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I = V __ = _____ 7.5 V = 0.136 mA R 55 kΩ The potential difference across the 33 kΩ resistor is therefore: V = IR = 0.136 mA × 33 kΩ = 4.5 V Alternatively, if you are good at ratios you might spot that the 7.5 V will split up in the ratio of the two resistors: V across 33 kΩ = 33/55 × 7.5 V = 4.5 V b) i) When the switch is closed, the voltmeter is connected in parallel with the 33 kΩ resistor. The resistance of this parallel combination will be less than 33 kΩ, so the voltage dropped across the combination will be less than the previous 4.5 V. ii) If the voltmeter reads 4.0 V, the potential difference across the 22 kΩ resistor must be (7.5 − 4.0) V = 3.5 V. The current in the 22 kΩ resistor, which is also the circuit current, will be: I = V __ = _____ 3.5 V = 0.159 mA R 22 kΩ The combined resistance of the parallel arrangement of the voltmeter and the 33 kΩ resistor will therefore be: R = V __ I = _________ 4.0 V = 25.1 kΩ 0.159 mA For this parallel arrangement: 1 __ = R 1 ___ + R33 1 ___ ⇒ RV 1 ___ = RV 1 __ – R 1 ___ = _______ 1 – ______ 1 R33 25.1 kΩ 33 kΩ 1 ___ = 0.0398 kΩ RV −1 − 0.0303 kΩ−1 = 0.0095 kΩ−1 R V = 106 kΩ c) If the voltmeter is rated as 10 V/100 μA, its resistance should be: R = V __ I = 10 V ___________ 100 × 10 −6 A = 1.0 × 105 Ω = 100 kΩ This differs by 6% from the experimental value. As each resistor has a tolerance of 5%, the experimental value is within the overall tolerance and so is compatible with the stated rating of the voltmeter. 9 a) As the temperature of the thermistor falls, its resistance increases as there will be less charge carriers per unit volume. If the resistance of the thermistor increases, the proportion of the supply voltage dropped across it will also increase and so the proportion of the supply voltage across the resistor R will decrease and V out will fall. b) i) Reading from the graph, at 0 8C the thermistor has a resistance of 18.0 kΩ. When V out is 5.0 V, the voltage across the thermistor and the potentiometer will be (9.0 − 5.0) V = 4.0 V. Assuming that the potentiometer is set at zero, the voltage across the thermistor will be 4.0 V. The current in the thermistor (and therefore the circuit current as it is a series circuit) will be given by: I = _______ 4.0 V = 0.22 mA 18.0 kΩ The value of R is then given by: R = V ____ out I = ________ 5.0 V = 22.5 kΩ 0.22 mA Full Answers to Review Questions: 15 Electric circuits Edexcel Physics for AS © Hodder Education 2009 24
The best value to use would therefore be the 22 kΩ resistor. ii) The purpose of the potentiometer is to fine tune the circuit. It is adjusted so that the lamp is switched on at exactly 0 8C. 16 Nature of light 1 a) First of all, 5% of 60 W = 3 W. We then need to use the relationship that intensity I at a distance r is given by: I = power ______ 4πr 3 W 4π × (1.5 m) 2 = 0.11 W m−2 = ___________ 2 The answer is A. b) We will need to use E = hf, so we must find the frequency corresponding to a wavelength of 660 nm. Rearranging c = fλ: f = c __ Then: = λ 3.00 × 108 m s−1 _____________ 660 × 10 −9 m = 4.55 × 1014 s −1 E = hf = 6.63 × 10 −34 J s × 4.55 × 10 14 s −1 = 3.01 × 10 −19 J Converting to electron-volts using 1 eV = 1.6 × 10 −19 J: E = 3.01 × 10−19 _______________ J −1 ≈ 2 eV 1.6 × 10 −19 J eV The answer is B. 2 a) Rearranging φ = hf 0 and remembering to convert eV into J: f0 = φ __ = h 2.28 eV × 1.6 × 10−19 J eV−1 _______________________ = 5.5 × 10 14 Hz The answer is C. 6.63 × 10 −34 J s b) We will need to use E = hf, so we must find the frequency corresponding to a wavelength of 447 nm. Rearranging c = fλ: f = c __ Then: = λ 3.00 × 108 m s−1 _____________ 447 × 10 −9 m = 6.71 × 1014 s −1 E = hf = 6.63 × 10 −34 J s × 6.71 × 10 14 s −1 = 4.45 × 10 −19 J Converting to electron-volts using 1 eV = 1.6 × 10 −19 J: E = 4.45 × 10−19 ______________ J −1 = 2.78 eV 1.6 × 10 −19 J eV The answer is C. c) In part b) we found that the energy of a photon of blue light is 2.78 eV. As the work function is 2.28 eV, the reverse potential difference that would have to be applied to just prevent photoemission would be: V = (2.78 − 2.28) V = 0.50 V The answer is A. 3 a) Using hf = E2 − E1 and remembering to convert the energy from eV to J: f = (1.51 – 0.54) × 1.6 × 10–19 _______________________ J 6.63 × 10 –34 J s = 2.34 × 1014 s−1 Rearranging c = fλ: Full Answers to Review Questions: 16 Nature of light Edexcel Physics for AS © Hodder Education 2009 25
Page 1 and 2: Full Answers to Review Questions 1
Page 3 and 4: 3 a) For example: 10d/mm 10d/mm mea
Page 5 and 6: 3 Rectilinear motion 1 a) a = b) s
Page 7 and 8: ) The gravitational potential energ
Page 9 and 10: 5 Ductile materials can be drawn in
Page 11 and 12: 2 To form a minimum on an interfere
Page 13 and 14: Full Answers to Review Questions: 1
Page 15 and 16: Full Answers to Review Questions: 1
Page 17 and 18: 4 a) e) The resistance of the whole
Page 19 and 20: iii) Rearranging ρ = RA ___ l : ρ
Page 21 and 22: 4 a) A 50 MΩ resistor in series w
Page 23: c) i) 2.2 P/mW 2.0 1.8 1.6 1.4 1.2
Page 27 and 28: f = c __ Then: = λ 3.00 × 108 m s
Page 29: c) The energy of the photon in the

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