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Sample Student Answers AS Unit G482: Electrons, waves and photons Module 3: DC circuits Question 1 Total marks: 10 (a) Figure 12 shows an electrical circuit. The cell has e.m.f. 1.5 V and internal resistance r. The current drawn from the cell is 0.60 A. Calculate the internal resistance of the cell. Marks available: 3 Student answer: (a) E = I(R + r) 1.5 = 0.60(r + 1.8) r = 0.70 Ω Examiner comments: See (b). © Pearson Education Ltd 2008 This document may have been altered from the original 1 (b) Figure 13 shows a filament lamp and two resistors connected to a d.c. supply. The d.c. supply has negligible internal resistance. (i) The filament lamp is rated as ‘12 V, 36 W’. Calculate the resistance of the filament lamp when used in this circuit. (ii) Calculate the total resistance of this circuit. (iii) Calculate the ratio current in the filament lamp current in the 20 Ω resistor Marks available: (i) 2 (ii) 3 (iii) 2 Student answer: (b) (i) P = V2/R 36 = 122/R R = 4.0 Ω (ii) R in series = 30 Ω Total R = 30 × 4.0 30 + 4.0 Total resistance = 3.53 Ω (iii) Current in lamp = 36/12 = 3.0 A Current in 20 Ω resistor = 12/30 = 0.40 A Ratio = 3.0 / 0.40 = 7.5 Examiner comments: All calculations correct. As always for calculations, make sure you follow the four steps: (1) write down equation (2) substitute figures (3) work out answer (4) put in the unit. © Pearson Education Ltd 2008 This document may have been altered from the original 2 Module 3: DC circuits Question 2 Total marks: 12 (a) State what is meant by (i) the electromotive force (e.m.f.) of a battery (ii) the internal resistance of a battery. Marks available: (i) 2 (ii) 2 Student answer: (a) (i) E.m.f. is the amount of chemical energy in a battery converted into electrical energy per unit charge. (ii) The internal resistance of a battery is the resistance within the battery. Examiner comments: (a) Correct for e.m.f., but for internal resistance it would be best to add that it prevents all the energy supplied by the battery from being used in the external circuit. (b) A student is asked to design a circuit to obtain potential differences of 4.0 V and 8.0 V from a 12 V supply of negligible internal resistance. A number of identical resistors are available. (i) Draw a complete circuit diagram to show how a potential-divider arrangement could be used. (ii) On your diagram show the connections made to the circuit for each potential difference. Marks available: (i) 2 (ii) 2 Student answer: (b) Examiner comments: (b) Correct. It would also have been possible to have two of the resistors in parallel and one in series with these other two. The 8 V would then have been across the single resistor. © Pearson Education Ltd 2008 This document may have been altered from the original 3 (c) Figure 14 shows a potential-divider circuit designed as a touch-sensor. The battery has negligible internal resistance and the voltmeter has infinite resistance. (i) Explain why the voltmeter reading is zero when there is nothing connected between the contacts X and Y. (ii) When the finger makes contact between X and Y, the voltmeter reading changes from 0 V to 3.4 V because of the electrical resistance of the skin. Use this information to calculate the electrical resistance of the skin between the two contacts. Marks available: (i) 1 (ii) 3 Student answer: (c) (i) The circuit is not complete and therefore the current cannot flow round it. (ii) V = ___R2___ × V0 R1 + R 2 3.4 = __168__ × 5.0 168 + R resistance = 79 kΩ Examiner comments: (c) (i) Correct, but it might also be worth pointing out that in this situation the p.d. between X and Y is 5 V. (ii) Correct. © Pearson Education Ltd 2008 This document may have been altered from the original 4