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. 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- 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: and 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. 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 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 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 R = 8.42 ! parallel 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 is a magnetic north pole and which end is a magnetic south pole.

The magnetic field around a solenoid has a shape The plane of the loop is parallel to the external similar to that of a bar magnet. magnetic field so the induced current is at a 50. The motor principle states that a current- maximum. carrying conductor that cuts across external 56. For coil B to generate the sane amount of magnetic field lines experiences a force electricity as coil A in a changing magnetic field, it perpendicular to both the magnetic field and the should have the same number of windings as coil direction of the . A, which has 300 windings. To determine the direction of the force on a current 300 – 150 = 150 carrying conductor placed in an external magnetic So coil B should have 150 windings added. field, point the fingers of your open right hand in 57. The transformer in Figure 5 has fewer the direction of the external magnetic field and secondary windings than primary windings so it your thumb in the direction of the conventional decreases the voltage in the secondary coil. Since current. Your palm will now face the direction of it decreases the voltage it is a step-down the force on the conductor. transformer. 51. Diagrams should resemble Figure 2 on page 58. Solve for I in the power equation P = VI and 567 of the Student Book. then substitute the values given for P and V to find 52. A split ring commutator was used to make the I, which is the amount of current produced: transition between a galvanometer and a DC P = VI motor. The split ring commutator works by P interrupting the current through the circuit when I = the wire loop in the DC motor is perpendicular to V the external magnetic field, and then allowing the 4 current to flow in the opposite direction through P = 1800 MW; V = 5.0 × 10 V the wire loop once the split ring comes in contact P I = with the circuit again. This allows the wire loop to V continuously rotate. 1800 MW = 53. Coil A will induce more current than coil B 5.0! 104 V because it has more windings, and the greater 9 1.8 ! 10 W number of windings in a coil, the more electric = 5.0 104 V current that can be induced for a given change in ! 4 the magnetic field. = 3.6 ! 10 A 54. Ground fault circuit interrupters will prevent I = 36 kA short circuits due to water. So the amount of current produced is 36 kA.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-6 59. Substitute the values given for V and I in the 0.88 ! Pin = 1500 MW power equation P = VI to find P, which is the 1500 MW amount of power produced: P = in 0.88 P = VI P = 1700 MW in V = 180 kV; I = 35 A The power wasted is Pin - Pout, and Pout is P = VI 1500 MW, so the power wasted is = (180 kV)(35 A) 1700 MW – 1500 MW = 200 MW

= (1.8 105 V)(35 A) ! The nuclear power plant has an efficiency of 30 % 6 = 6.3 ! 10 W and produces 1500 MW of electrical power, so P = 6.3 MW 30 % of the input power, Pin, is 1500 MW. This is So the amount of power produced is 6.3 MW. 0.30 × Pin = 1500 MW, which can be used to solve 2 60. Use the power equation in the form P = I R: for Pin: 2 P = I R 0.30 ! Pin = 1500 MW = (3.0 kA)2 (0.40 !) 1500 MW Pin = = (3.0 ! 103 A)2 (0.40 !) 0.30 6 P = 5000 MW = 3.6 ! 10 W in The power wasted is P - P , and P is P = 3.6 MW in out out So the total power loss due to transmission through 1500 MW, so the power wasted is the wire is 3.6 MW. 5000 MW – 1500 MW = 3500 MW 61. First determine the amount of power lost in transmission in watts: The difference in the amounts of power wasted is 0.60 % of 2100 MW = 12.6 MW (one extra digit 3500 MW – 200 MW = 3300 MW. carried) So, the nuclear power plant wastes 3300 MW more So the amount of power lost in transmission, P, is power than the hydroelectric power plant. 12.6 MW. 63. The coal power plant has an efficiency of 47 % and produces 3500 MW of electrical power, so Now solve for R in the power equation in the form 47 % of the input power, Pin, is 3500 MW. This is P = I2R and then substitute the value given for I 0.47 × Pin = 3500 MW, which can be used to solve and the value found for P to find R, which is the for Pin: total resistance in the transmission wire: 0.47 ! Pin = 3500 MW P = I 2 R 3500 MW P = P in 0.47 R = 2 P = 7447 MW (two extra digits carried) I in 12.6 MW = (5.0 kA)2 The coal power plant with carbon capture 7 technology installed has an efficiency of 41 % and 1.26 ! 10 W = still has an input of 7447 MW of electrical power, (5.0! 103 A)2 so 41 % of 7447 MW, is Pout. This is R 0.50 ! = 0.41 × 7447 MW = Pout, which can be used to So the total resistance in the transmission wire is solve for Pout: 0.50 Ω. 0.41! 7447 MW = P out P = 3053 MW (two extra digits carried) Analysis and Application out 62. The hydroelectric power plant has an efficiency of 88 % and produces 1500 MW of The difference in the amount of output power is electrical power, so 88 % of the input power, Pin, is 3500 MW – 3053 MW = 450 MW. 1500 MW. This is 0.88 × Pin = 1500 MW, which So the amount of extra power lost to the carbon can be used to solve for Pin: capture technology is 450 MW.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-7 64. Given: P = 90 % of 60.0 W; ΔE = 3.0 × 103 J 67. (a) First find the total resistance of the circuit. Required: Δt Start by finding the equivalent resistance for the !E parallel part of the circuit. Analysis: P = 1 1 1 !t = + Solution: Rparallel R2 R3 P = 90 % of 60.0 W 1 1 1 = + = 0.90 ! 60.0 W Rparallel 20.0 ! 30.0 ! P 54.0 W = R = 12 ! parallel

!E P = Now find the total resistance. !t R is in series with R , so !E parallel 1 !t = R = R + R P total 1 parallel 3.0 " 103 J = 5.0 ! + 12 ! = J R = 17 ! 54.0 total s = 55.56 s V Now find Isource using Ohm’s law written as I = . !t = 56 s R Statement: It takes 56 s to charge the battery. Vsource 65. (a) Voltmeter should be placed in parallel to Isource = Rsource R1. 15 V (b) Ammeter should be placed in series with R2. = 17 ! 66. Start by finding Rtotal. Isource = 0.8824 A (two extra digits carried) Rtotal = R1 + R2

= 40 ! + 15 ! The amount of charge in coulombs passing R = 55 ! total Q through the circuit can now be found using I = . !t V Now find Isource using Ohm’s law written as I = . Q R I = !t Vsource Q I t I = = ! source R source = (0.8824 A)(8.0 s) 30.0 V Q = 7.1 C = 55 ! So after 8.0 s, 7.1 C of charge passes through the I = 0.55 A source circuit. (b) The source and resistor 1 are in series. Using The amount of charge in coulombs passing the value found for Isource in part (a) and KCL for a through the circuit in 10.0 s can now be found series circuit, you can find I1: Q I = I using I = . source 1 !t I = 0.8824 A (two extra digits carried) 1 Q I = t ! You can now find V1 using Ohm’s law written as Q = I!t V = IR. = (0.55 A)(10.0 s) V1 = I1R1 Q = 5.5 C = (0.8824 A)(5.0 !) So, in 10.0 s, 5.5 C of charge that passes through V = 4.412 V (two extra digits carried) the circuit. 1

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-8 Apply KVL to the complete pathway involving the V R = 3 source, resistor 1, and resistor 2 to find V2. 3 I3 V = V + V source 1 2 8.0 V V = V ! V = 2 source 1 0.50 A = 15 V ! 4.412 V R = 16 ! 3 V = 10.59 V (two extra digits carried) 2 You can now find I2 using Ohm’s law written as You can now find I using Ohm’s law written as V 2 I = . V R I = . R V2 I = V 2 R I = 2 2 2 R 8.0 V 2 = 10.59 V 40.0 ! = I = 0.20 A 20.0 ! 2 I = 0.5295 A 2 So the value of R3 is 16 Ω and the value of I2 is 0.20 A. The amount of charge in coulombs passing (b) First use KCL for a parallel circuit to find Q Iparallel. Note that in part (a), I2 was found to be through R2 can now be found using I = . 0.5000 A. !t I I I Q parallel = 2 + 3 I = !t = 0.20 A + 0.50 A Q = I!t I = 0.70 A parallel = (0.5295 A)(10.0 s) Q = 5.3 C The source and Iparallel are in series. Using KCL for So after 10.0 s, 5.3 C of charge passes through R2. a series circuit, you can find Isource.

68. (a) Apply KVL to the complete pathway Isource = Iparallel involving the source, resistor 1, and resistor 2 to Isource = 0.70 A find V2.

V = V + V source 1 2 The time it will take for 15 C of charge to pass V V V 2 = source ! 1 Q through the circuit can now be found using I = . = 20.0 V ! 12.0 V !t V = 8.0 V Q 2 I = !t Apply KVL to the complete pathway involving the Q !t = source, resistor 1, and resistor 2 to find V3. I V = V + V 15 C source 1 3 = 0.70 A V3 = Vsource ! V1 !t = 21 s V = 20.0 V ! 12.0 V 3 So it takes 21 s for 15 C of charge to pass through V = 8.0 V 3 the circuit. 69. (a) The resistance of the bulb would be You can now find R3 using Ohm’s law written constant if the temperature remained the same. V Since the bulb heats up when it is on, the as R = . I resistance of the bulb changes, which is why the resistance values are not the same for the bulb on and off. (b) If the filament breaks, then the circuit is broken. I would expect no resistance value.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-9 70. Since the sum of the four resistances is 74 Ω, rise slope = they cannot all be in series. If the first and fourth run are in parallel and the second and third are in !V parallel, then R = 16.5 Ω: m = total !I 1 1 1 21 V "10 V + = = 6.0 18 4.5 "4 "4 ! ! ! 8.4 # 10 A " 4.1# 10 A 1 1 1 4 + = m = 2.6 # 10 ! 20 30 12 4 ! ! ! So the resistance of the circuit is 2.6 × 10 Ω. 4.5 ! + 12 ! = 16.5 ! 72. (a) First apply KVL to the determine V2, V3, and V4. Since the first two resistors are in parallel, V1 = V2, so V2 = 3.0 V.

Vsource = V1 + V3 V = V ! V 3 source 1 = 12 V ! 2.0 V V = 9.0 V 3 Since the second pair of resistors are in parallel, V3 = V4, so V4 = 9.0 V.

V 71. (a) Now find I3 using Ohm’s law in the form I = . R

V3 I3 = R3 9.0 V = 250 ! I = 0.036 A 3

V Now find I4 using Ohm’s law in the form I = . R

V4 I4 = R4 (b) The slope of the line connecting the data points 9.0 V represents the resistance. For example, the line = 300 ! passes through the data points (0.41 mA, 10 V) I = 0.03 A and (0.84 mA, 21 V). First convert the current to 4 amperes to find the resistance in ohms: The total current entering the second pair of 1 A resistors is: I1 = 0.84 mA ! 0.036 A + 0.03 A = 0.066 A 1000 mA By KCL for a series circuit, this is the same "4 I1 = 8.4 ! 10 A amount of current entering the first pair of resistors, and the current divides equally between 1 A them because they have the same resistance. I = 0.41 mA ! 5 I + I 1000 mA I = 3 4 1 2 I = 4.1! 10"4 A 5 0.066 A The two data points (12 V, 9.7 × 10-5 A) and = 2 (16 V, 1.29 × 10-4 A) can be used to find the slope: I = 0.033 A 1

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-10 V (b) Your right hand would be flat in a vertical Now find R1 using Ohm’s law in the form R = . I plane, with your thumb pointing to the left and V your fingers pointed downward. R = 1 (c) Since the direction of the force is toward you, it 1 I 1 is directed out of the page. 3.0 V = 78. In an ammeter, a galvanometer is placed in 0.033 A parallel with a resistor with lower resistance than R = 91 ! 1 the galvanometer. In a voltmeter, a galvanometer is placed in series Use the relationship between R1 and R2 to find R2. with a resistor with a very high resistance. R = R 79. Answers may vary. Students’ answers should 2 1 name a household object, such as a electric lock, R = 91 ! 2 speaker, or doorbell, state whether the magnet it So the resistance of R1 and R2 are both 91 Ω. uses interacts with another magnet or produces a (b) As found in part (a), I3 is 0.036 A. force on a conductor, and also whether it is a 73. (a) The magnetic field lines are in the incorrect permanent magnet or an electromagnet. direction; they should travel from north to south. 80. Answers may vary. Sample answer: (b) The magnetic field decreases (not increases) in The stove element of an induction cooker strength as you move away from the magnet. The generates a rapidly changing magnetic field. When magnetic field lines should be drawn farther apart a metal pot is placed on the element, an electric (instead of closer together) as you move away current is induced in the metal by the changing magnet. magnetic field, in accordance with the law of 74. The aurora are caused by charged particles electromagnetic induction. Metal has a high emitted from the Sun that interact with Earth’s resistance, so the induced electric current in the pot magnetic field and cause gases in the atmosphere produces a large amount of thermal energy that to become excited and emit energy in the form of cooks the food. light. This only happens at the poles because that is Metal detectors also work by using a rapidly where Earth’s magnetic field is strongest. changing magnetic field. When the changing 75. An electromagnet is useful because it can be magnetic field is brought in the region of metal, by switched on and off. This makes it useful in the law of electromagnetic induction it must electronic devices such as speakers, subwoofers, induce an electric current in that metal. The and electric bells. The electromagnet’s strength induced electric current creates a magnetic field, can also be controlled by the amount of current which is then detected by the metal detector. The and number of coils, which is useful in a DC metal detector only detects metal because electric motor. The polarity can also be switched, unlike in current cannot be induced in non-conductors. a permanent magnet. 81. First determine the total number of windings in 76. Earth’s magnetic field lines normally point the length of coil type A that the student has: from south to north, since the south pole is 20 windings magnetically north, and the north pole is ! 12 cm = 240 windings 1 cm magnetically south. For this case the field lines have reversed so that facing westward reads east The length, L, of coil type B must have the same on a compass. Since the cables are buried below number of windings as coil type A to produce the the car, using the right-hand rule determines that same amount of current in a changing magnetic current must be travelling north. field. So the length, L, of coil type B must have 77. (a) To determine the direction of the force on a 240 windings and L can now be found: current carrying conductor placed in an external 15 windings ! L = 240 windings magnetic field, point the fingers of your open right 1cm hand in the direction of the external magnetic field 1 cm L = 240 windings ! and your thumb in the direction of the 15 windings conventional current. Your palm will now face the direction of the force on the conductor. L = 16 cm So the student needs 16 cm of coil type B.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-11 82. (a) Answers may vary. Sample answer: So the ratio of the primary windings to secondary Yes it is possible to create an AC motor. It works windings in the transformer is 2 : 1. just like an AC generator but in reverse. If the 85. (a) Given: Vp = 80.0 V; Np = 100; Ns = 160 V; wires from the AC generator in Figure 1, Section Is = 10.0 A 13.4, were plugged into an outlet in North America Required: Vs it would spin at a frequency of 60 Hz. V N Analysis: p = p (b) Answers may vary. Sample answer: V N Yes, this set up would be practical for any device s s V N that uses a motor and can be plugged into an AC V = p s source. Any device that rotates in place could use s N p this motor. An example of this could be a blow Solution: dryer or a fan. V N 83. As the shaded side of the armature moves away V = p s s N from the south pole of the external magnet, Lenz’s p law determines the left side of the armature to be a (80.0V)(160) = north magnetic pole. Using the right-hand rule for 100 a coil, the direction of the conventional current is Vs = 128V up across the front of the coil. Statement: The voltage of the secondary circuit is 84. (a) Given: Vp = 250 V; Ip = 5.0 A; Is = 10.0 A 128 V. Required: Vs (b) Use the relevant equation related to I V Analysis: s = p transformers to find the current of the primary I V circuit, I : p s p Solve for V : I N p s s = Vp Ip Ip Ns Vs = Is I N I = s s p Solution: N p V I p p (10.0A)(160) Vs = = Is 100 (250V)(5.0 A) I = 16.0A = p 10.0 A So the current of the primary circuit, Ip, is 16.0 A. 2 V = 125V (one extra digit carried) 86. (a) Given: Vp = 3.0 × 10 V; Np = 150; s 1 Statement: The voltage in the secondary circuit is Vs = 6.0 × 10 V 120 V. Required: Ns V N (b) Substitute the values given for V and the value p p p Analysis: = found in part (a) for Vs in the relevant equation Vs Ns related to transformers to find the ratio of the Vs N p windings: Ns = Vp Vp N p = Solution: V N s s V N N V s p p p Ns = = Vp Ns Vs 1 (6.0 ! 10 V )(150) Vp = 250 V; Vs = 125 V = 2 N V 3.0 ! 10 V p = p Ns = 30 Ns Vs Statement: The secondary circuit has 30 coils. 250 V = 125 V N p = 2 N s

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-12 V V (b) First use Ohm’s law in the form I = and the (b) First use Ohm’s law in the form I = and the R R values given for Vp and Rp to find Ip, the current in values given for Vp and Rp to find Ip, the current in the primary circuit: the primary circuit:

V Vp I = p I = p p R Rp p 3.0 ! 102 V 3.0kV = = 10.0! 750! 3 I = 3.0 ! 101 A 3.0! 10 V p = 750! So the current in the primary circuit, Ip, is 1 I = 4.0A 3.0 × 10 A. p So the current in the primary circuit, Ip, is 4.0 A. Now use the relevant equation related to transformers to find the current in the secondary Now use the relevant equation related to circuit, Is: transformers to find the current in the secondary I N circuit, Is: s = p I N I V p s s = p I N Ip Vs I = p p s N I V s I = p p 1 s (3.0 ! 10 A)(150) Vs = 30 (4.0 A)(3.0kV) = I = 150A 2 s 1.0 ! 10 V So the current in the secondary circuit, Is, is 150 A. (4.0A)(3.0! 103 V ) = 87. (a) Given: Vp = 3.0 kV; Rp = 750 Ω; 2 2 1.0 ! 10 V Vs = 1.0 × 10 V; Ns = 60 I = 120 A Required: Np s V N So the current in the secondary circuit, Is, is Analysis: p = p V N 120 A. s s Solve for Np: V V N Now use Ohm’s law in the form R = and the N = p s I p V s value found for Is and the value given for Vs to find Solution: Rs, the resistance in the secondary circuit:

V N Vs N = p s R = p s I Vs s (3.0kV)(60) 1.0 ! 102 V = = 1.0 ! 102 V 120 A 3 R = 0.83 ! (3.0! 10 V)(60) s = 1.0 ! 102 V So the resistance in the secondary circuit is 0.83 Ω. 88. First determine the amount of power lost in N = 1800 p transmission in watts: Statement: The primary circuit has 1800 coils. 5 0.80 % of 14 MW = 1.12 × 10 W (one extra digits carried) So the amount of power lost in transmission, P, is 5 1.12 × 10 W.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-13 Now determine the current that the power is 2 " P% transmitted with using the power equation in the 0.0060 ! P = R $V ' form P = VI and the values given for P and V: # & 2 P = VI P 2 R = 0.0060 ! P P V I 2 = P P V R = 0.0060 ! 14 MW V 2 P P = 160 kV P R = 0.0060 14! 106 W V 2 = 1.6 ! 105 V 0.0060V 2 P = R I = 87.5 A (one extra digit carried) So the power is transmitted with a current of 87.5 A. V = 240 kV; R = 0.50 Ω 0.0060V 2 P = Now solve for R in the power equation in the form R 2 P = I R and then substitute the value found for P 0.0060(240 kV)2 = and I to find R, which is the total resistance in the 0.50 ! transmission wire: 0.0060(2.4 ! 105 V)2 2 = P = I R 0.50 ! P P = 690 MW R = 2 I So the dam produces 690 MW of power. 1.12 ! 105 W 90. First find the potential difference of the = (87.5 A)2 primary circuit, Vp, using the relevant equation related to transformers: R = 15 ! V V So the total resistance in the transmission wire is p = s 15 Ω. N p Ns 89. First find an equation relating the power V N generated by the dam, the current that the power is V = s p p N transmitted with, and the total resistance. This s equation is: (250 kV)(1000) = 0.60 % of P = I 2 R 6000 4 2 V = 4.2 ! 10 V 0.0060 ! P = I R p So the potential difference of the primary circuit is Now rearrange for I in the power equation P = VI 4.2 × 104 V. and substitute the expression for I into the previous equation: Now use the power equation in the form P = VI P = VI with the value found for V and the value given for P I to find P, the power that the nuclear plant I = V produces: 0.0060 ! P = I 2 R P = VI 4 2 = (4.2 ! 10 V)(30.0 kA) " P% 0.0060 ! P = R = (4.2 104 V)(3.00 104 A) #$ V &' ! ! 9 = 1.3 ! 10 W Now solve for P in this equation and substitute the P = 1300 MW values given for V and R to determine P, which is So the power plant produces 1300 MW of power. the total power generated by the dam:

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-14 Evaluation Reflect on Your Learning 91. (a) To make an ammeter, place a galvanometer 95. Answers may vary. Sample answer: in parallel with a resistor with lower resistance I leave my computer on 24 hours a day at 450 W. than the galvanometer (to keep the current away (24 h)(450 W) = 10 800 Wh or 10.8 kWh. from the galvanometer). The galvanometer My computer uses 11 kWh of electricity in a day. measures the current passing through the resistor. (b) To make a voltmeter, place a galvanometer in 7(10.8 kWh) = 75.6 kWh series with a resistor with a very high resistance (to My computer uses 76 kWh in a week. keep the current away from the galvanometer). Then, knowing the resistance of the path the 3657(10.8 kWh) = 394 275.6 kWh galvanometer is on, calculate the voltage by My computer uses 3.9 ! 105 kWh in a year. multiply the current by the resistance. 96. Conventional current in a circuit is directed 92. (a) Household circuits are set up with an AC from positive to negative. Now we know that power source and the devices plugged in are in electrons are the actual particles moving in a wire parallel. and that they travel from negative to positive. This (b) Since the circuit is in parallel and all power convention did not matter when scientists were strips and additional plugs create more parallel developing electrical devices. It just set the circuits, as more devices are plugged in the overall standard for the right-hand rule conventions in resistance decreases. determining many of the formulas relating (c) The voltage of an outlet remains constant, so electromagnetic phenomena. using Ohm’s Law, as the resistance decreases there 97. Answers may vary. Students should explain the is a corresponding increase in current, which could key concepts of the electrical power grid, including melt or fray the wires. To prevent this, houses have power supply and distribution so that the electrical safety devices such as circuit breakers, fuses, needs of everyone on the grid are met. GFCIs, and AFCIs. 93. Yes, it is possible to create a DC motor without Research permanent . Instead of magnets, add two 98. Answers may vary. Students may choose to solenoids to the circuit. Orient the solenoids so write about James Watt, , they have north and south magnetic poles in the Charles-Augustin de Coulomb, , former position of the magnets. Resistors are André-Marie Ampère, or Georg Simon Ohm. needed to control the amount of current in the Biographies should include details about when and circuit. If all three components are in parallel, then where the scientist lived and what contributions he they will have the same voltage. By controlling the made to science. amount of current with the resistors you can 99. (a) Answers may vary. The United States, control the strength of the magnetic fields and China, and Japan rank among the world’s top power of the motor. energy consumers with most of the energy being 94. (a) The current induced from a magnet in a coil derived from fossil fuels. The United States uses or loop cannot result in a magnetic field that 22 % of the total global energy, China uses 20 %, attracts the magnet because this violates the laws while Japan uses 7 %. Students may mention of conservation of energy. If the current created an France, which gets 80 % of its power from nuclear attraction to the magnet then work could be done energy but only uses 3 % of the total global without any work put in. For example, we could energy. lift a bar magnet through a horizontal loop and (b) Answers may vary. Switzerland and Denmark then the induced current would pull the weight of rank among the greenest industrialized nations by the magnet upward without any additional work experts at Yale and Columbia universities. These being put in. rankings are based on carbon emission reductions, (b) We can control the current in a coil to attract or reforestation efforts, and use of hydropower and repel a magnet because in both cases the energy geothermal energy. Japan is ranked as one of the required to do so and thus the work done is most energy efficient. Other countries students supplied by an external power source. The power may mention include Hong Kong, Ireland, and the source must do work to create the potential U.K. difference necessary to maintain the current in the coil.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-15 100. Answers may vary. Students should research at least two methods of improving coal power plants, such as coal washing, wet scrubbers, electrostatic precipitators, and gasification. Emphasis should be put on benefits and cost to upgrade existing coal power plants.

101. Answers may vary. Students should describe how fuel cells use stored chemical energy converted to electrical energy, primarily by using hydrogen. Students should describe how the energy is converted and how the fuel cells get recharged, and include the costs of buying and operating a fuel cell vehicle. 102. Answers may vary. Students should describe how photovoltaic cells work. Light is absorbed by semiconducting materials, which then generate a current. Students should discuss reasons why this is so limited, including the impurities in the material and different wavelengths of light. They should also report on any new developments that might make solar power a more economic energy source and give estimates as to how much energy might be obtained. 103. Answers may vary. Students should research Tesla’s ideas on wireless electricity, his successful experiments, and the lack of funding for his Wardenclyffe Tower that could supply electricity but not keep track of how to bill for its use. 104. Answers may vary. Students should describe how most transformers used in power transmission have multiple step down ratios so that the transformer can be used for different purposes. They should explain how the winding-to-voltage ratios still hold, as does conservation of power. They should include a diagram of a transformer as an example.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-16 Unit 5 Self-Quiz, pages 624–625 1. (c) 2. (c) 3. (d) 4. (d) 5. (b) 6. (c) 7. (a) 8. (d) 9. (b) 10. (a) 11. (c) 12. (d) 13. (d) 14. (a) 15. (d) 16. (a) 17. (b) 18. False. Electrical energy for buildings is measured in kilowatt hours. 19. False. Voltmeters should be placed in parallel to measure the voltage across a load. 20. True 21. True 22. False. Loads that are placed in parallel all have the same potential difference. 23. True 24. False. Earth’s magnetic field exists both inside and surrounding Earth. 25. False. When two unlike magnetic poles are brought close to one another, they will attract. 26. True 27. False. Oersted found that a charge moving through a straight conductor produces a circular magnetic field around the conductor. 28. False. Using your right hand, if your thumb points along the direction of current in a wire, then your fingers curl in the direction of the magnetic field. 29. True 30. True 31. True 32. True 33. False. Most appliances in Canadian homes require a 120 V energy source. 34. True 35. False. Step-up transformers increase the amount of voltage in the secondary circuit. 36. False. Transformers are used to step up the voltage in order to minimize power loss.

Copyright © 2011 Nelson Education Ltd. Unit 5: Electricity and Magnetism U5-2