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GCSE Physics Exam Doctor

GCSE Physics Exam Doctor. Electricity – Energy in Circuits. Question 2. Question 3. Question 4. Question 1. Question 5. GCSE Physics Exam Doctor. Electricity – Energy in Circuits. Question 1.

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GCSE Physics Exam Doctor

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  1. GCSE Physics Exam Doctor Electricity – Energy in Circuits Question 2 Question 3 Question 4 Question 1 Question 5

  2. GCSE Physics Exam Doctor Electricity – Energy in Circuits Question 1

  3. In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains. a) i) What is the name for this type of circuit? (1 mark) ii) How much electrical energy does the 60W lamp convert in 5 minutes? (3 marks) iii) Not all the electrical energy is converted into light. What happens to the wasted energy? (2 marks)

  4. b) The lamps are now incorrectly connected as shown in the circuit below. Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer. (3 marks)

  5. In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains. a) i) What is the name for this type of circuit? Parallel circuit (1 mark) ii) How much electrical energy does the 60W lamp convert in 5 minutes? Power is energy / time so energy = 60 x 5 = 300J (3 marks) iii) Not all the electrical energy is converted into light. What happens to the wasted energy? It is converted to heat energy (2 marks)

  6. b) The lamps are now incorrectly connected as shown in the circuit below. Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer. It will not give out as much energy because the first lamp is taking some of the electricity. They have to share the electricity (3 marks)

  7. Mark scheme In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains. a) i) What is the name for this type of circuit? Parallel circuit (1 mark) ii) How much electrical energy does the 60W lamp convert in 5 minutes? Power is energy per second, so the time must be converted into seconds Energy = Power x time = 60 x 5 x 60 = 18kJ (3 marks) iii) Not all the electrical energy is converted into light. What happens to the wasted energy? It is converted to heat energy The energy becomes spread out among many molecules and therefore irrecoverable, or “dissipated”. (2 marks)

  8. Mark scheme b) The lamps are now incorrectly connected as shown in the circuit below. Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer. It will not give out as much energy. The voltage across each lamp is now reduced and also the current through both lamps is reduced because the resistance has increased. (3 marks)

  9. In the circuit shown below, a 60W lamp and a 100W lamp are connected correctly across the mains. a) i) What is the name for this type of circuit? Parallel circuit (1 mark) ii) How much electrical energy does the 60W lamp convert in 5 minutes? Power is energy / time so energy = 60 x 5 = 300J Power is energy per second, so the time must be converted into seconds Energy = Power x time = 60 x 5 x 60 = 18kJ (3 marks) iii) Not all the electrical energy is converted into light. What happens to the wasted energy? 4 6 It is converted to heat energy Sufficient for 1 mark, but for the second mark needs a reference to the energy being spread out among many molecules and therefore irrecoverable, or “dissipated”. (2 marks)

  10. b) The lamps are now incorrectly connected as shown in the circuit below. Will the 60W lamp still convert the same amount of energy in 5 minutes, as before? Explain your answer. It will not give out as much energy because the first lamp is taking some of the electricity. They have to share the electricity Just scores the first mark, but does not show understanding that the voltage across each lamp is now reduced and also the current through both lamps is reduced because the resistance has increased. (3 marks) 1 3

  11. GCSE Physics Exam Doctor Electricity – Energy in Circuits Question 2

  12. A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second? (2 marks) ii) In what form is the energy stored in the cell? (1 mark)

  13. iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell? (1 mark) b) When the cell is connected across two identical lamps, each of resistance 2W, it delivers a current of 0.3A What will be the rate of energy conversion in each lamp? (3 marks)

  14. A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second? Energy = V x I = 1.5 x 0.2 = 0.3J (2 marks) ii) In what form is the energy stored in the cell? Electrical energy (1 mark)

  15. iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell? Heat (1 mark) b) When the cell is connected across two identical lamps, each of resistance 2W, it delivers a current of 0.3A What will be the rate of energy conversion in each lamp? Energy = V x I = 1.5 x 0.3 = 4.5J (3 marks)

  16. Mark scheme A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second? V x I = power (energy per second) 1.5 x 0.2 = 0.3W (0.3J per second) (2 marks) ii) In what form is the energy stored in the cell? Chemical energy (1 mark)

  17. Mark scheme iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell? Thermal energy is dissipated in the cell. (1 mark) b) When the cell is connected across two identical lamps, each of resistance 2W, it delivers a current of 0.3A What will be the rate of energy conversion in each lamp? I2R = 0.09 x 2 = 0.18J/s in each lamp. (3 marks)

  18. A current of 0.2A is drawn from a new cell, which is operating at its stated voltage of 1.5V. a) i) How much energy will be converted into electrical energy per second? Energy = V x I = 1.5 x 0.2 = 0.3J V x I is actually power (energy per second) not total energy, but this is what the question asked for. (2 marks) ii) In what form is the energy stored in the cell? Electrical energy The cell converts from chemical energy to electrical energy. (1 mark) 2 3

  19. iii) In practice, the cell may not be operating at 1.5V. How might energy be wasted in the cell? Heat Not an ideal answer, but adequate for 1 mark. Thermal energy is dissipated in the cell. (1 mark) b) When the cell is connected across two identical lamps, each of resistance 2W, it delivers a current of 0.3A What will be the rate of energy conversion in each lamp? Energy = V x I = 1.5 x 0.3 = 4.5J Scores 1 mark for V x I as rate of energy production, but the previous V is no-longer applicable. The cell is not necessarily giving 1.5V now, and each lamp has half the P.D. across it. V x I (with the appropriate, unknown, V) becomes I2R = 0.09 x 2 = 0.18J/s in each lamp. (3 marks) 2 4

  20. GCSE Physics Exam Doctor Electricity – Energy in Circuits Question 3

  21. In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A. a) i) How much energy is converted in the lamp in 2 minutes? (3 marks)

  22. ii) Calculate the resistance of the motor. (3 marks) b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not. (3 marks)

  23. In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A. a) i) How much energy is converted in the lamp in 2 minutes? Energy = 4.5 x 0.25 x 2 = 2.25J (3 marks)

  24. ii) Calculate the resistance of the motor. V = I x R R = 4.5/0.25 = 18 Ohms (3 marks) b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not. The motor is stronger, it would take all the energy. The lamp needs 0.25 not 0.1 like the motor. (3 marks)

  25. Mark scheme In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A. a) i) How much energy is converted in the lamp in 2 minutes? Energy = 4.5 x 0.25 x 2 x 60 = 135J (3 marks)

  26. Mark scheme ii) Calculate the resistance of the motor. V = IR R = 4.5/0.1 = 45W (3 marks) b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not. The motor has higher resistance, therefore would have a greater proportion of the P.D. across it. The combined resistance would lower the current, leaving not enough power (V x I) to light the lamp. (3 marks)

  27. In the circuit shown, the motor takes a current of 0.1A, and the lamp a current of 0.25A. a) i) How much energy is converted in the lamp in 2 minutes? Energy = 4.5 x 0.25 x 2 = 2.25J Scores for method, but time should be in seconds. Energy = 4.5 x 0.25 x 2 x 60 = 135J (3 marks) 2 3

  28. ii) Calculate the resistance of the motor. V = I x R R = 4.5/0.25 = 18 ohms Method is correct, but the wrong current has been used. The current in the motor branch is 0.1A. R = 4.5/0.1 = 45W (3 marks) b) If the lamp and the motor were connected in series across the cell, the lamp would not light, explain why not. The motor is stronger, it would take all the energy. The lamp needs 0.25 not 0.1 like the motor. Stronger is not an appropriate word. The motor has higher resistance, therefore would have a greater proportion of the P.D. across it. The combined resistance would lower the current, leaving not enough power (V x I) to light the lamp. (3 marks) 2 6

  29. GCSE Physics Exam Doctor Electricity – Energy in Circuits Question 4

  30. A solar cell is used to power a small electric motor. • This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V. i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell. (3 marks)

  31. ii) Calculate the rate, in J/s, at which energy is supplied to the motor. (1 mark) iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why. (2 marks) b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this? (2 marks)

  32. A solar cell is used to power a small electric motor. • This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V. i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell. Energy = VIt = 12/3 x 200 = 0.02 (3 marks)

  33. ii) Calculate the rate, in J/s, at which energy is supplied to the motor. Power = V x I = 3 x 0.02 = 0.06 J/s (1 mark) iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why. Energy is lost as heat (2 marks) b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this? One cell provides 0.06J per s therefore 7000/0.06 cells are needed = about 117 000 (2 marks)

  34. Mark scheme A solar cell is used to power a small electric motor. • This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V. i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell. Energy = VIt I = Energy/Vxt I = 12/(3 x 200) = 0.02A (3 marks)

  35. Mark scheme ii) Calculate the rate, in J/s, at which energy is supplied to the motor. Power = V x I 3 x 0.02 = 0.06 J/s (1 mark) iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why. Energy is wasted as heat. Low efficiency. (2 marks) b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this? One cell provides 0.06J per s therefore 7000/0.06 cells are needed = about 117 000 (2 marks)

  36. A solar cell is used to power a small electric motor. • This circuit is set up outside in bright sunlight. The reading on the voltmeter is 3V. i) When the circuit is switched on for 200s, the motor is supplied with 12J of energy. Calculate the current in the cell. 2 3 Energy = VIt = 12/3 x 200 = 0.02 (3 marks) Although the calculation is badly expressed, marks have been awarded for a correct calculation; however, if the calculation had been only partly correct, the way it is expressed could have caused loss of a mark which would have otherwise been awarded. Units have been omitted

  37. ii) Calculate the rate, in J/s, at which energy is supplied to the motor. Power = V x I = 3 x 0.02 = 0.06 J/s (1 mark) The answer could also be obtained from 12J of energy in 200s, so energy per s = 12J/200s = 0.06J/s iii) The rate at which energy is supplied to the solar cell is much greater than the rate at which energy is supplied to the motor. Explain why. Energy is lost as heat Scores for heat, but energy is not “lost”; it is wasted. Reference to efficiency could also score. (2 marks) b) A pupil suggests that solar cells could be used to power an electric shower, which needs to be supplied with energy at the rate of 7000J/s. How many of the solar cells would be needed to do this? 4 5 One cell provides 0.06J per s therefore 7000/0.06 cells are needed = about 117 000 (2 marks)

  38. GCSE Physics Exam Doctor Electricity – Energy in Circuits Question 5

  39. A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done? (2 marks) ii) How much gravitational potential energy does the load gain? (1 mark) • i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second? (2 marks)

  40. ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether? (1 mark) iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor (2 marks)

  41. A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done? Work done = Force x distance = 2.5 x 0.2 = 0.5 Nm (2 marks) ii) How much gravitational potential energy does the load gain? 0.5 J (1 mark) • i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second? 6 X 0.1 = 0.6 (2 marks)

  42. ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether? 0.6 x 4 = 2.4J (1 mark) iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor 0.5/2.4 = 0.21% (2 marks)

  43. Mark scheme A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done? Work done = Force x distance = 2.5 x 0.2 = 0.5 Nm (2 marks) ii) How much gravitational potential energy does the load gain? Energy gained = work done. e.c.f from part i) would be allowed. Units can be Nm or J for either part. 0.5 Nm (1 mark) • i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second? 6 X 0.1 = 0.6J (2 marks)

  44. Mark scheme ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether? 0.6 x 4 = 2.4J (1 mark) iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor 0.5/2.4 = 0.21 or 21% (2 marks)

  45. A motor is used to lift a load. a) i) If it lifts a load of 2.5N through a height of 0.2m, how much work is done? Work done = Force x distance = 2.5 x 0.2 = 0.5 Nm (2 marks) ii) How much gravitational potential energy does the load gain? 0.5 J Energy gained = work done. e.c.f from part i) would be allowed. Units can be Nm or J for either part. (1 mark) • i) The motor has a P.D. of 6V across it and takes a current of 0.1A. How much energy is transformed from electrical energy each second? 4 5 6 X 0.1 = 0.6 Units are omitted. (2 marks)

  46. ii) If the motor takes 4s to lift the load through the 2m how much energy is transformed from electrical energy altogether? 0.6 x 4 = 2.4J (1 mark) iii) Not all of the energy transformed from electrical energy has been transferred to the load as g.p.e. Calculate the efficiency of the motor 0.5/2.4 = 0.21% Efficiency can be expressed as a fraction e.g. 0.21 or as a percentage 21%, but it is not 0.21% (2 marks) 2 3

  47. GCSE Physics Exam Doctor Electricity – Energy in Circuits End of questions

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