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Colonel Frank Seely School

Exampro A-level Physics (7407/7408) 3.5.1.5 Potential divider

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Time: 280

Marks: 237

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Q1.(a) Complete the circuit diagram below for a difference amplifier with a voltage gain of 100. Label the inputs and output.

(6)

(b) The difference amplifier is used with strain gauge sensors to measure the strain on a metal girder in a bridge.

A strain gauge sensor consists of very fine wires enclosed in a plastic case, as shown below. When the sensor is stretched its resistance increases and when compressed its resistance decreases. The changes in resistance are very small.

Two strain gauge sensors are glued to the girder, one to the top and one to the bottom, as shown below.

When the girder bends, the resistance of sensor 1 decreases and the resistance of sensor 2 increases.The sensors are connected to the circuit shown below.


(i) What is the voltage at point A?

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(ii) If sensor 1 and sensor 2 are both unstrained and each has a resistance of 200 Ω, what will be the difference in voltage between points A and B?

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(c) The inputs to the difference amplifier in part (a) are connected to points A and B. When the metal girder bends a small amount, the resistance of sensor 1 decreases by 1% and that of sensor 2 increases by 1%.

Estimate, showing your calculation, the output voltage of the difference amplifier under these conditions.

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(Total 11 marks)

Q2.(a) Figure 1 and Figure 2 show two circuits that may be used for controlling the voltage across a 3.0 Ω resistor. In each circuit the supply has an e.m.f. E of 10 V and


negligible internal resistance.

Figure 1 Figure 2

(i) Calculate the minimum voltage which can exist across the 3.0 Ω resistor using the circuit shown in Figure 1.

(3)

(ii) State one advantage of using the circuit shown in Figure 2 for controlling the voltage across the 3.0 Ω resistor.

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(iii) The total resistance of the potentiometer wire in Figure 2 is 30 Ω. Explain why the voltage across the 3.0 Ω resistor would not be half of the maximum when the slider of the potentiometer is half-way along the wire, as shown in Figure 2.

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(iv) Label with a letter P the approximate position of the slider in Figure 2, when the voltage across the 3.0 Ω resistor is about half the maximum possible.

(1)


(b) The circuit in Figure 3 is used to balance the power dissipated by two components that have different resistances. This is achieved by adjusting the position of S.

Figure 3

(i) Show that for the power dissipation to be the same, the ratio V1/V2 = 3/2.(2)

(ii) Calculate the power dissipated by one of the components when they are balanced.

(1)(Total 10 marks)

Q3.The graph in Figure 1 shows how the resistance of a thermistor varies with temperature.


Figure 1

(a) Explain why the resistance decreases at higher temperatures.

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(b) The thermistor is included in the circuit shown in Figure 2.

Figure 2

The thermistor has to be maintained at a temperature of 60°C.

Calculate:

(i) the potential difference across the thermistor;(3)


(ii) the power that has to be removed from the thermistor to maintain the temperature at 60°C.

(2)

(c) (i) Sketch below a possible variation of resistance with temperature for a material that becomes superconducting at a temperature of –80°C.

(1)

(ii) State one application of superconductors and explain briefly the advantage of superconductors over ordinary conductors in the application you have chosen.

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(Total 10 marks)

Q4.The diagram shows two wires, P and Q, of equal length, joined in series with a cell. A voltmeter is connected between the end of Q and a point X on the wires. The p.d. across the cell is V. Wire Q has twice the area of cross-section and twice the resistivity of wire P. The variation of the voltmeter reading as the point X is moved along the wires is best shown by


(Total 1 mark)

Q5. In the circuit shown in the diagram below cell X has an emf of 12 V and a negligible internal resistance. The resistances of RA and RB are 10 Ω and 15 Ω respectively.


(a) Calculate the potential difference across RB.

Potential difference .......................................(2)

(b) Cell X is replaced by cell Y that has an emf of 12 V and an internal resistance of7.5 Ω. Calculate the terminal potential difference across cell Y.

Potential difference ......................................(3)

(Total 5 marks)

Q6. The diagram below shows a potential divider consisting of a resistor in series with a light dependent resistor. The voltmeter connected in parallel with the light dependent resistor has an infinite resistance. The battery has an emf of 16V with a negligible internal resistance.


(a) Calculate the reading on the voltmeter when the light dependent resistor has a resistance of 1200 Ω.

Voltmeter reading .................................................(2)

(b) The light intensity in the room is increased. State and explain what happens to the resistance of the LDR and the reading on the voltmeter.

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(Total 5 marks)

Q7. The circuit shown in the diagram below can be used as an electronic thermometer. The battery has negligible internal resistance.


The reading on the digital voltmeter can be converted to give the temperature of the thermistor T which is used as a temperature sensor.

(a) Explain why the reading on the voltmeter increases as the temperature of the thermistor increases.

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(b) When the thermistor is at 80.0 °C the voltmeter reading is 5.0 V. Show that the resistance of the thermistor at this temperature is 4.0 Ω.

(1)

(c) When the thermistor is at 20.0 °C its resistance is 24.5 Ω. Calculate the reading on the voltmeter.


Voltmeter reading ..............................(2)

(d) The battery is replaced with another having the same emf but an internal resistance of 3.0 Ω.

(i) Calculate the new voltmeter reading when the thermistor temperature is 80.0 °C.

Voltmeter reading ................................(2)

(ii) State and explain the effect, if any, on the measured temperature when the thermistor is at 20.0 °C.

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(Total 8 marks)

Q8. Figure 1 shows the resistance against temperature characteristic for a thermistor.


Figure 1

(a) Suggest the range of temperatures for which the resistance change of the thermistor is most sensitive to changes in temperature.

(1)

Temperature range from ...........°C to ..........°C

(b) Explain, in terms of charge carriers, why the resistance of the thermistor falls as the temperature rises.

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(c) Figure 2 shows a circuit in which the thermistor is connected in series with a 100 kΩ fixed resistor and a 12 V battery of negligible internal resistance.


Figure 2

Calculate the potential difference across the thermistor at a temperature of –30 °C.

Potential difference = ..................................(4)

(Total 8 marks)

Q9. The figure below shows a simple light sensing circuit. When the output voltage V falls below 2.0 V, this acts as a signal which switches on a safety lamp. The LDR has a resistance of 1.25 kΩ when it is fully illuminated and 105 kΩ when it is in the dark. The battery has an emf of 6.0 V and negligible internal resistance.

(a) (i) Show that the safety light will come on when it is dark if resistor R has the value 50 kΩ.


(ii) Calculate V when the LDR is fully illuminated and the value of R is 50 kΩ.

V = .......................................................(5)

(b) Draw a diagram for a circuit, using the same LDR and battery, which would produce an output signal which increases to 3.0 V when the LDR is in the dark.

(2)

(Total 7 marks)

Q10.


Figure 1 Figure 2

(a) The current flowing through a torch bulb can be controlled by a variable resistor using either of the two circuit arrangements shown above. Figure 1 is called a potential divider arrangement and Figure 2 may be called a rheostat arrangement. For each of these two methods explain one advantage and one disadvantage.

potential divider

advantage ......................................................................................................

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disadvantage ..................................................................................................

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rheostat

advantage ......................................................................................................

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disadvantage ..................................................................................................

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(b) In Figure 1, the variable resistor has a total resistance of 16 Ω. When the slider of the variable resistor is set at X, exactly mid-way along AB, the bulb works according to its specification of 2.0 V, 500 mW. Calculate

(i) the current through section XB of the variable resistance,

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(ii) the current through section AX of the variable resistance.

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(Total 6 marks)

Q11.(a) (i) Give the equation which relates the electrical resistivity of a conducting material to its resistance. Define the symbols in the equation.

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(ii) A potential difference of 1.5 V exists across the ends of a copper wire of length 2.0 m and uniform radius 0.40 mm. Calculate the current in the wire.

resistivity of copper = 1.7 × 10–8 Ω m

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(b) In the circuit shown, each resistor has the same resistance. The battery has an e.m.f. of 12 V and negligible internal resistance.


(i) Calculate the potential difference between A and B.

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(ii) Calculate the potential difference between B and C.

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(iii) A high resistance voltmeter is connected between A and C. What is the reading on the voltmeter?

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(Total 10 marks)

Q12.In the circuit shown, the battery has negligible internal resistance.

(a) (i) If the emf of the battery = 9.0 V, R1 = 120 Ω and R2 = 60 Ω, calculate the current I flowing in the circuit.

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(ii) Calculate the voltage reading on the voltmeter.

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(b) The circuit shown in the diagram acts as a potential divider. The circuit is now modified by replacing R1 with a temperature sensor, whose resistance decreases as the temperature increases.

Explain whether the reading on the voltmeter increases or decreases as the temperature increases from a low value.

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(Total 7 marks)

Q13.The circuit diagram shows a light emitting diode (LED) connected in series with a resistor, R, and a 3.0 V battery of negligible internal resistance.

(a) The LED lights normally when the forward voltage across it is 2.2 V and the current


in it is 35 mA.

Calculate

(i) the resistance of R,

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(ii) the number of electrons that pass through the LED each second.

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(b) The LED emits light at a peak wavelength of 635 nm.

(i) Calculate the energy of a photon of light of this wavelength.

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(ii) Estimate the number of photons emitted by the LED each second when the current through it is 35 mA. Assume all the photons emitted by the LED are of wavelength 635 nm and that all the electrical energy produces light.

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(Total 8 marks)

Q14.In the circuit shown, an ideal operational amplifier is used as a voltage comparator.

(a) The voltage Vin is steadily increased from 0 V. Calculate Vin when the sign of Vout changes from negative to positive.

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(b) (i) The op-amp is required to operate an LED. Add to the circuit an LED and its limiting resistor so that the LED lights when Vin is less than the value calculated in part (a).

(ii) Explain why the LED functions in the position you have drawn it.

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(iii) Calculate the minimum value for the limiting resistor with the LED. Assume


that the LED has a voltage drop of 2.0 V across it when emitting and a maximum current of 25 mA through it.

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(c) An LDR is now connected between the points A and B in the circuit. The characteristic of the LDR is shown below.

Determine the light intensity at which the LED switches.

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(Total 10 marks)

Q15. (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)


Q16. 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

....................................................................(1)

(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)

Q17. 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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(3)

(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)

Q18. (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.

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(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.

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

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(Total 9 marks)

Q19. Figure 1 shows a circuit that can be used to sense temperature changes. Sensing is possible because the potential difference across the thermistor changes as the temperature changes.

Figure 1

The power supply has a negligible internal resistance and the resistor R has a resistance of 67 Ω.

(a) When the thermistor is at a high temperature the potential difference across it is 4.5 V.

(i) Calculate the potential difference across R.

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potential difference ............................. V(1)

(ii) Calculate the current in the circuit.

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current ....................................... A(2)

(b) (i) The temperature of the thermistor changes to 25 °C and its resistance becomes 360 Ω.Show that the potential difference across the thermistor at 25 °C is about 10 V.

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(ii) Calculate the power dissipated in the resistor R when the thermistor temperature is 25 °C, giving an appropriate unit for your answer.

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power dissipated ..................................................................................(4)

(c) The circuit is modified as shown in Figure 2. A resistor of resistance 570 Ω is connected in parallel with the thermistor.

Figure 2


For the circuit in Figure 2 calculate the current in the 67 Ω resistor when the thermistor temperature is 25 °C.

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current in 67 Ω resistor .......................................... A(4)

(d) Explain, in terms of charge carriers, why the resistance of the thermistor decreases as the temperature rises.

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(Total 17 marks)

Q20. Figure 1 shows an electrical circuit that contains a 4.0 Ω resistor, a 0 – 8.0 Ω variable resistor and a 12 V power supply with negligible internal resistance.

Figure 1


(a) State the name given to this type of circuit.

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(b) Calculate the minimum and maximum potential differences that can be obtained across XY. State the corresponding values of the resistance of the variable resistor.

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minimum pd .......... V when the resistance of the variable resistor is ........ Ω

maximum pd ......... V when the resistance of the variable resistor is ........ Ω(4)

(c) (i) Complete the circuit diagram to show how a potentiometer (variable resistor) can be connected so that the relative loudness of the sound from two loudspeakers can be adjusted.


(1)

(ii) Explain the advantage of using a potentiometer over the two-resistor arrangement in Figure 2 for this purpose.

Figure 2

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(Total 8 marks)

Q21. (a) Explain why the resistance of an NTC thermistor decreases when its temperature increases.

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Figure 1 shows the variation of resistance with light level for a light dependent resistor


(LDR).

Figure 1

(b) An LDR is used in the circuit in Figure 2 to monitor light levels.

Figure 2


(i) Calculate the output voltage of the circuit when the light level at the LDR is 300 lux and the resistance of the variable resistor is 150Ω. Assume that the battery has no internal resistance.

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output voltage ................................................................... V(3)

(ii) State and explain the effect on the output voltage of an increase in light level at the LDR.

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(Total 8 marks)


Q22. The circuit shown below shows a thermistor connected in a circuit with two resistors, an ammeter and a battery of emf 15V and negligible internal resistance.

(a) When the thermistor is at a certain temperature the current through the ammeter is 10.0 mA.

(i) Calculate the pd across the 540 Ω resistor.

answer = ..................................... V(1)

(ii) Calculate the pd across the 1200 Ω resistor.

answer = ..................................... V(1)

(iii) Calculate the resistance of the parallel combination of the resistor and the thermistor.


answer = ..................................... Ω(2)

(iv) Calculate the resistance of the thermistor.

answer = ..................................... Ω(2)

(b) The temperature of the thermistor is increased so that its resistance decreases.State and explain what happens to the pd across the 1200 Ω resistor.

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(Total 9 marks)

Q23. X and Y are two lamps. X is rated at 12 V 36 W and Y at 4.5 V 2.0 W.

(a) Calculate the current in each lamp when it is operated at its correct working voltage.


X .................................................. A

Y .................................................. A(2)

(b) The two lamps are connected in the circuit shown in the figure below. The battery has an emf of 24 V and negligible internal resistance. The resistors, R1 and R2 are chosen so that the lamps are operating at their correct working voltage.

(i) Calculate the pd across R1.

answer ......................................... V(1)

(ii) Calculate the current in R1.

answer ......................................... A(1)

(iii) Calculate the resistance of R1.

answer ......................................... Ω(1)


(iv) Calculate the pd across R2.

answer ......................................... V(1)

(v) Calculate the resistance of R2.

answer ......................................... Ω(1)

(c) The filament of the lamp in X breaks and the lamp no longer conducts. It is observed that the voltmeter reading decreases and lamp Y glows more brightly.

(i) Explain without calculation why the voltmeter reading decreases.

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(ii) Explain without calculation why the lamp Y glows more brightly.

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(Total 11 marks)

Q24. The figure below shows two resistors, R1 and R2, connected in series with a battery of emf 12 V and negligible internal resistance.


(a) The reading on the voltmeter is 8.0 V and the resistance of R2 is 60 Ω.

(i) Calculate the current in the circuit.

answer = ...................................... A(2)

(ii) Calculate the resistance of R1.

answer = ..................................... Ω(1)

(iii) Calculate the charge passing through the battery in 2.0 minutes. Give an appropriate unit for your answer.

answer = .............................................. unit = ...............................(2)

(b) In the circuit shown in the figure above R2 is replaced with a thermistor. State and explain what will happen to the reading on the voltmeter as the temperature of the thermistor increases.

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(Total 8 marks)

Q25.(a) Define the volt.

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(b) To test the potential differences in a potential divider circuit, a student sets up the circuit of Figure 1. R1 is the resistance of section AB and R2 that of section BC of the potential divider. The battery has an emf of 9.0 V and negligible internal resistance

Figure 1

(i) Calculate the voltmeter reading when R1 = 2.2 k and R2 = 1.8 k. Assume that the voltmeter has infinite resistance.


voltmeter reading ........................................... V(2)

(ii) State the benefit of using a high value of resistance in potential divider circuits.

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(iii) An 8.0 k resistor is connected in the circuit to replace the voltmeter in Figure 1.This is shown in Figure 2.

Figure 2

Calculate the potential difference across this resistor when the sliding contact B is in the position shown in Figure 2.

potential difference ........................................... V(3)


(iv) The 8.0 k resistor is now connected in a circuit with a 4.0 k variable resistor as shown in Figure 3.

Figure 3

Compare this arrangement for controlling the current in the 8.0 k resistor with the potential divider arrangement in Figure 2.

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(Total 9 marks)

Q26.The circuit diagram below shows a 12 V battery of negligible internal resistance connected to a combination of three resistors and a thermistor.


(a) When the resistance of the thermistor is 5.0 kΩ

(i) calculate the total resistance of the circuit,

total resistance = ......................................... kΩ(3)

(ii) calculate the current in the battery.

current = ........................................ mA(1)

(b) A high-resistance voltmeter is used to measure the potential difference (pd) between points A-C, D-F and C-D in turn.Complete the following table indicating the reading of the voltmeter at each of the three positions.

voltmeter position pd / V


A-C

D-F

C-D

(3)

(c) The thermistor is heated so that its resistance decreases. State and explain the effect this has on the voltmeter reading in the following positions.

(i) A–C........................................................................................................

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(ii) D–F........................................................................................................

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(Total 11 marks)

Q27.The circuit diagram below shows a 6.0 V battery of negligible internal resistance connected in series to a light dependent resistor (LDR), a variable resistor and a fixed resistor, R.


(a) For a particular light intensity the resistance of the LDR is 50 kΩ. The resistance ofR is 5.0 kΩ and the variable resistor is set to a value of 35 kΩ.

(i) Calculate the current in the circuit.

current...........................................A(2)

(ii) Calculate the reading on the voltmeter.

voltmeter reading ...........................................V(2)

(b) State and explain what happens to the reading on the voltmeter if the intensity of the light incident on the LDR increases.

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(c) For a certain application at a particular light intensity the pd across R needs to be 0.75 V. The resistance of the LDR at this intensity is 5.0 kΩ.


Calculate the required resistance of the variable resistor in this situation.

resistance ........................................... Ω(3)

(Total 9 marks)

Q28.In the circuit shown in the diagram the cell has negligible internal resistance.

What happens to the reading of both meters when the resistance of R is decreased?

Reading of ammeter Reading of voltmeter

A increases increases

B increases decreases

C decreases increases

D unchanged decreases

(Total 1 mark)

Q29.Figure 1 shows a circuit including a thermistor T in series with a variable resistor R. The battery has negligible internal resistance.

Figure 1


The resistance–temperature (R−θ) characteristic for T is shown in Figure 2.

Figure 2

(a) The resistor and thermistor in Figure 1 make up a potential divider.

Explain what is meant by a potential divider.

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(b) State and explain what happens to the voltmeter reading when the resistance of R is increased while the temperature is kept constant.

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(c) State and explain what happens to the ammeter reading when the temperature of the thermistor increases.

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

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

(d) The battery has an emf of 12.0 V. At a temperature of 0 °C the resistance of the thermistor is 2.5 103 Ω.

The voltmeter is replaced by an alarm that sounds when the voltage across it exceeds 3.0 V.

Calculate the resistance of R that would cause the alarm to sound when the temperature of the thermistor is lowered to 0 °C.

resistance = ............................... Ω(2)

(e) State one change that you would make to the circuit so that instead of the alarm coming on when the temperature falls, it comes on when the temperature rises above a certain value.

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

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

........................................................................................................................(1)

(Total 9 marks)


M1.(a) Feedback resistor to inverting inputSeries inverting input resistorSeries non-inverting input resistorResistor from non- inverting input to 0V – same value as Rf

All resistors in range 1kΩ to 1MΩMust have negative feedback then ratio of gain resistors = 100

6

(b) (i) 3V1

(ii) 0VNo differenceThe same

1

(c) e.g.New resistor values of 198Ω and 202Ω (1% change)

New output voltage = 0.03 × 100 = 3V3

[11]

M2.(a) (i) current = emf / total resistance or or equivalentC1

current = 0.77 AC1

V = 2.3(1) VA1

(or alternative approach using potential divider formula)(3)


(ii) wider range of voltages available

from 0 to 10 V or since lower minimum voltageB1

(1)

(iii) resistance of lower part is reduced

(since 3 Ω resistor is in parallel with lower half of potentiometer)

or calculation of resistance = 2.5 ΩB1

so voltage across lower section is reduced or calculation of p.d. = 1.43 VB1

allow 1 for calculation of p.d using 3 / (3 + 15) × 10 = 1.67 V(2)

(iv) point labelled at least of the way toward the top of the potentiometer

(correct position is just over of the way up)B1

(2)

(b) (i) power = V2 / RB1

V12 / 9 = V2

2 / 4 with correct clear manipulationA1

or calculates V1 (6 V) and V2 (4 V)C1

calculates power in each resistor (4 W) showing them to be equalA1

(2)

(ii) 4(.0) W (c.a.o.)B1

NB not half of (102 / 13)(1)

[10]


M3.(a) at higher temperatures more charge carriers / electrons are liberated or more electrons moving in the material or more electrons become mobile not electrons move faster or have more energy

B1

(for a given p.d.) more charge carriers flow increasing current or use of I = nAvq to explain increased current or there are more electrons in the conduction band or this more than makes up for increased resistance due to lattice vibrations

B1(2)

(b) (i) at 60 °C, R = 340 Ω (allow if 340 is seen used in equations)

B1

quotes and attempts to apply the potential divider formula or tries to determine the circuit current using emf/ total resistance (0.011 A) and hence the pd across the thermistor using V = IR

C1

3.7(4) V allow ecf for incorrect R for the thermistor 320 Ω gives 3.6(3) V; 380 Ω gives 3.4 V

A1(3)

(ii) power generated = power removed or P = I2R; V2 / R; VIC1

0.040 W

allow e.c.f. from (b)(i) 320 Ω 3.63 V gives 0.041W; 380 Ω, 3.4 V gives 0.038 W

A1(2)

(c) (i) graph decreasing in any shape until –80 °C where it decreases rapidly to zero graph must show a transition temperature at –80 °C

B1(1)

(ii) acceptable application


as connectors in amplifiers / electronic circuits/computers to produce higher memory capacity in computers to produce strong magnets (condone maglev or MRI scanners) to make high power motors to transmit electrical power in transformers

B1

explanation of advantage because amplifiers would have low thermal noise because large currents can be produced because little or no (thermal) energy is generated because no energy is transferred in conductors

B1(2)

[10]

M4.B[1]

M5. (a) 12 * 15/25

C1

= 7.2 V

A12

(b) total R now 32.5

C1

12 * 7.5/32.5 = 2.7[7] V or calculates I = 0.369 A

C1

terminal p.d. 12 - 2.8 = 9.2 V or V = 0.369 × (10 + 15) = 9.2 V

A13

[5]


M6. (a) [V1 = V × R1/(R1 + R2)]

C1

= 16 × 1200/2000 = 9.6 V

A12

(b) LDR resistance drops

B1

voltmeter reading decreases

B1

because more conduction electrons/chargecarriers released

B13

[5]

M7. (a) as the temperature of T increases its resistance decreases/more charge carriers are released

B1

increasing the current in the circuit/changing the ratio of resistance/reducing pd across T

B1

(so that so that the pd across the resister increases)2

(b) T/ 20.0 = 1.0/5.0 OR 5.0/6.0 = 20/(20+T) OR equivalent(Therefore T = 4.0 ohms)

Note T = (1/5)20 just ok but T = 20/5 not enough


M11

(c) Use of Vout = R1/(R1 + R2) × Vin OR I = 6/44.5 = 0.135 A

C1

V = 2.7 V

A12

(d) (i) V/6.0 = 20.0/(20.0+4.0+3.0) OR I = 0.222 A

C1

V = 4.4V

A12

(ii) The measure temperature would be lower becausethe pd across the resistor would be less (ie 2.53V)

B11

[8]

M8. (a) region from –50°C → max of –40°C

B11

(b) electrons/charge carriers released

B1

more charge carriers/electrons available for conduction

B1

this effect more than compensates for increased (rate of)


collision allow compensation of one mark for increasedrate of collision argument => increase of resistance

B13

(c) (117 ± 2) kΩ

B1

total resistance correctly calculated or potential dividerformula shown

C1

current correctly calculated for candidate’s data or correctsubstitution into potential divider formula

C1

6.41 V – 6.52 V (ecf for 120 Ω etc)

A14

[8]

M9. (a) (i) V/50 = 6/155 or equivalent method

B1

V = 1.94 V

B1

or R/105 = 2/4

or equivalent method

B1

R = 52.5 kΩ

B15

correct conclusion based on correct Ror V (V < 2, R > 50)


B1

(ii) V/6.0 = 50/(50 + 1.25)

C1

V = 5.85 V

A1

(b) potential divider circuit with

battery, LDR and 105 kΩ in series

B1

Vout clearly shown across the LDR

B12

[7]

M10.(a) potential divider: advantage: better control from 0 to maximum (1) disadvantage: power wasted because lower half of resistor always carries current (or top half of resistor must be capable of carrying lower half current and bulb current) (1)

rheostat: advantage: easier to connect (1) disadvantage: minimum current through bulb never zero (1)

(4)

(b) (i) VXB = Vlamp = 2.0 V ∴ IXB = = 0.25 A (1)


(ii) IAX = IXB + Ilamp, Ilamp = IXB = 0.25 A, ∴ IAX = 0.5 A (1)(2)

[6]

M11.(a) (i) resistivity defined by ρ = (1)

symbols defined

(ii)

= 0.068 (Ω) (1) (0.0676 Ω)

I = = 22 A (1) (22.2 A)

(allow e.c.f. from value of R)(5)

(b) (i) pdAB = × 12 = 8 V (1) (1)

(ii) pdBC = ( × 12) = 4 V (1)

(iii) pdAC = potential at A – potential at C (1) = (8 – 4) = 4 V (1)(allow e.c.f. from (i) and (ii))

(5)[10]


M12.(a) (i) (use of V = IR gives) V = I(R1 + R2) (1)

= 50 mA

(ii) Vout (= IR2) = 0.05 × 60 = 3 V (1) (allow C.E. for value of I from (i))

4

(b) (temperature increases, resistance decreases), total resistance decreases (1) current increases (1) voltage across R2 increases (1) [or R2 has increased share of (total) resistance (1) new current is same in both resistors (1) larger share of the 9 V (1)]

[or Vout = Vin (1) R1 decreases (1) Vout decreases (1)]3

[7]

M13.(a) (i) pd across resistor (= 3.0 – 2.2) = 0.8 (V) (1)

(use of V = IR gives) R = = 23 Ω (1) (22.9 Ω)

(ii) charge flow in 1 s = 0.035 (C) (1)

no. of electrons (in 1 s) = 2.2 × 1017 (1) (2.19 × 1017)4

(b) (i) (use of E = hf = gives) E = (1)

= 3.1(3) × 10–19 J (1)


(ii) (use of P = VI gives) P (= 2.2 × 0.035) = 0.077 (W)

[or use of P = I2R with R = 63 (Ω)]

maximum no. of photons emitted per sec. =

= 2.5 × 1017 (1) (2.48 × 1017)(allow C.E. for value of E from (i) and value of P from (ii))

4[8]

M14.(a) V– = 12 × (1)

= 7.8 V (1)2

(b) (i) between Vout and 0 V (1) (or from +12 V to Vout) correct direction and resistor (1)

(ii) (since Vin Vout = – 12 V (12 V across LED) (1) (or alternative)

(iii) voltage across R = (12 - 2) = 10 (V) (1) 10 = 25 × 10-3 × R gives R = 400 Ω (1) (or alternatively 22 = 25 × 10-3 to give R = 880 Ω)

5

(c) to switch LED voltage at B = 7.8 (V) (1)

RLDR given by 7.8 = or

RLDR = 25.(3) kΩlight level = 30 lux (1)

max 3[10]


M15. (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]

M16. (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]

M17. (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]

M18. (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]

M19. (a) (i) pd across resistor = 12 – 4.5 = 7.5 V

C11

(ii) I = (answer to (a) (i))/67 (allow 12/7.5/4.5 for this mark)

C1

0.110/0.112 (A)

A12

(b) (i) 360 + 67 (= 427) seen

C1

V = 12 × 360/(360 + 67)


C1

10.1 V

A13

(ii) substitution P = V2/R allow 360 Ω/67 Ω;10 V, 10.1 V, 1.9 V, 2 V

C1

1.92/67

C1

0.053

C1

W or J s–1

A14

(c) 1/R = 1/570 + 1/360

C1

220 [Ω]

C1

total R = 287 Ω

C1

42/41.7 mA 4.2 × 10–2/4.17 × 10–2

A14

(d) extra charge carriers released as temperature rises

B1

increased thermal agitation of atoms resists flow ofcharge carriers

B1


1st effect overwhelms 2nd

A13

[17]

M20. (a) potential divider or potentiometer

B11

(b) minimum 4 (V)

B1

variable R 8 Ω

B1

max I 12 V

B1

variable R 0 (Ω)

B14

(c) (i)

B11


(ii) full amplitude/voltage/volume range in each speaker/onlylimited range in Figure 2

B1

Figure 2 circuit would have different ranges in eachspeaker

B1

allow arguments related to relative values of resistors andspeakers

2[8]

M21. (a) more charge carriers available

B1

(internal) energy used to liberate electrons

B1

more electrons is more significant than additional vibrationof lattice ions

B13

(b) (i) (resistance of LDR is) 400 (Ω) seen – may be ondiagram

C1

their 400 + 150 seen

C1

2.45 or 2.5 V

A13

(ii) reduces resistance of LDR/correct example data fromgraph


B1

increase share of voltage taken by variable resistor(owtte)

B12

[8]

M22. (a) (i) voltage = 0.01 × 540 = 5.4 V (1)1

(ii) voltage = 15 – 5.4 = 9.6 V (1)1

(iii) (use of resistance = voltage/current)

resistance = 9.6/0.01 (1) = 960 Ω (1)

or RT = 15/0.01 = 1500 Ω (1)

R = 150 – 590 = 960 Ω (1)

or potential divider ratio (1)(1)2

(iv) (use of 1/R = 1/R1 + 1/R2)

1/960 = 1/200 + 1/R2 (1)

1/R2 = 1/960 – 1/1200

R2 = 4800 Ω (1)2

(b) (voltage of supply constant)

(circuit resistance decreases)

(supply) current increases or potential divider argument (1)

hence pd across 540 Ω resistor increases (1)


hence pd across 1200 Ω decreases (1)

or resistance in parallel combination decreases (1)

pd across parallel resistors decreases (1)

pd across 1200 Ω decreases (1)3

[9]

M23. (a) (use of P = V/l)

l = 36/12 = 3.0 A

l = 2.0/4.5 = 0.44 A 2

(b) (i) pd = 24 − 12 = 12 V 1

(ii) current = 3.0 + 0.44 = 3.44 A 1

(iii) R1 = 12/3.44 = 3.5 Ω 1

(iv) pd = 12 − 4.5 − 7.5 V 1

(v) R2 = 7.5/0.44 = 17 Ω 1

(c) (i) (circuit) resistance increases


current is lower (reducing voltmeter reading)

or correct potential divider argument2

(ii) pd across Y or current through Y increases

hence power/rate of energy dissipation greater or temperature of lampincreases

2[11]

M24. (a) (i) (use of V = IR)

I = (12-8) / 60 = 0.067 Or 0.066(A) 2

(ii) (use of V = IR)

R = 8/0.067 = 120 (Ω) 1

(iii) (use of Q = It)

Q = 0.067 × 120 = 8.0 C 2

(b) reading will increase

resistance (of thermistor) decreases (as temperature increases)

current in circuit increase (so pd across R1 increases) OR correct potential divider argument

3[8]


M25.(a) 1 joule per coulomb (or equivalent)

B1allow watt per amp

1

(b) (i) Use of potential divider formula

C1allow 1 for 4.05 (V) or current of 2.25 (mA)

4.95 (V)

A12

(ii) reduced current

B11

(iii) use of parallel resistor formula

C1

leading to 1.72 (kΩ)

C1

pd = 4.4 (V)

A13

(iv) potential divider can provides sensitive control of current (from 0 - 1.1 mA)

B1allow pot div can provide zero current and variable resistor gives larger current

variable resistor can provide larger current but cannot get near 0 A owtte

B12


[9]

M26.(a) (i) 1/R total = 1/(40) +1/(10+5) = 0.09167 R total = 10.9 kΩ

3

(ii) I = 12 / 10.9 k = 1.1 mA 1

(b)

position pd / V

AC 6.0

DF 4.0

CD 2.0

C.E. for CD3

(c) (i) AC: no change constant pd across resistors / parallel branches(AE)

no CE from first mark2

(ii) DF: decreases as greater proportion of voltage across fixed / 10 k Ω resistor

no CE from first mark2

[11]

M27.(a) (i) (use of I = V / R)first mark for adding resistance values 90 k Ω


I = 6.0 / (50 000 + 35 000 + 5000) = 6.7 × 10−5A accept 7 × 10−5 or dotted 6 × 10−5

but not 7.0 × 10 −5 and not 6.6 × 10 −5

2

(ii) V = 6.7 × 10−5 × 5000 = 0.33 (0.33 − 0.35) V ORV = 5 / 90 × 6 = 0.33( V)

CE from (i)BALD answer full credit0.3 OK and dotted 0.3

2

(b) resistance of LDR decreases need first mark before can qualify for second

reading increase because greater proportion / share of the voltage across R OR higher current

2

(c) I = 0.75 / 5000 = 1.5 × 10−4 (A) (pd across LDR = 0.75 (V))pd across variable resistor = 6.0 − 0.75 − 0.75 = 4.5 (V) R = 4.5 / 1.5 × 10−4 = 30 000 Ω orI = 0.75 / 5000 = 1.5 × 10−4 (A) RtotalI = 6.0 / 1.5 × 10−4 = 40 000 Ω R = 40 000 − 5000 − 5000 = 30 000 Ω

3[9]

M28.B[1]

M29.(a) A combination of resistors in series connected across a voltage source (to produce a required pd)

Reference to splitting (not dividing) pd1


(b) When R increases, pd across R increases

Pd across R + pd across T = supply pd

So pd across T / voltmeter reading decreases Alternative:

Use of V=

Vtot and R2 remain constant So V increases when R1 increases

3

(c) At higher temp, resistance of T is lower 1

So circuit resistance is lower, so current / ammeter reading increases 1

(d) Resistance of T = 2500 Ω

Current through T = V / R = 3 / 2500 = 1.2 × 10−3 A (Allow alternative using V1/R1 = V2/R2)

pd across R = 12 − 3 = 9 VThe first mark is working out the current

1

Resistance of R = V / I = 9 / 1.2 × 10−3 = 7500 ΩThe second mark is for the final answer

1

(e) Connect the alarm across R instead of across T allow: use a thermistor with a ptc instead of ntc.

1[9]


E1.(a) Most candidates produced creditworthy responses with a significant number gaining full marks. A common error was to give resistor values below 1kohm, which would cause errors when the op-amp was producing large output voltages.

(b) (i) Many correct responses were seen for this question.

(ii) Many correct responses were seen for this question.

(c) The responses from those candidates who attempted this section were pleasing, with a significant number gaining full credit. A common error was for candidates to calculate the new resistances of the sensor and then use these values in the difference amplifier formula instead of calculating the new voltage from the sensor potential divider.

E2.(a) (i) Many were unable to make any progress with this part. The only slight complication was the requirement to understand that the voltage is a minimum when the rheostat is set to 10 Ω but this usually seemed to be the least of the candidates’ concerns. The formula quoted was often that for terminal p.d. that led nowhere. Some quoted the potential divider formula and were then unable to apply it. Other candidates worked out the circuit current and gave this as the minimum voltage.

(ii) Most answers to this part were very vague and relatively few acknowledged the fact that it was possible to vary the voltage from 0 V – 10 V. Some stated only that it ‘increased the range’ without further qualification. Some focused attention on the resistor giving ‘finer control’ but this would depend on the structure of the potentiometer (e.g. the resistance per unit length).

(iii) Although many stated that the 3 Ω resistor would be in parallel with 15 Ω of the potentiometer relatively few continued the argument to a satisfactory conclusion. Many assumed that the final situation was 3 Ω in series with 15 Ωand drew a conclusion from this which gained some credit. Although only a correct qualitative argument was required for full marks, the best candidates actually calculated the voltage correctly.

(iv) Many placed the new position toward the bottom of the potentiometer even


after making progress to the correct explanation in part (iii). Generally this was not well done.

(b) (i) Many assumed the currents in the two components to be equal. Some who quoted the correct starting formula (V2 / R) tried to determine the power dissipated and prove them to be the same. This was an acceptable approach but candidates were expected to use the actual voltages in their explanations rather than the ratio given.

(ii) There was a good proportion of correct answers but often, even when part (i) was correct, this part was wrong. A significant proportion calculated the total power using 102 / 13 and then halved it.

E3.(a) This is a fairly standard question at this level. Most were able to gain some credit for stating that more electrons become free at higher temperatures but far fewer went on to explain that this led to a higher rate of flow of charge and hence a higher current. A significant proportion thought that free electrons would be given more energy so that they moved more quickly. Some discussed the increase in the amplitude of lattice vibrations and hence an increase in resistance which was the opposite of what they were asked to explain.

(b) (i) Incorrect graph reading (320 Ω instead of 340 Ω) was not uncommon but there were many correct responses to this part. Another common error was calculating current from 5 / 120 A and then using this to determine the voltage across the thermistor.

(ii) Most were able to quote a correct formula for power and there was a good proportion of correct answers (allowing errors carried forward in many cases). Assuming that the p.d. across the thermistor was 5 V was a not uncommon error.

(c) (i) Few showed a clear transition temperature at which the resistance suddenly dropped to zero at −80°C. A common response was a graph showing the resistance falling continuously and reaching zero at this temperature.


(ii) Good applications and reasons were rare but the use in transmission of electrical power was a frequent correct response. ‘Use in power stations’ without reference to what part of the power station or ‘to produce electromagnets’ without reference to the fact that the magnets could be made very strong were common as was ‘in computers’ or in ‘electronic circuits’ without mention of what part they would be useful for. That large currents could be produced or that there would be little or no thermal energy generated in the application were the anticipated advantages. Some thought that electrical signals would travel more quickly through superconductors and some seemed to be confusing superconductivity with semiconductors.

E5. This question was answered well by many. Those with low scores rarely gave the examiners much of a hint of what they were thinking. Too often a set of fairly random electrical symbols would appear with no underlying physical thread.Only the better candidates were able to find the terminal p.d. of the cell. Commonly, an equation from the formula sheet would be used with an incorrect current. A substantial number were able to find the ‘lost’ p.d. but did not then go on to subtract this from the e.m.f. of the cell.

E6. (a) This question was well answered by large numbers of the entry.

(b) Thinking was confused in this question, however. About half the candidates could not correctly state or explain how the resistance of a light-dependent resistor varies with light intensity. Those who knew this often could not go on to state correctly how the voltmeter reading varies. Only rarely could a candidate write fluently in terms of the ratio of the resistance values in the potential divider. Explanations were usually on a much lower level and rarely went beyond: ‘resistance goes down so the voltage drops’.

E7. Another difficult question when judged by the candidates. responses: more candidates failed to score on this question than any other. The calculations proved to be beyond most candidates. abilities.


Many good answers to part (a) were seen, and the better candidates scored on part (b) as well, although the idea of showing something mathematically is unfamiliar with many.

It was quite clear that very few candidates understood what was being asked in part (d)(ii).

E8. (a) Examiners accepted the range from -40o C to -50o C as being the most extreme one to gain credit; smaller ranges to include -50o C were preferable. A significant number of candidates agreed with this but a considerable number quoted far greater ranges than this.

(b) Most candidates realised that as the temperature of a semi-conductor increases more charge carriers will be released. In many cases answers suggested that candidates had no idea what charge carriers were or how they related to current. Few candidates explained that the higher temperature meant charge carriers would also collide with ions more frequently but that the release of the extra charge carriers was the dominant effect.

(c) The allowed resistance range for the thermistor was (117 ± 2) kW. Many candidates gained credit for this but a considerable number opted for 120 kW or 120 W and were penalised one mark for this. Most could use the potential divider formula correctly (or calculated the current) and went on to gain the remaining three marks.

E9. Disappointingly, a large number of candidates either did not attempt this question or gained only one to two marks from their attempt.

(a) In part (i), a correct calculation of R for V = 2.0 V, or a correct calculation of F when R = 50 kQ was not always followed by a clearly stated conclusion as is expected in a show that question. Most candidates who were able to attempt part (i) also gained credit in part (ii).

(b) Very few candidates gained both marks here. Often the value of the resistance was not stated and/or the output signal was not shown to be across the LDR. In some answers the LDR and resistor were shown connected in parallel.


E11.Giving the relevant equation for resistivity in part (a)(i) was straightforward for the majority of candidates. Errors were quite common however in the calculation in part (ii). The usual error occurred in calculating the area of cross-section, e.g. by omitting the π or the factor 10−3 or not squaring the radius. These errors however were allowed to be carried forward and credit was given for subsequent calculations and although it often resulted in currents of the order of 107 A very few candidates made any comment on this unlikely value.

Part (b) was probably the worst answered section in the paper and very few candidates gained full credit. The basic problem seemed to be that candidates were under the impression that the 12 V was somehow split, resulting in 6 V across each chain of three resistors. Thus answers of 4 V and 2 V respectively for parts (i) and (ii) were common. Many candidates attempted to calculate the various potential differences but their mathematics was so confused that it was difficult to decide what approach they were taking. Answers to part (iii) usually bore no relation to the previous answers and was very often a matter of guesswork.

E12.Part (a) provided the candidates with a reasonably easy four marks, and very few failed completely on the calculation. Usual errors such as units and arithmetic errors occurred but, in general, the candidates knew how to proceed with the calculation.

Part (b) required clear, logical thinking and sadly, the majority of candidates failed to gain the full three marks. Having been told in the question that the resistance of the sensor decreased with increasing temperature, many candidates simply wrote that the reading of the voltmeter would increase. Such a statement, although in itself correct, without any reasoning did not gain marks. Many candidate realised that the current in the circuit would decrease, but failed to go any further. The best approach seemed to be using the potential divider equation and candidates who tackled the question from this angle were usually awarded two or three marks.

E13.Part (a) proved to be very accessible and many candidates scored full marks. Most candidates calculated the resistor pd as 0.8 V and then calculated the resistance, as expected. Other candidates however, calculated the total circuit resistance, then the diode resistance and obtained the required resistance by subtraction. In this particular problem some candidates used an incorrect pd and were not awarded any credit. Many clear and correct answers were seen in part (ii).

The energy of the photon was calculated correctly in part (b) by many candidates, but some failed to score because the wavelength was taken as 1 / f or because the energy


was taken to be ½QV. The general principle behind the question in part (ii) was understood by most candidates and many correct answers were seen. A small minority of candidates however, calculated and used the power supplied to the resistor and not the diode.

E14.A large number of candidates calculated the correct switch over voltage in part (a). In part (b) it was pleasing to see that a majority of candidates had drawn the LED between the output and 0 V or between the output and the 12 V supply. In addition, the direction of the diode was usually correct and also the calculation for the value of the series resistor.

Part (c) proved to be more difficult, but many candidates successfully carried out the calculation giving the resistance of the LDR and subsequently read correctly the light intensity from the graph. Other candidates did not know where to start and made a guess at the resistance. If no effort had been made to calculate the resistance, there was no carry forward error for the value of the intensity.

E15. 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.

E16. 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.

E17. 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.

E18. 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.

E19. The level of attainment in this question that tested the electrical areas of the specification was very poor. Some candidates are clearly not comfortable with even the simplest problems tested here.

The pd in part (a) (i) was correctly identified by most candidates.

Answers to part (a) (ii) were very mixed and in general poor. Explanations were not adequate and the application of V = IR confused.

20% of candidates did not attempt part (b) (i) and only 10% gave completely accurate answers. Both the understanding and application of the potential divider by candidates as poor.

Again many used poor physics that did not relate to the circuit in part (b) (ii). There was a misunderstanding as to which pd to use in the equation.

One-fifth of candidates offered no explanations to part (c). The essence of the question was a computation of the total resistance of a parallel resistor network. About half of candidates were able to negotiate this with comfort, but the remainder were usually able to quote the appropriate equation but not to get further with it.

Only the more able candidates gave a complete picture of the situation in part (d). To gain full credit candidates needed to recognise both the change in atomic vibration and the increase in charge carrier density as a result of temperature increase, but then to go on to state that the second effect outweighs the first.

E20. In part (a), few than half the candidates identified this as potential divider circuit.


There were many correct responses to part (b) fairly straightforward question. A small minority gave the answers the wrong way round. Some knew that 12 V was likely to be the maximum without knowing when this would occur and some knew the resistor setting but were unable to calculate the voltages.

It was disappointing that more than half of the candidates were unable to complete the circuit in part (c) (i) correctly.

Few could make progress with part (c) (ii). Many thought that the circuit in Figure 6 would only enable the amplitude of one speaker to be changed. They did not appreciate that the voltage could not be made zero in the left hand loudspeaker or that the range of voltages available would be different in the two loudspeakers. Using the potentiometer both loudspeakers have the same range of voltages from zero to the maximum.

E21. Many candidates managed a partial explanation of this standard phenomenon in part (a). Most recognised the importance of the increased numbers of charge carriers. Acceptable comments about how electrons are liberated or about the increased lattice vibrations were rare. The calculations in parts (b) (i) and (b) (ii) tended to be done well. Data was correctly extracted and well processed. Some candidates got the potential divider voltages the wrong way round.

E22. This question proved to be very discriminating with only the more able candidates able to score high marks. The calculations involved in part (a) proved too challenging for many candidates. Part (a) (i) and (ii) generated the most correct responses, but the remainder of the analysis was only accessible to the more able candidates.

Part (b) required analysis without calculation and the majority of explanations seen were confused and not self consistent. Many candidates stated that more current goes through the thermistor and therefore the pd across it falls, resulting in the pd across the parallel 1200 Ω resistor increasing. Another common misunderstanding was the effect that the decreasing thermistor resistance had on the current through the battery. Many thought that the current remained constant and, although this still led them to deduce that the pd fell, their arguments frequently contained contradictions.

E23. With the exception of part (a), students found this question particularly challenging.

The calculations in part (b) were very structured but this did not seem make the analysis


of the circuit straightforward. In part (b) (i), less than half the students were able to calculate the pd across the resistor correctly with many not appreciating that the pd across the a parallel network was the same as the pd across lamp X. Part (b) (ii) produced better answers, although a significant proportion of students did not appreciate that they simply needed to add together the two currents calculated in part (a).

Part (b) (iii) was answered well, although many students benefited from being allowed to use incorrect answers from parts (b) (i) and (ii). The remainder of the circuit analysis did cause problems due to many students not realising that the pd across R2 was simply the difference in the pd’s across the lamps or that the current through R2 was the same as the current in lamp Y.

Part (c) required students to consider the effect of lamp X ceasing to conduct. In part (c) (i) they had to explain the effect on the voltmeter reading. This was not answered well with a significant proportion of students thinking the voltmeter reading would increase. This was mainly due to the mistaken assumption that the current in the circuit would increase. Part (c) (ii) generated more correct responses because many students stated that the current through lamp Y would increase, although it was clear from their answers many thought that this was due to the current from lamp X now going through lamp Y. It was not commonly appreciated that although the overall current in the circuit had decreased, the current through R2 and lamp Y was higher than it had been when lamp X was conducting.

E24. The majority of students were able to analyse the circuit correctly although surprisingly a significant minority had problems with (a)(i) because they did not appreciate that the pd across R2 was 4.0 V. This did not affect their subsequent responses however, as the answer they gave was carried forward to subsequent calculations. The qualitative aspect of the question presented students with a greater challenge. Many incorrectly stated that the voltmeter reading would decrease as the thermistor resistance falls seemingly forgetting that the voltmeter was connected across R1.

E25.Few candidates were able to define the volt correctly – most tried to use V = IR. In (b)(i) most candidates were able to apply the potential divider formula correctly to calculate the voltmeter reading.

A minority of candidates suggested that the benefit of using high resistances in potential dividers was to result in low currents in (b)(ii).

A high proportion of candidates answered (b)(iii) correctly by calculating the resistance of the parallel arrangement and then either applying the potential divider formula or else


calculating the current and consequent potential difference. A significant minority of candidates failed to make any progress in this part and there were a number of blank answers.

In (b)(iv) few could make any creditworthy comment on the relative merits of a potential divider and series arrangement when controlling current. Only a very limited number of candidates were able to calculate the current range in the two circuits and thereby inform their comments – it was clear that this topic was not well understood.

E26.In this question candidates were required to analyse a bridge network of resistors. The calculation of the circuit resistance in part (a) proved to be reasonably straightforward with over two thirds of candidates scoring full marks. The only common error in weaker scripts was the combining of all the resistors as parallel resistors instead of combining the series branches first. The calculation of current in part (a) (ii) was done well and with consequential error applied, the majority of candidates were able to do this successfully.

Part (b) was not well answered and very few candidates were able to give correct answers for the voltmeter reading in the three positions. The position that proved the most challenging was the pd between C and D and it is clear that many candidates did not appreciate that this was found by subtracting the pd across D and F from the pd across C and E.

Part (c) was a qualitative question and previous papers suggest that candidates find these difficult. Only the very best candidates managed to get full marks in this section and it was the explanations of the effect on the voltmeter that proved to be the most challenging. For example over 60% of candidates appreciated that the pd across the thermistor decreased but only about 14% managed to explain why. A common mistake was to try and use current in explanations and this led them to conclude incorrectly that if current goes up then so does pd or that the increase in current cancels out the decrease in resistance. Very few used the constant 12 V across the parallel branches to justify their conclusions.

E27.This question on a potential divider circuit was a mixture of qualitative and quantitative. As is often the case with questions involving electric circuits, candidates coped better with quantitative parts. This was particularly true in part (a) where the calculation involved more than one stage.

Part (b) was not well done and only the strongest candidates manage to relate the changing light intensity to the voltmeter reading. A significant proportion of candidates were under the impression that increasing the light intensity increases the ldr resistance.

Part (c) did involve a calculation but this was much more challenging than part (a) because there were no intermediate stages. Only a third of candidates were able to calculate a correct value for the resistance of the variable resistor. The majority of those


who were successful calculated the value using a ratio method rather than calculating the current and then using this value with the correct pd to find the resistance.

animated science · web view2021/03/03 · some discussed the increase in the amplitude of lattice...

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