Fig shows a battery connected across a negative temperature coefficient (NTC) thermistor. Fig. 24.1

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1 Answer all the questions. 1. Fig shows a battery connected across a negative temperature coefficient (NTC) thermistor. Fig The battery has electromotive force (e.m.f.) 3.0 V and negligible internal resistance. The ammeter has negligible resistance and the voltmeter has a very large resistance. The thermistor has resistance 100 Ω at room temperature and a cross-sectional area of m 2. The number density of the free electrons within the thermistor is m 3. i. Calculate the mean drift velocity v of the free electrons in the thermistor. v =... m s ii. The thermistor is now heated using a naked flame. Describe and explain the effect on the ammeter and voltmeter readings. [3] OCR You may photocopy this page. Page 1 of 11 Created in ExamBuilder

2 2. A small thin rectangular slice of semiconducting material has width a and thickness b and carries a current I. The current is due to the movement of electrons. Each electron has charge e and mean drift velocity v. A uniform magnetic field of flux density B is perpendicular to the direction of the current and the top face of the slice as shown in Fig Fig. 2.1 Here are some data for the slice in a particular experiment. number of conducting electrons per cubic metre, n = m 3 a = 5.0 mm b = 0.20 mm I = 60 ma B = T Use this data to calculate i. the mean drift velocity v of electrons within the semiconductor v =... m s 1 [3] ii. the potential difference V between the shaded faces of the slice. V =... V [1] OCR You may photocopy this page. Page 2 of 11 Created in ExamBuilder

3 3. A student monitors the temperature in a room by using a potential divider circuit containing a negative temperature coefficient (NTC) thermistor. The student sets up the circuit shown in Fig Fig. 4.2 The battery has an e.m.f. of 6.0 V and negligible internal resistance. i. When the temperature of the thermistor is 12 C the thermistor has a resistance of 6.8 kω. The resistance of the variable resistor is set to a value of 1.4 kω. Calculate the reading V on the voltmeter. V =... V [2] ii. Explain how the reading on the voltmeter will change when the temperature of the thermistor increases. OCR You may photocopy this page. Page 3 of 11 Created in ExamBuilder

4 [4] 4. Fig. 4.1 shows the I-V characteristic of a blue light-emitting diode (LED). The energy of each photon emitted by the LED comes from an electron passing through the LED. The energy of each blue photon emitted by the LED is J. i. Calculate the energy of a blue photon in electron volts. energy =... e OCR You may photocopy this page. Page 4 of 11 Created in ExamBuilder

5 ii. Explain how your answer to (i) is related to the shape of the curve in Fig [2] 5. A chemical cell is connected across a resistor. i. The terms electromotive force (e.m.f.) and potential difference (p.d.) are terms associated with the circuit. State one similarity and one difference between e.m.f. and p.d. similarity: difference: [2] ii. The resistor is cylindrical in shape. It has cross-sectional area m 2 and length m. In this resistor there are free electrons. Calculate the mean drift velocity v of the electrons when the current in the resistor is 3.0 ma. OCR You may photocopy this page. Page 5 of 11 Created in ExamBuilder

6 v =... m s 1 [3] 6. The maximum power input to a domestic fan heater is 2.6 kw when connected to the 230 V mains supply. The electric circuit of the fan heater consists of two heating elements (resistors) rated at 1.5 kw and 1.0 kw, a motor rated at 100 W and three switches. i. Show that the resistance of the 1.5 kw heating element is about 35Ω. ii. The 1.5 kw heating element is made from a wire of cross-sectional area m 2 and resistivity Ωm. Calculate the length of the wire. length = Two resistors of resistances R1 and R2 are connected in parallel. Show that the total resistance R of this combination is given by the equation OCR You may photocopy this page. Page 6 of 11 Created in ExamBuilder

7 [3] END OF QUESTION paper OCR You may photocopy this page. Page 7 of 11 Created in ExamBuilder

8 Mark scheme Question Answer/Indicative content Marks Guidance current = (A) C1 1 i (I = Anev); = v Examiner s Comments Almost all candidates were familiar with the equation I = Anev. The modal score here was two marks. v = (m s 1 ) Most scripts had well-structured answers. The final answer was often quoted to the correct number of significant figures and written in standard form. A very small number of candidates incorrectly calculated the current using current = VR = = 300 A ; this scored zero because of incorrect physics. Allow V = IR (any subject) and V = constant Allow more electrons / more charge carriers Allow voltmeter reading stays 3.0 (V) The resistance (of the thermistor or circuit) decreases Examiner s Comments ii Current / I / ammeter reading increases because I 1/R or number density (of charge carriers) increases This question on the heating of a thermistor favoured the top-end candidates. Most candidates recognised that the resistance of the NTC thermistor decreased as its temperature was increased. The Voltmeter reading does not change (because there is no internal resistance) explanation of why the current increased lacked robustness. Some correctly gave the explanation as increased number density of free electrons or successfully showed that current was inversely proportional to the resistance. The fate of the voltmeter reading baffled many candidates. The answer was simple, the voltmeter reading remained unchanged because the battery had no internal resistance. For many, the voltmeter reading increased because p.d. was proportional to the current. Total 5 OCR You may photocopy this page. Page 8 of 11 Created in ExamBuilder

9 I = naev; C1 2 i v = / x10 6 v = 3.1 (m s 1 ) C1 allow any subject ecf (b)(i); allow 1.2 mv; (V) ii V = = (V) Examiner s Comments This exercise of choosing a formula, substituting values in correct units and evaluating was done well with about three quarters of the candidates gaining full marks. Total 4 Allow Allow i (V) C1 Examiner s Comments Candidates who use the potential divider equation invariably gained the correct answer of 0.71 V. Alternatively, some candidates correctly determined the current and then determined the voltmeter reading. Examiner s Comments Candidates were expected to explain how the As temperature of thermistor increases, resistance of thermistor decreases voltmeter reading would change as the temperature Total resistance of circuit decreases or current increases of the thermistor increased. Good answers used a step-by-step approach. Candidates needed to ii Greater proportion of p.d. across fixed resistor or p.d. across fixed resistor increase M1 explain how the potential difference of across the fixed resistor would change. It was essential that clearly defined terms were used often candidates referred to V1, R2, or p.d. and resistance without Reading on the voltmeter will increase indicating explicitly the meaning of V1, R2, or explaining which p.d. or resistance was being referred to. Total 6 4 i energy in ev = 4.1 x /1.6 x = 2.6 (ev) expect 2.56 ev ii LED strikes at 2.6 V / only conducts above 2.6 V M1 ii an electron must pass through a p.d. of 2.6 V to lose energy as ii a photon of blue light / AW. Examiner's Comments Almost all managed to calculate the energy of a blue photon and then recognised that it was related to the onset of current in the LED. Total 3 OCR You may photocopy this page. Page 9 of 11 Created in ExamBuilder

10 Allow both defined as energy (transformed) per unit 5 i Similarity same unit (AW) charge or both defined as work done per unit charge Allow any pair from: e.m.f. p.d. Energy (transformed) Energy from electrical or (transformed) Energy to electrical (transformed) to heat /other forms Charges gain energy Charges lose energy i Difference For e.m.f, energy is transformed from chemical / other forms to electrical and for p.d., energy is transformed to heat / other forms from electrical Work done on charges Work done by charges Examiner's Comments Most candidates knew that e.m.f. and p.d. were both measured in volts (V). A small number of candidates thought that volt was the same as voltage. This question benefitted those who taken time to revise thoroughly. The modal mark was one, but a significant number of candidates scored two marks for their flawless answers. ii C1 (I = Anev) ii = v C1 Note Any subject for this equation Allow 1 mark for 1.6(3) 10 5 (m s 1 ); n = used Examiner's Comments Almost all candidates were familiar with the equation I = Anev. However, only the top-end candidates ii v = (m s 1 ) realised that the number density of the charge carriers (electrons) had to be calculated from the number of electrons given and the volume of the resistor. The majority of candidates incorrectly assumed n to be m 3 when it should have been m 3. Examiners awarded one mark for those candidates who arrived at the answer m s 1 using the incorrect value of n. Total 5 6 i P = V 2 /R = /R = 1500 C1 accept I = P/V = 6.52 A and R = 230/6.52 OCR You may photocopy this page. Page 10 of 11 Created in ExamBuilder

11 i R = 35.3 Ω allow 52900/1500 = 35 Ω, i.e. some working shown ii use of p = RA/l or R = pl/a C1 formula ii l = / C1 correct substitution answer (2.48) Examiner's Comments ii l = 2.5 (m) Most candidates were able to complete this section successfully with the majority scoring full marks. Only the weakest stumbled here. Total 5 7 I = I1 + I2 M1 V is the same (for each resistor) M1 leading to correct expression Total 3 OCR You may photocopy this page. Page 11 of 11 Created in ExamBuilder