Unit 5 Review, pages 626–633 Understanding Knowledge 29. The nuclear power plant has an efficiency of 1. (b) 32 %, so 32 % of the total power is transformed 2. (a) into electrical energy. 32 % of 14 000 MW is: 3. (d) 0.32 × 14 000 MW = 4500 MW 4. (c) So, the power plant produces 4500 MW of 5. (b) electrical power. 6. (c) 30. Given: P = 35 W; Δt = 220 h 7. (b) Required: ΔE 8. (c) !E Analysis: P = 9. (b) !t 10. (a) ΔE = PΔt 11. (c) Solution: Convert time to seconds to get the 12. (b) answer in joules: 13. (d) 3600 s !t = 220 h " 14. (b) 1 h 15. (c) !t = 792 000 s 16. False. Voltage is a measure of the amount of electrical potential energy associated with each charge. !E = (35 W)(792 000 s) 17. False. In a circuit, electrons flow from negative = 2.772 " 107 Wi s to positive. !E = 2.772 " 107 J (two extra digits carried) 18. True To find the answer in kilowatt hours, convert from 19. True joules: 20. True 1 kWh 21. False. Then two like magnetic poles are 7 2.772 ! 10 J ! 6 = 7.7 kWh brought close one another, they repel. 3.6 ! 10 J 22. True Statement: The light bulb requires 7.7 kWh or 23. False. Adding a soft-iron core will increase the 28 MJ of energy to operate for 220 h. strength of a DC motor. 31. Given: ΔE = 1400 J; Δt = 7.0 min 24. True Required: P 25. True !E Analysis: P = 26. False. A step-up transformer increases the !t voltage in the secondary circuit. Solution: First convert time to seconds to get the OR answer in joules per second or watts: A step-up transformer decreases the current in the " 60 s % secondary circuit. !t = 7 min #$ 1 min&' 27. True 28. (a) (v) !t = 420 s (b) (viii) (c) (i) !E P = (d) (ii) !t (e) (iii) 1400 J = (f) (ix) 420 s (g) (vii) P = 3.3 W (h) (iv) Statement: The amount of power required to (i) (vi) charge the battery is 3.3 W. Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-3 32. Given: V = 120 V; ΔE = 540 J Q !t = Required: Q I !E 4.0 C Analysis: V = = Q 0.50 A !E !t = 8.0 s Q = V Statement: It takes 8.0 s for the charge to pass !E through the resistor. Solution: Q = V 35. Answers may vary. Sample answer: 540 J = 120 V Q = 4.5 C Statement: The total amount of charge moved across the terminals is 4.5 C. 36. (a) The current in a series circuit is constant 33. Given: Q = 0.65 C; Δt = 1.5 min and the same as the source current. The source and Required: I lamp 1 are in series, and I1 = 8.5 mA. Using these Q Analysis: I = values and KCL, you can find Isource: !t Isource = I1 Solution: Convert time to seconds to get the I = 8.5 mA answer in coulombs per second, or amperes: source 60 s !t = 1.5 min " The amount of current entering a junction is equal 1min to the amount of current exiting the junction. This !t = 90 s can be used to find I3: Iparallel = I2 + I3 Q I = 8.5 mA = 2.1 mA + I !t 3 I = 6.4 mA 0.65 C 3 = 90 s So I3 is 6.4 mA. #3 (b) The current in a series circuit is constant and I = 7.2 " 10 A the same as the source current. From part (a), Convert the current to milliamperes: Isource = 8.5 mA. The amount of current entering a junction is equal to the amount of current exiting 1000 mA I = 7.2 ! 10"3 A ! the junction. Letting I = I , this can be used to find 1A 2 3 I3: I = 7.2 mA I = I + I Statement: The amount of current in the wire is parallel 2 3 I I I 7.2 mA. parallel = 3 + 3 2 34. Given: Q = 4.0 C; I = 5.0 × 10 mA Iparallel = 2I3 Required: Δt Iparallel Q I = Analysis: I = 3 2 t ! 8.5 mA Q = !t = 2 I I = 4.2 mA Solution: Convert current to amperes to get the 3 answer in coulombs per ampere, or seconds: So I3 is 4.2 mA. 1 A I = 5.0 ! 102 mA ! 1000 mA I = 0.50 A Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-4 37. Given: R = 9.0 × 102 Ω; A = 0.72 mA. 40. Magnet diagrams should resemble the bar Required: V magnet and horseshoe magnet in Figure 3 on page V 549 of the student book. Analysis: R = The bar magnet diagram most closely resembles I Solution: Convert the current to amperes to get the the magnetic field lines of Earth. answer in volts: 41. A magnetic field exerts a force on an iron filing. Iron filings are light and can be moved 1 A I = 0.72 mA ! independently, so when placed in the presence of a 1000 mA magnetic field the filings are easily forced into I = 7.2 ! 10"4 A position along the magnetic field lines. 42. (a) By the right-hand rule for a straight V conductor, the direction of the current is into the R = page. I (b) By the right-hand rule for a straight conductor, V = IR the direction of the magnetic field is counter- "4 2 = (7.2 ! 10 A)(9.0 ! 10 !) clockwise. = 6.48 ! 10"1 V (c) By the right-hand rule for a straight conductor, 1000 mV the magnetic field is coming out of the page. = 6.48 ! 10"1 V ! 1 V 43. Oersted placed a compass near a conducting wire that was aligned with Earth’s north and south V 650 mV = magnetic poles. The compass was aligned with the Statement: The potential difference across the wire. When current was turned on the compass resistor is 650 mV. needle pointed perpendicular to the wire. When the 38. The resistors R2 and R3 are in parallel and can current was switched off the compass needle be replaced with an equivalent resistance: returned to normal. 1 1 1 = + 44. Answers may vary. Diagrams should resemble Rparallel R2 R3 Figure 4 on page 560 of the Student Book. 1 1 1 45. Conventional current is directed from positive = + R 11.4 ! 32.2 ! to negative, which is in the opposite direction to parallel the flow of electrons in a circuit. For conventional Rparallel = 8.42 ! current, use the right hand to determine the direction of the magnetic field produced by the The resistor R1 and the equivalent resistance current; for electron flow, use the left hand. The Rparallel are in series and can be replaced with an magnetic field points in the same direction equivalent resistance: regardless of which model is used. R = R + R 46. Using the right-hand rule for a straight total 1 parallel conductor, my right fingers curl around the = 7.0 ! + 8.42 ! conductor in a clockwise direction and my thumb R = 15 ! total points into the page, which is the direction of the Statement: The total resistance of the circuit is conventional current. So the direction of the 15 Ω. current is into the page. 39. Earth’s magnetic field causes a force that is 47. Ampère performed an experiment with two nearly parallel to the Earth’s surface at positions parallel current-carrying wires. He found that they not near its magnetic poles and is directed from the repelled each other when the currents were in south magnetic pole to the north magnetic pole. At opposite directions and that they attracted each positions near its magnetic poles, it causes a force other when the currents flowed in the same that is nearly perpendicular to the Earth’s surface direction. and is directed toward the Earth at its south 48. (a) For the two parallel wires to experience a magnetic pole and away from the Earth at its north magnetic force of repulsion the magnetic field magnetic pole. Earth’s gravitational field causes a lines between them must point in the same force that is always directed toward Earth’s centre. direction. For this to happen, the currents in the wires must be in opposite directions. Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-5 (b) If both currents in the wires were reversed the 55. The rotation of the loop in a clockwise currents would still be in opposite directions so the direction in the magnetic field causes a magnetic force would still be repulsion. conventional current in the loop in the direction (c) If the currents were increased the wires would shown: experience more repulsion. (d) If one current were switched off there would only be magnetic field lines around the other wire. There would be no interacting magnetic field lines so there would be no magnetic force between the wires. 49. The right-hand rule for a solenoid states that if the fingers of your right hand wrap around a coil in the direction of the conventional current, your thumb will point in the direction of the north magnetic pole of the coil. The right-hand rule helps to understand the operation of a solenoid by showing how to determine which end of the electromagnet is a magnetic north pole and which end is a magnetic south pole.