimensions had changed, but most appreciated that there was a very large decrease in the value of the resistance.

E17. The response to this question was very disappointing, especially in view of the fact that this topic has been examined several times previously, including questions on the graphical nature of the quantities involved. Rearranging the equation in part (a) was intended as a guide to drawing the graph in part (b). The majority of candidates did rearrange the equation correctly, but some candidates failed to do this and ended up with a quotient.

Sketching the I-V graph in part (b) was, quite literally, a disaster area. The large majority of candidates drew a straight line of positive gradient passing through the origin. Obviously this was the easiest line to draw without applying any thought to the question. No marks were awarded for such attempts. If a line of positive gradient was drawn, and did not pass through the origin, then 1 mark was awarded. A large number of curved lines, some starting at zero, others at a positive value of V and decreasing to zero, were also presented. Of the candidates who drew a straight line with a negative gradient, many lost marks by either extending the line into negative values of V, or negative values of I. It must be pointed out that when carrying out an experiment to obtain this graph, it is not possible to obtain zero values of V or I. However, since some textbooks do show the graph extending to the V axis, this was accepted, but graphs extending to the I axis were not.

Most candidates gained at least one mark in part (c), but the impression gained was that candidates had learned the answers parrot fashion with no reference to the graph. The gradient

of a curved graph was often given as the answer.

E18. Candidates are by now well used to questions on resistors in series and in parallel and part (a) contained no hidden terrors. Invariably, correct answers were gained for both circuits. Examiners are concerned however at the trend of using an unusual nomenclature for a combination of resistors in series and in parallel, e.g. the combination of resistors in the second circuit would be given by candidates in some

centres as RT = . The fact that they subsequently gave the correctanswer of 60 Ω showed that the candidates had worked out the parallel section first and then added the series section. Such a system is not to be encouraged since it serves to confuse and if a wrong answer is given, it does not help the examiner to find out where the error occurred and then perhaps award a consequential error mark. The other point concerning this section, was that answers to part (i) were given in many instances as 131/3 Ω or 13.3 Ω (i.e. recurring). These were treated as significant figure errors. Answers given as fractions are not accepted.

Part (b) was more difficult and many candidates failed to understand the physics of the circuit. Comparatively few candidates gained full marks. An error which cropped up continually in part (i) was correctly calculating the resistance of three resistors in parallel (2Ω) but then using 12 V to calculate the current, not realising that the effect of the other three resistors halved the pd. There were also many candidates who calculated the correct current from the supply, but split this equally between the six resistors and not three. In part (ii), although most candidates calculated the current as 0.33 A and thus correctly concluded that the current was less than that in the part (i), they failed to capitalise on this and merely said that ‘therefore the heater was less effective’. In order to gain the final mark it was necessary to mention the heat/power generated by the current.

E19. This was the first time in this Specification that a question on charge flow had been set. Many candidates gained full marks and examiners were pleased by the response. It gave many candidates a good start to the paper. There were however, a significant number of candidates who failed to give Q = It in part (i), but were still able to gain the remaining marks as consequential errors. There was usually no problem with the units of charge. In part (ii) many candidates, because they multiplied instead of dividing by the electronic charge, obtained answers of 10–19 of an electron. It is also surprising, at this level, how many candidates failed to convert 40 mA into amperes.

It was pleasing to see that the majority of candidates were familiar with the equation V = W/Q in part (iii) and obtained the correct answer. Using the equations P = VI and E = VIt as starting points was equally popular. As expected, the use of Ohm‘s law in part (iv) was well known and candidates scored well. What did concern the examiners was that candidates had no feel for the magnitude of the answers, e.g. in part (ii) many candidates gave the answer as 11.5 × 10–19 for the number of electrons; this raised no comment. Again, the answer to part (iii), resulting from a consequential error, could be many millions of volts; no comment would be made.

E20. The circuit diagrams produced in part (a) were generally disappointing. Few candidates showed the voltmeter correctly connected to the tapered conducting paper and many had the positive pole of the battery connected to the broad end of the strip and the negative pole to the narrow end. The voltmeter was often shown in parallel with the full length of the conducting strip or else connected between the probe and the narrow end of the conducting strip.

Equally disappointing were the explanations of the non-linearity of voltage with distance along the strip in part (b). Most candidates realized that the increase of potential was non-linear because the strip was tapered, but few scored any further marks. Candidates who knew the correct relationship between resistance, length and cross-sectional area often failed to mention that the cross-sectional area was the product of the thickness and the width of the strip, although others did realize that the increase of resistance was due to the decrease of area as well as the increase of length. Only a few candidates stated that the current was constant or that the pd was proportional to the resistance.

In the graphical section of part (c) most candidates completed the table satisfactorily and proceeded to plot an adequate graph. Some candidates failed to score full marks as a result of careless errors such as omitting the unit of potential from the graph axis or failing to choose suitable scales. Candidates were usually able to relate the given equation to y = mx + c and thereby demonstrate that their graph confirmed to the given equation. Both marks were usually gained in part (iii) although there were some who plotted the potential on the x-axis thereby often losing the final mark because they failed to realize that the gradient was now equal to 1/(1.44 Vl).

E21. This question gave the most consistently correct answers in the whole paper but there were some poor efforts in part (a), e.g. leaving the voltage at Vrms, not converting to V0, or just calculating the value of the peak current.

The drawing of the oscilloscope trace in part (b) was usually correct, although many candidates, having worked out the value of the period T correctly, failed to translate this to the correct number of squares in the trace.

E22. Very few candidates failed to give the correct equation and definition of the symbols in part (a). Likewise in part (b) (i); calculations on resistivity are well understood and very few incorrect answers were seen. The usual error occurred in the calculation of the resistance of the wire, but the incorrect value would be carried forward as a consequential error. However, if the final answer then turned out to be a ridiculous value, such as 2 0 × 10–6 m, it was not accepted.

Part (b) (ii) gave rise to a few problems when deciding which combination of resistors gave an output of 1 kW. There were many very good answers, with the candidates showing that the output from the parallel combination gave the required output. A common error was stating that in the series combination, since the same current went through each element, then this gave the required output. Another common misconception was stating that when the resistors were in series, all, or almost all the voltage would be used up at the first resistor, leaving very little for the other resistor.

E23. Reading the values of the two resistances from the graph in part (a) did not prove to be difficult although many candidates lost the mark by not looking at the graph carefully and also by drawing a thick pencil line across, so that it covered the required point. This was especially true for the resistance of the thermistor.

The calculation in part (b) (i) was very accessible and the majority of candidates obtained the correct value for the voltage across AB. In part (b) (ii), many candidates gained the mark by simply writing down 50 °C. Others would fill up the available three lines with voltage calculations before arriving at the correct answer.

The calculation in part (c) (i) was usually correct, but answers to part (c) (ii) frequently failed to be awarded full marks. Candidates find this type of question, where the answer requires a logical sequence of steps, difficult. Usually, candidates would note that the lamp and thermistor were connected in parallel, thus reducing the resistance across AB. The next step of stating that the total resistance in the circuit would also decrease was usually ignored. The final steps of relating the decrease in resistance to a decrease in voltage across the thermistor was rarely done correctly. Candidates who attempted the answer in terms of the potential divider equation, or in terms of the ratio of resistances usually fared better.

E24. Answers to part (a) (i) produced a variety of circuit diagrams, ranging from those which were correct and well drawn, gaining maximum marks, to those having no diode or hardly anything else. The error which occurred most frequently was drawing the diode in the reverse bias position. Many candidates did not draw the symbol for the diode, but merely had a labelled square box. No marks were given for this since the correct symbol was expected. The role of the data logger continues to baffle many candidates. A large number drew the circuit correctly with leads from the ammeter and voltmeter to the data logger. Others merely broke the circuit and indicated that this went to a data logger to measure the current. The difficulty was obviously that although many candidates are familiar with the use of a data logger, others are not and the circuits seemed to suggest that they had never seen one in use. The description in part (ii) was, in general, quite well done, with the variable resistor (when included) being put to good use. Many wrote at length regarding the connection of the data logger to the computer and how the characteristic was displayed. This was not required. Part (iii) was usually correct, with either the diode or the battery being reversed. Answers such as, “…diode in reverse bias…” were judged not to have answered the question, since it asked for changes to the circuit.

The characteristic of the filament lamp, required in part (b) was usually drawn correctly, although a few candidates extended the curve so far as to show the current decreasing. The explanations were also very well done, with at least two of the relevant marking points being given. The point that always seems difficult for candidates to describe is when the rate of increase of current decreases because of the increasing resistance.

E25. The majority of candidates found this question straightforward and gained the maximum number of marks. Others, however, were not sure of the effect on the circuit of having the switch open or closed. A considerable number of candidates reversed the calculations for parts (a) (i) and (ii). Several candidates, in the situation when the switch was closed, i.e. effectively shorting out the 60Ω, resorted to adding up the two resistances using the expression for parallel resistors.

In part (b) the majority of candidates realized that a voltmeter of infinite resistance had the same effect on the circuit as an open switch and proceeded accordingly.

E26. Although the circuit presented in part (a) was, in general, correct, many candidates were not familiar with the symbol for the thermistor, which was necessary in order to gain full marks. Candidates who drew a potential divider often ended up losing a mark, because the ammeter would be in the wrong part of the circuit. It should also be noted that in order to gain full marks, it was necessary to label the component that enabled the current to be varied.

Although there were some very good answers to part (b), the general response was poor. In part (b) (i) it was common to find that candidates who had obtained the correct data from the graph, had an incorrect final answer because of their lack of understanding of what mA meant. Many took it to mean 103, others 10–6, while others just ignored it. This is a point that needs attention. Part (b) (i) also produced a glut of significant figure errors, especially since the answer to the second graph point was 28.3(recurring) × 103 Ω. Many candidates ended the answer with a fraction. It has been pointed out before that fractions are not acceptable and are treated as significant figure errors.

Answers to part (b) (ii) were also disappointing, primarily because candidates were aware of how the resistance of the thermistor varied with temperature and just wrote down what they had been taught without reading the question carefully and referring to the answers obtained in part (b) (i). Another part of the answer which gave rise to concern was stating that ‘as the temperature of the thermistor increases, then the current increases’. This is not correct and it should be emphasised that it is an increasing current that produces an increasing temperature. A large number of candidates only gained 1 mark for their answer and a great deal of candidates failed to score any marks.

E27. This question proved to be more difficult than the previous two questions and few candidates gained full marks. However, all candidates scored. In part (a), the large majority of candidates were aware that the internal resistance of the battery played an important role in the emf being greater than the voltmeter reading. Unfortunately, that is as far as most candidates went in the argument. To gain the remaining two marks it was essential to state that a current flowed through the circuit. Most candidates failed to mention this. Once the presence of a current had been established it was quite easy then to gain the third mark.

The circuits drawn in part (b) were, generally speaking, disappointing. Many candidates, rather than drawing two separate cells, drew a battery, which was not acceptable. In most cases the symbol for the internal resistance was not included. Many candidates lost marks by drawing the

resistance symbol, but marking it as r instead of 2r. Quite a large number of the calculations in part (ii) were correct, except that candidates who had marked the resistance as being r instead of 2r ended up with twice the required value, thereby losing a mark.

A question along the lines of that set in part (c) has been asked many times and yet many answers were poor. Very few candidates took the trouble to give the standard equation in the form of V= –Ir + , thereby helping themselves to realise that that the y intercept was equal to eand the gradient was r, and not the other way round. Many candidates gave the intercept as R.The other common feature, which was not acceptable, was to take values off the graph and calculate the required data from the equation.

E28. This is the first time in these series of examinations that candidates have been required to draw their own arrangement of resistors. The majority of candidates gave the correct answers in part (a), although some did try an arrangement of resistors similar to that in part (b). There were a few incorrect calculations in part (a) (ii) even though the three resistors were in series. The usual error in part (iv) was calculating correctly the value of \IR but then forgetting to invert to obtain R.

In part (b) the calculation for the total resistance was usually correct although there was some

concern amongst the examiners to see the expression RT = + 4 occurring quite frequently.

Invariably this resulted in the wrong answer, because candidates would not invert the value for the parallel resistors. The occurrence of this ‘system’ of calculating resistance was brought to the attention of teachers in the last report, but it seems to be more common than before. Part (ii) was not answered well, with candidates just writing numbers down without any reasoning and in the end confusing themselves. Candidates who just gave an answer of 4 V with no working shown were not credited, because it was possible to obtain that answer by incorrect physics. Candidates should be trained to give some explanation of what they are attempting in such calculations. It was also sad to see candidates obtaining the (correct) answer of 4.0 V across the parallel resistors, but then shooting themselves in the foot by assuming that the voltage across the 6.0 Ω was different to that across the 3.0 Ω.

E29. All candidates were able to gain a reasonable number of marks for this question, and many were awarded full marks. The answer to part (a) (i) was usually correct, but there were some problems with part (a) (ii). The most common error was obtaining the correct value of the current in the circuit, as required, but then dividing this value by the number of lamps. The resistance of each lamp, in part (iii), was usually calculated correctly, although some candidates made heavy weather of the calculation when using the expression VIR for power. Several recurring errors in part (iv) resulted in this part not being answered as well as the others. These errors would be calculating the energy used by one lamp instead of by the set, or omitting the factor of 2 (hours) when calculating the number of seconds, or omitting the conversion from minutes into seconds.

Incorrect answers in part (a) were allowed to be carried forward into part (b), which resulted in part (b) performing quite well. The usual error in part (i) was carrying out the calculations for 10 lamps, instead of 56. Answers to part (ii) were usually correct, candidates realising that the

greater the current, the greater the brightness.

E30. Explaining what is meant by emf in part (a) is still beyond the capabilities of a very large number of candidates. The most popular acceptable definition was that of the voltage across the terminals when no current flowed. Many candidates attempted to define emf in terms of the energy produced in the battery, but either forgot, or did not know, that it was the energy per unit charge. In part (a)(ii), the majority of candidates were aware that the reason involved the internal resistance, but merely quoting internal resistance on its own was not sufficient to gain a mark. There must be some reference to the voltage or pd across this resistance when a current flows.

The graph section in part (b) was answered well by the large majority of candidates, who drew excellent graphs. Some candidates missed out by using almost impossible scales in their bid to use the full page of the exam paper. Again, most candidates, having produced the correct equation in (i), knew how to obtain ∈and r from the graph. Some candidates, having produced an acceptable value for∈, proceeded to insert values in their equation. This was not acceptable. Another incorrect method was to obtain ∈ from the area under the graph.

E31. Questions requiring a description of an experiment invariably perform well and part (a) was no exception. Many candidates produced clear, concise and logical answers. This could be partly due to the fact that measuring the resistivity of a wire is a popular experiment for coursework. In the circuit diagrams submitted, the only point that attention needs to be drawn to is when candidates use a potentiometer rather than a variable resistance. It is important that the ammeter should be placed in the correct position to measure the current through the wire, and not measure the current through the battery. Many of the circuit diagrams did not include the resistance wire, the candidates seemingly being confused between the resistance wire and the variable resistance. Unfortunately, candidates who produced such a circuit invariably penalised themselves not only in part (i) but also in part (ii).

When listing, in part (ii), the measurements which needed to be taken, a mark was usually lost by candidates who stated that the area of cross section, rather than the diameter, had to measured. Several candidates referred to the diameter as the thickness of the wire, a term which was not acceptable. The other point which was missed quite frequently, was the need to make a series of measurements of voltage and current by altering the variable resistor. Another point which needs to be stressed is that many candidates stated that the length of the wire was changed, contradicting the evidence of their own circuit diagram, which showed a fixed length resistance wire.

In part (iii), the majority of candidates gave the correct equation relating resistivity to the other parameters and stated how to calculate A from the diameter and how to calculate the resistanceR of the wire. Most candidates used a graph to obtain R, with those who used a variable length using a graph of length against R, calculating ρ from the gradient.

Very few candidates failed to obtain the correct value for the length of wire in part (b).

E32. Although the type of question set in part (a) has appeared in previous examinations, many candidates lost marks unnecessarily due to the use of incorrect significant figures in part (i), the usual answer being given as 0.04 A or as a fraction or as a repetitive number. None of these are acceptable. The type of answer required in part (a) (ii) should, by now, be familiar to candidates. The question requires a candidate to reason through the problem without the aid of a calculation. It was pleasing to see how many candidates realised that not only did the resistance in the parallel resistance section decrease, but that this also produced a decrease in the total resistance of the circuit. This step in the answer enabled many candidates to gain full marks. A surprising number of candidates did carry out a calculation and were duly penalised.

The calculation in part (b) was usually correct, but the majority of candidates either ignored the request in the question to state the assumption made or were not aware that a voltmeter has a very large, or infinite resistance. The ideas of short circuiting a section in part (c), is now well understood and the large majority of candidates obtained the correct answer.

E33. This proved to be a comparatively easy question, with many candidates gaining full marks. Most of the errors in part (a) were due to the equation relating charge, time and current being incorrect. The necessary conversion of the time involved into seconds seemed to be well understood, but the same point in part (b), namely the conversion of hours into seconds, proved to be more open to errors. There were very few arithmetical errors. A consequential error in part (b)(ii) ensured that even if the incorrect answer was obtained in part (b)(i), candidates could still gain the allocated mark for the power of the bulb.

E34. The calculation to determine frequency in part (a) surprisingly brought difficulties to the surface. Many candidates simply did not know how to obtain frequency from the period, but many others had trouble in converting the waveform to an actual value of T. Calculating the rms voltage in part (ii) was usually carried out correctly, although many candidates just used the voltage sensitivity of 15 V per division as the peak voltage.

Although the majority of candidates had part (b) correct, a significant number became completely confused and converted back to peak values, under the impression that they were rms values.

Part (c) introduced a new type of problem for candidates, where they had to select a particular voltage sensitivity to suit the output voltage across the resistor. The general failing, even for good candidates, was failing to convert the answer of part (b) back to peak voltage, thereby losing a mark. Given that the answer to part (b) was treated as a consequential error in part (c), most candidates then chose the correct voltage sensitivity, but failed to support their choice by stating that the other sensitivities would either give a trace that was off the screen or else a trace that was too small for any meaningful measurements to be made.

E35. Part (a) effectively tested the use and understanding of basic relationships such as Q = Itand, to a very large extent, most candidates coped well with this section and scored high marks. In fact the maximum mark of seven was a common occurrence. Very few candidates failed to recognise that 7.2 kW was the same as 7.2 × 103 W, but conversely comparatively few candidates gave the answers to parts (iii) and (iv) in scientific notation, preferring to give the answer to part (iii), for example, as 288000 C. There was very little difficulty with realising that two hours had to be expressed in seconds, and the only real problem with the whole section was using the expression ½ QV for the energy. This was considered to be a basic error in physics and both marks for part (iv) were withheld.

Part (b) proved to be more difficult and comparatively few candidates gained the maximum marks. The effect of the internal resistance on the current, i.e. reducing it, was usually deduced correctly, but the effect on the time took a little more thought. Candidates who deduced correctly that the time increased usually approached the problem from energy or charge considerations.