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OpenStax College Physics Instructor Solutions Manual Chapter 23 CHAPTER 23: ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES 23.1 INDUCED EMF AND MAGNETIC FLUX 1. What is the value of the magnetic flux at coil 2 in Figure 23.56 due to coil 1? Solution Using the equation Φ BA cos , we can calculate the flux through coil 2, since the coils are perpendicular: Φ BA cos BA cos 90 0 2. What is the value of the magnetic flux through the coil in Figure 23.56(b) due to the wire? Solution Φ BA cos BA cos 90 0 23.2 FARADAY’S LAW OF INDUCTION: LENZ’S LAW 3. Referring to Figure 23.57(a), what is the direction of the current induced in coil 2: (a) If the current in coil 1 increases? (b) If the current in coil 1 decreases? (c) If the current in coil 1 is constant? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenz's Law. Solution (a) CCW (b) CW (c) no current in coil 2 4. Referring to Figure 23.57(b), what is the direction of the current induced in the coil: (a) If the current in the wire increases? (b) If the current in the wire decreases? (c) If the current in the wire suddenly changes direction? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenz’s Law. Solution (a) CCW (b) CW (c) CW 5. Referring to Figure 23.58, what are the directions of the currents in coils 1, 2, and 3 (assume that the coils are lying in the plane of the circuit): (a) When the switch is first closed? (b) When the switch has been closed for a long time? (c) Just after the switch is opened? OpenStax College Physics Instructor Solutions Manual Chapter 23 Solution (a) (i) CCW (ii) CCW (iii) CW (b) (i) no current (ii) no current (iii) no current (c) (i) CW (ii) CW (iii) CCW 6. Repeat the previous problem with the battery reversed. Solution (a) (i) CW (ii) CW (iii) CCW (b) (i) no current (ii) no current (iii) no current (c) (i) CCW (ii) CCW (iii) CW 7. Verify that the units of Φ / t are volts. That is, show that 1 T m 2 / s 1 V . Solution The units of ΔΦ will be: Δt ΔΦ T m 2 N A m m 2 N m N m V so that 1 T m 2 /s 1 V s A s Δt s C 8. Solution 9. Solution Suppose a 50-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.250 m 2 . It is stretched to have no area in 0.100 s. What is the direction and magnitude of the induced emf if the uniform magnetic field has a strength of 1.50 T? EN ΔΦ NBA (50)1.50 T 0.250 m 2 187.5 V CW 188 V CW Δt t 0.100 s (a) An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner’s 2.00 T field with his fingers pointing in the direction of the field. Find the average emf induced in his wedding ring, given its diameter is 2.20 cm and assuming it takes 0.250 s to move it into the field. (b) Discuss whether this current would significantly change the temperature of the ring. ΔΦ N B A (1)2.00 T 1.10 10-2 m 2 3.04 10-3 V 3.04 mV Δt t 0.250 s 2 V (b) Since power goes as P , the smaller the resistance the larger the power and R the more heat transferred to the ring. So, as a lower limit on the ring, estimate R 1.00 m . In this case, the power would be: P 9.24 mW . This power exists for 1 of a second, so the heat transferred will be: 4 Q Pt 9.24 10 3 W 0.250 s 2.31 mJ . This is not a significant amount of heat. (a) E N 2 OpenStax College Physics 10. Solution Instructor Solutions Manual Chapter 23 Referring to the situation in the previous problem: (a) What current is induced in the ring if its resistance is 0.0100 ? (b) What average power is dissipated? (c) What magnetic field is induced at the center of the ring? (d) What is the direction of the induced magnetic field relative to the MRI’s field? V 3.04 10 3 V 0.304 A R 0.0100 Ω (b) P I 2 R 0.304 A 2 0.0100 Ω 0.924 mW (a) I (c) B 0 I 4π 10 T.m/A 0.304 A 1.74 10 21.10 10 m 7 5 T 2r (d) Looking in the direction of the magnetic field, the induced current is counterclockwise. 11. Solution 12. Solution 13. Solution 2 An emf is induced by rotating a 1000-turn, 20.0 cm diameter coil in the Earth’s 5.00 10 5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 10.0 ms? EN ΔΦ NBA (1000) 5.00 10 -5 T 10.0 10 -2 m Δt t 10.0 10 -3 s 2 0.157 V A 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V. EN ΔΦ NBA Et (10,000 V) 4.17 10-3 s B 0.425 T Δt t NA 500 0.250 m2 Approximately how does the emf induced in the loop in Figure 23.57(b) depend on the distance of the center of the loop from the wire? The magnetic field due to an infinite straight wire is emf induced in the loop will be proportional to wire. μ0 I , so the flux and hence the 2π r 1 , where r is the distance from the r OpenStax College Physics Instructor Solutions Manual Chapter 23 14. A lightning bolt produces a rapidly varying magnetic field. If the bolt strikes the earth vertically and acts like a current in a long straight wire, it will induce a voltage in a loop aligned like that in Figure 23.57(b). What voltage is induced in a 1.00 m diameter loop 50.0 m from a 2.00 106 A lightning strike, if the current falls to zero in 25.0 μs ? (b) Discuss circumstances under which such a voltage would produce noticeable consequences. Solution (a) We know E 0 NΦ , where the minus sign means that the emf creates the t current and magnetic field that opposes the change in flux, and Φ BA πr 2 B . Since the only thing that varies in the magnetic flux is the magnetic field, we can I 2 then say that Φ r B . Now, since B 0 the change in magnetic field 2d I occurs because of a change in the current, or B 0 . Finally, substituting 2d NΦ these into the equation E 0 gives: t Nr 2 0 I 0 Nr 2 I E0 2dt 2dt 4 10 T m/A 10.500 m 2.00 10 6 A 251 V 250.0 m 2.50 10 5 s (b) An example is the alternator in your car. If you were driving during a lightning storm and this large bolt of lightning hit at 50.0 m away, it is possible to fry the alternator of your battery because of this large voltage surge. In addition, the hair at the back of your neck would stand on end because it becomes statically charged. 7 2 23.3 MOTIONAL EMF 15. Use Faraday’s law, Lenz’s law, and RHR-1 to show that the magnetic force on the current in the moving rod in Figure 23.11 is in the opposite direction of its velocity. I Solution l R I v Because the flux through the loop (into the page) is increasing, a magnetic field out of the page is induced. This requires the CCW current in the loop. The force on the current in the rod, from the B field (into the page), is to the left. OpenStax College Physics Instructor Solutions Manual Chapter 23 16. If a current flows in the Satellite Tether shown in Figure 23.12, use Faraday’s law, Lenz’s law, and RHR-1 to show that there is a magnetic force on the tether in the direction opposite to its velocity. Solution The flux through the loop (into the page) is increasing because the loop is getting larger and enclosing more magnetic field. Thus, a magnetic field (out of the page) is induced to oppose the change in flux from the original field. Using RHR-2, point your fingers out of the page within the loop, then your thumb points in the counterclockwise direction around the loop, so the induced magnetic field is produced by the induction of a counterclockwise current in the circuit. Finally, using RHR-1, putting your right thumb in the direction of the current and your fingers into the page (in the direction of the magnetic field), your palm points to the left. This means the magnetic force on the wire is to the left (in the direction opposite to its velocity). 17. (a) A jet airplane with a 75.0 m wingspan is flying at 280 m/s. What emf is induced between wing tips if the vertical component of the Earth’s field is 3.00 10 5 T ? (b) Is an emf of this magnitude likely to have any consequences? Explain. Solution 18. Solution (a) E Blv 3.00 10 5 T 75.0 m280 m/s 0.630 V (b) No, this is a very small emf. (a) A nonferrous screwdriver is being used in a 2.00 T magnetic field. What maximum emf can be induced along its 12.0 cm length when it moves at 6.00 m/s? (b) Is it likely that this emf will have any consequences or even be noticed? (a) E Blv 2.00 T0.120 m6.00 m/s 1.44 V (b) No. 19. At what speed must the sliding rod in Figure 23.11 move to produce an emf of 1.00 V in a 1.50 T field, given the rod’s length is 30.0 cm? Solution E Blv v 20. The 12.0 cm long rod in Figure 23.11 moves at 4.00 m/s. What is the strength of the magnetic field if a 95.0 V emf is induced? Solution E Blv B 21. Prove that when B , , and v are not mutually perpendicular, motional emf is given by emf Bv sin . If v is perpendicular to B , then is the angle between and B . If is perpendicular to B , then is the angle between v and B . E 1.00 V 2.22 m/s Bl 1.50 T 0.300 m E 95.0 V 197.9 T 198 T lv 0.120 m 4.00 m/s OpenStax College Physics Solution Instructor Solutions Manual Chapter 23 Case i lB B B B v Side view of Figure 23.11 with angled field. Case ii vB v End view of Figure 23.11 with angled field. ΔΦ BA cos90 Δx in either case E B sin l Blv sin Δt t Δt 22. In the August 1992 space shuttle flight, only 250 m of the conducting tether considered in Example 23.2 could be let out. A 40.0 V motional emf was generated in the Earth’s 5.00 105 T field, while moving at 7.80 103 m/s . What was the angle between the shuttle’s velocity and the Earth’s field, assuming the conductor was perpendicular to the field? Solution 40.0 V E Blv sin sin 1 24.2 5 3 5.00 10 T 250 m 7.80 10 m/s 23. Integrated Concepts Derive an expression for the current in a system like that in Figure 23.11, under the following conditions. The resistance between the rails is R , the rails and the moving rod are identical in cross section A and have the same resistivity . The distance between the rails is l, and the rod moves at constant speed v perpendicular to the uniform field B . At time zero, the moving rod is next to the resistance R . Solution From R l A the resistance of the rails is 2 vt . A l . The total resistance, R tot , is then A 2vt l RA E BlvA ; E Blv I A Rtot 2vt l RA The resistance of the moving bar is Rtot OpenStax College Physics Instructor Solutions Manual Chapter 23 24. Integrated Concepts The Tethered Satellite in Figure 23.12 has a mass of 525 kg and is at the end of a 20.0 km long, 2.50 mm diameter cable with the tensile strength of steel. (a) How much does the cable stretch if a 100 N force is exerted to pull the satellite in? (Assume the satellite and shuttle are at the same altitude above the Earth.) (b) What is the effective force constant of the cable? (c) How much energy is stored in it when stretched by the 100 N force? Solution 2.00 10 4 m 100 N LA LF (a) F A 2.10 1011 N/m 2 π 1.25 10 3 m L F 100 N (b) F kL k 51.54 N/m L 1.94 m 1 2 (c) v kL2 0.551.54 N/m 1.94 m 97.0 J 2 25. Solution 2 1.94 m Integrated Concepts The Tethered Satellite discussed in this module is producing 5.00 kV, and a current of 10.0 A flows. (a) What magnetic drag force does this produce if the system is moving at 7.80 km/s? (b) How much kinetic energy is removed from the system in 1.00 h, neglecting any change in altitude or velocity during that time? (c) What is the change in velocity if the mass of the system is 100,000 kg? (d) Discuss the long term consequences (say, a week-long mission) on the space shuttle’s orbit, noting what effect a decrease in velocity has and assessing the magnitude of the effect. (a) F IlB 10.0 A2.00 10 4 m5.00 10 5 T 10.0 N (b) W Fx Fvt 10.0 N 7.80 10 3 m/s 3600 s 2.81 108 J (c) 1 mv' 2 1 mv 2 W ; 2 2 2W v ' v 2 m 12 2 2 2.81 10 8 J 7.79964 10 3 m/s 7.80 10 3 m/s 1.00 10 5 kg Δv v-v' 0.36 m/s 0; Note that five digits are needed to see the difference between v and v . (d) For a week-long mission (168 hours), the change in velocity will be 60 m/s, or approximately 1%. In general, a decrease in velocity would cause the orbit to start spiraling inward because the velocity would no longer be sufficient to keep the circular orbit. The long term consequences are that the shuttle would require a little more fuel to maintain the desired speed, otherwise the orbit would spiral slightly inward. 12 OpenStax College Physics Instructor Solutions Manual Chapter 23 23.4 EDDY CURRENTS AND MAGNETIC DAMPING 26. Make a drawing similar to Figure 23.14, but with the pendulum moving in the opposite direction. Then use Faraday’s law, Lenz’s law, and RHR-1 to show that magnetic force opposes motion. Solution Pivot v B in v F F As the plate enters from the right, flux increases, so an eddy current is set up (Faraday’s Law) in the counter-clockwise direction (Lenz’s Law). The force on the left side of the current loop is to the right. When the plate leaves the field the eddy current is now clockwise and again there is a force to the right on the current. 27. A coil is moved through a magnetic field as shown in Figure 23.59. The field is uniform inside the rectangle and zero outside. What is the direction of the induced current and what is the direction of the magnetic force on the coil at each position shown? Solution (a) The magnetic field is zero and not changing, so there is no current and therefore no force on the coil. (b) The magnetic field is increasing out of the page, so the induced magnetic field is into the page, created by an induced clockwise current. This current creates a force to the left. (c) The magnetic field is not changing, so there is no current and therefore no force on the coil. (d) The magnetic field is decreasing out of the page, so the induced magnetic field is out of the page, created by an induced counterclockwise current. This current creates a force to the right. (e) The magnetic field is zero and not changing, so there is no current and therefore no force on the coil. 23.5 ELECTRIC GENERATORS 28. Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field. OpenStax College Physics Solution Instructor Solutions Manual Chapter 23 2π rad 1 min 377 rad/s; 1 rev 60 s 3600 rev/min E0 NAB 200π 0.050 m 0.800 T 377 rad/s 474 V 2 29. At what angular velocity in rpm will the peak voltage of a generator be 480 V, if its 500-turn, 8.00 cm diameter coil rotates in a 0.250 T field? Solution E0 NAB 30. Solution E0 480 V 763.9 rad/s 7.30 103 rpm 2 2 NR B 500π 0.0400 m 0.250 T What is the peak emf generated by rotating a 1000-turn, 20.0 cm diameter coil in the Earth’s 5.00 105 T magnetic field, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 10.0 ms? π/2 rad 157.08 rad/s t 0.0100 s 2 E0 NAB 1000π 0.100 m 5.00 10 -5 T 157.08 rad/s 0.247 V 31. What is the peak emf generated by a 0.250 m radius, 500-turn coil that is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Solution Using the information given in Exercise 23.12: 1 1 rev 2 rad and t 4.17 10 3 s, 4 4 2 N 500; A r 2 0.250 m ; and B 0.425 T, 1 / 42 rad 376.7 rad/s. Therefore, we get: t 4.17 10 -3 s 2 E0 NAB 500 0.250 m 0.425 T376.7 rad/s 1.57 10 4 V 15.7 kV 32. (a) A bicycle generator rotates at 1875 rad/s, producing an 18.0 V peak emf. It has a 1.00 by 3.00 cm rectangular coil in a 0.640 T field. How many turns are in the coil? (b) Is this number of turns of wire practical for a 1.00 by 3.00 cm coil? Solution 4 2 (a) A 0.0100 m 0.0300 m 3.00 10 m E 18.0 V E0 NAB N 0 50.0 -4 2 ABω 3.00 10 m 0.640 T 1875 rad/s (b) Yes, 50 turns is practical for a coil of this size. OpenStax College Physics 33. Solution Instructor Solutions Manual Chapter 23 Integrated Concepts This problem refers to the bicycle generator considered in the previous problem. It is driven by a 1.60 cm diameter wheel that rolls on the outside rim of the bicycle tire. (a) What is the velocity of the bicycle if the generator’s angular velocity is 1875 rad/s? (b) What is the maximum emf of the generator when the bicycle moves at 10.0 m/s, noting that it was 18.0 V under the original conditions? (c) If the sophisticated generator can vary its own magnetic field, what field strength will it need at 5.00 m/s to produce a 9.00 V maximum emf? (a) v r 8.00 10 3 m1875 rad/s 15.0 m/s E' ' v' 10.0 m/s (b) E0 NAB 0 E '0 E0 18.0 V 12.0 V E0 v 15.0 m/s rE0 NABv 8.00 10 3 m 9.00 V (c) E0 NAB B 0.960 T r NAv 50 3.00 10 4 m 2 5.00 m/s 34. (a) A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 5.00 by 8.00 cm rectangular coil rotates in an adjustable magnetic field so that it can produce sufficient voltage even at low rpms. What is the field strength needed to produce a 24.0 V peak emf? (b) Discuss how this required field strength compares to those available in permanent and electromagnets. Solution 1 min 41.89 rad/s; (a) 400 rev/min 2 rad/rav 60 s E 24.0 V E0 NAB B 0 0.477 T NAω 3000.0500 m 0.0800 m 41.89 rad/s (b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet. 35. Show that if a coil rotates at an angular velocity , the period of its AC output is 2 / . Solution The output voltage of the coil will be positive for half of a revolution and negative for the other half. So one period corresponds to one revolution 1 2 2f f ;T 2 f 36. A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25 T field, starting with the plane of the coil parallel to the field. (a) What is the peak emf? (b) At what time is the peak emf first reached? (c) At what time is the emf first at its most negative? (d) What is the period of the AC voltage output? OpenStax College Physics Solution Instructor Solutions Manual Chapter 23 B B B t B E E0 sin and 90 t (a) E0 NAB 75 0.0500 m2 1.25 T8.00 rad/s 5.89 V (b) E E0 sin E0 sin 90 ωt E0 cos t. We know this is the correct expression for E since E must change sign when t , E E0 when t 0 peak emf is when t 0 (just after the 2 generator is turned on). π π 0.393 s (c) E E0 when ωt π t ω 8.00 rad/s 2 2π (d) T 0.785 s 8.00 rad/s 37. Solution 38. (a) If the emf of a coil rotating in a magnetic field is zero at t 0 , and increases to its first peak at t 0.100 ms , what is the angular velocity of the coil? (b) At what time will its next maximum occur? (c) What is the period of the output? (d) When is the output first one-fourth of its maximum? (e) When is it next one-fourth of its maximum? T 2 2π 0.100 ms T 0.400 ms , so that 1.57 10 4 rad/s 4 4 T 4.00 10 s (b) 0.100 ms T 0.500 ms (c) T 0.400 ms E 1 (d) E 0 E0 sin ωt sin ωt ωt 0.2527 rad 4 4 0.25272 rad t 1.6110 5 s 4 1.57 10 rad/s (e) t T 1.6110 5 s 4.00 10 4 s 4.16 10 4 s (a) Unreasonable Results A 500-turn coil with a 0.250 m 2 area is spun in the Earth’s 5.00 10 5 T field, producing a 12.0 kV maximum emf. (a) At what angular velocity must the coil be spun? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? OpenStax College Physics Solution Instructor Solutions Manual Chapter 23 (a) E0 NAB E 1.20 10 4 V 0 1.92 10 6 rad/s 1.83 10 7 rpm NAB 500 0.250 m 2 5.00 10 5 T (b) This angular velocity is unreasonably high, higher than can be obtained for any mechanical system. (c) The assumption that a voltage as great as 12.0 kV could be obtained is unreasonable. 23.6 BACK EMF 39. Suppose a motor connected to a 120 V source draws 10.0 A when it first starts. (a) What is its resistance? (b) What current does it draw at its normal operating speed when it develops a 100 V back emf? Solution (a) R 40. A motor operating on 240 V electricity has a 180 V back emf at operating speed and draws a 12.0 A current. (a) What is its resistance? (b) What current does it draw when it is first started? Solution (a) R 41. What is the back emf of a 120 V motor that draws 8.00 A at its normal speed and 20.0 A when first starting? Solution V 120 V 6.00 , so that I 20 A 120 V 8.00 A 6.00 48.0 V V 120 V 48.0 V 72.0 V 42. The motor in a toy car operates on 6.00 V, developing a 4.50 V back emf at normal speed. If it draws 3.00 A at normal speed, what current does it draw when starting? V 120 V 6.00 I 20.0 A V 120 V 100 V (b) I 3.33 A R 6.00 V 240 V 180 V 5.00 I 12.0 A V 240 V (b) I 48.0 A R 5.00 R OpenStax College Physics Instructor Solutions Manual Chapter 23 V 6.00 V 4.50 V 0.500 , so that I 3.00 A V 6.00 V I 12.0 A R 0.500 Solution R 43. Integrated Concepts The motor in a toy car is powered by four batteries in series, which produce a total emf of 6.00 V. The motor draws 3.00 A and develops a 4.50 V back emf at normal speed. Each battery has a 0.100 internal resistance. What is the resistance of the motor? Solution Since the resistors are in series, we know the total internal resistance of the batteries is R 40.100 . Therefore, E V E V 6.00 V 4.50 V I , so that R R R 40.100 Ω 0.100 Ω R R I 3.00 A 23.7 TRANSFORMERS 44. A plug-in transformer, like that in Figure 23.29, supplies 9.00 V to a video game system. (a) How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 400 turns? (b) What is its input current when its output is 1.30 A? Solution V 9.00 V (a) N s N p s 400 30.0 V 120 V p N 30 2 (b) I p I s s 1.30 A 9.75 10 A 400 Np 45. An American traveler in New Zealand carries a transformer to convert New Zealand’s standard 240 V to 120 V so that she can use some small appliances on her trip. (a) What is the ratio of turns in the primary and secondary coils of her transformer? (b) What is the ratio of input to output current? (c) How could a New Zealander traveling in the United States use this same transformer to power her 240 V appliances from 120 V? Solution (a) Np Ns Ip Vp Vs Np 240 V 2.00 120 V 0.500 Is Ns (c) By hooking it up backwards, that is, by using input as output and vice versa. (b) OpenStax College Physics Instructor Solutions Manual Chapter 23 46. A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 200 mA. (a) What is the current input? (b) What is the power input? (c) Is this amount of power reasonable for a small appliance? Solution I p N p Vp (a) I N V , so that s s s Vp 12.0 V 2 I p I s 0.200 A 2.00 10 A 20.0 mA 120 V Vs (b) Pin I pVp 0.200 A 120 V 2.40 W (c) Yes, this amount of power is quite reasonable for a small appliance. 47. (a) What is the voltage output of a transformer used for rechargeable flashlight batteries, if its primary has 500 turns, its secondary 4 turns, and the input voltage is 120 V? (b) What input current is required to produce a 4.00 A output? (c) What is the power input? Solution N 4 (a) Vs Vp s 120 V 0.96 V 500 Np N 4 2 (b) I p I s s 4.00 A 3.20 10 A 32.0 mA 500 Np (c) P IV 3.20 10 2 A120 V 3.84 W 48. (a) The plug-in transformer for a laptop computer puts out 7.50 V and can supply a maximum current of 2.00 A. What is the maximum input current if the input voltage is 240 V? Assume 100% efficiency. (b) If the actual efficiency is less than 100%, would the input current need to be greater or smaller? Explain. Solution V N s Vs 7.50 V I p I s s 2.00 A 0.0625 A 0.063 A I s N p Vp 240 V Vp (b) If the efficiency is less than 100%, the input current would need to be greater because more power would need to be input into the transformer to produce the required power output. (a) 49. Ip A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 240 V to a primary coil of 280 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages? (b) If the maximum input current is 5.00 A, what are the maximum output currents (each used alone)? OpenStax College Physics Instructor Solutions Manual Chapter 23 Solution V 5.60 V (a) N s N p s 280 6.53; 240 V Vp 280 12 V 14.0; 280 480 V 560 240 V 240 V Vp 240 V (b) I s I p 5.0 A 214 A 5.60 V Vs 5.0 A 240 V 100 A; 5.0 A 240 V 2.50 A 12 V 480 V 50. A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 335 kV. The secondary of this transformer is being replaced so that its output can be 750 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with the old secondary? (b) What is the ratio of new current output to old output (at 335 kV) for the same power? (c) If the upgraded transmission lines have the same resistance, what is the ratio of new line power loss to old? Solution (a) N s Vs N 's V 's N' V ' 750 kV ; s s 2.2 N p Vp N p Vp Ns Vs 335 kV (b) I 's Vp V 's Vs 335 kV 0.45 I s Vp Vs V 's 750 kV (c) Ploss I 2 R 51. P' I ' 2 0.20 P I2 If the power output in the previous problem is 1000 MW and line resistance is 2.00 , what were the old and new line losses? Solution (a) P 1.000 109 W 2985 A V 3.35 105 V 2 2985 A 2.00 Ω 1.78 10 7 W 17.8 MW Old : P IV I Ploss New : Uses 7.50 105 V I 2985 A, Ploss 3.56 MW 52. Unreasonable Results The 335 kV AC electricity from a power transmission line is fed into the primary coil of a transformer. The ratio of the number of turns in the secondary to the number in the primary is N s / N p 1000 . (a) What voltage is induced in the secondary? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? OpenStax College Physics Solution Instructor Solutions Manual Chapter 23 N (a) Vs Vp s 3.35 10 5 V 1000 335 MV Np (b) The result is way too high, well beyond the breakdown voltage of air over reasonable distances. (c) The input voltage is too high. 23.8 ELECTRICAL SAFETY: SYSTEMS AND DEVICES 54. Integrated Concepts A short circuit to the grounded metal case of an appliance occurs as shown in Figure 23.60. The person touching the case is wet and only has a 3.00 k resistance to earth/ground. (a) What is the voltage on the case if 5.00 mA flows through the person? (b) What is the current in the short circuit if the resistance of the earth/ground wire is 0.200 ? (c) Will this trigger the 20.0 A circuit breaker supplying the appliance? Solution (a) V IR 5.00 10 3 A 3000 Ω 15.0 V V 15.0 V 75.0 A (b) I R 0.200 Ω (c) Yes. 23.9 INDUCTANCE 55. Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is switched off in 1.00 ms, inducing a 9.00 V emf in the other. What is their mutual inductance? Solution Using the equation E2 M I1 , where the minus sign is an expression of Lenz’s t law, we can calculate the mutual inductance between the two coils: t 1.00 10 3 s M E2 9.00 V 1.80 mH I 1 5.00 A 56. Solution 57. If two coils placed next to one another have a mutual inductance of 5.00 mH, what voltage is induced in one when the 2.00 A current in the other is switched off in 30.0 ms? E2 M I1 5.00 10 3 H 2.00 A 0.333 V t 3.00 10 2 s The 4.00 A current through a 7.50 mH inductor is switched off in 8.33 ms. What is the emf induced opposing this? OpenStax College Physics Solution 58. Solution 59. Solution EL Instructor Solutions Manual Chapter 23 I 7.50 10 3 H 4.00 A 3.60 V t 8.33 10 3 s A device is turned on and 3.00 A flows through it 0.100 ms later. What is the selfinductance of the device if an induced 150 V emf opposes this? EL I t 150 V 1.00 10 4 s LE 5.00 mH t I 3.00 A Starting with emf 2 M E1 M I1 , show that the units of inductance are (V s)/A s . t t E1 s V /A I 2 M I 2 t from V IR V A Ω M s Ω 60. Solution Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time must the 0.100 A current through a 2.00 mH inductor be switched on or off to induce a 500 V emf? EL I LI 2.00 10 3 H 0.100 A t 4.00 10 -7 s t E 500 V 61. A large research solenoid has a self-inductance of 25.0 H. (a) What induced emf opposes shutting it off when 100 A of current through it is switched off in 80.0 ms? (b) How much energy is stored in the inductor at full current? (c) At what rate in watts must energy be dissipated to switch the current off in 80.0 ms? (d) In view of the answer to the last part, is it surprising that shutting it down this quickly is difficult? Solution (a) Using the equation E L I , we have t 100 A 3.125 10 4 V 31.3 kV I EL 25.0 H t 8.00 102 s 1 1 (b) Using Eind LI 2 25.0 H 100 A 2 1.25 10 5 J 2 2 E (c) Using the equation P ,we have t E 1.25 105 J P 1.563 10 6 W 1.56 MW -2 t 8.00 10 s (d) No, it is not surprising since this power is very high. OpenStax College Physics 62. Solution Instructor Solutions Manual Chapter 23 (a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid having 1000 loops. (b) How much energy is stored in this inductor when 20.0 A of current flows through it? (c) How fast can it be turned off if the induced emf cannot exceed 3.00 V? (a) L 0 N 2 A 0 N 2 r 2 4π 10 7 T m/A 1000 π0.0500 m 19.7 mH 0.500 m 2 2 l l 1 2 (b) Eind LI 2 0.5 1.974 10 2 H 20.0 A 3.95 J 2 I LI 1.974 10 2 H 20.0 A (c) E L t 0.132 s t 0.132 s t E 3.00 V 63. Solution A precision laboratory resistor is made of a coil of wire 1.50 cm in diameter and 4.00 cm long, and it has 500 turns. (a) What is its self-inductance? (b) What average emf is induced if the 12.0 A current through it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz AC)? (c) What is its inductance if it is shortened to half its length and counter-wound (two layers of 250 turns in opposite directions)? (a) L 0 N 2 A ; A r 2 0.00750 m 1.767 10 4 m 2 2 l 2 4π 10 7 T m/A 500 1.77 10 4 m 2 L 1.39 mH 0.0400 m I 1.388 10 -3 H 12.0 A (b) E L 3.33 V t 5.00 10 -3 s (c) L 0 . Two 250 turn coils in opposite directions will cancel each other. 64. Solution The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What is their total self-inductance assuming they act like a single solenoid? (b) How much energy is stored in them when 6.00 A flows? (c) What average emf opposes shutting them off if this is done in 5.00 ms (one-fourth of a cycle for 50 Hz AC)? (a) L 0 N 2 A 0 N 2 r 2 l l 4π 10 7 T m/A 400 π 4.00 10 3 m 1.00 m 2 2 1.0110 5 H 1 2 2 LI 0.5 1.01 10 5 H 6.0 A 1.82 10 4 J 2 I 1.01 10 5 H 6.0 A (c) E L 12.12 mV t 5.00 10 3 s (b) Eind OpenStax College Physics 65. Solution Instructor Solutions Manual Chapter 23 When the 20.0 A current through an inductor is turned off in 1.50 ms, an 800 V emf is induced, opposing the change. What is the value of the self-inductance? EL I Et 800 V 1.50 10 3 s L 60.0 mH t I 20.0 A 66. How fast can the 150 A current through a 0.250 H inductor be shut off if the induced emf cannot exceed 75.0 V? Solution EL 67. Integrated Concepts A very large, superconducting solenoid such as one used in MRI scans, stores 1.00 MJ of energy in its magnetic field when 100 A flows. (a) Find its selfinductance. (b) If the coils “go normal,” they gain resistance and start to dissipate thermal energy. What temperature increase is produced if all the stored energy goes into heating the 1000 kg magnet, given its average specific heat is 200 J/kg C ? Solution 2 Eind 2 1.00 10 6 J 1 2 200 H (a) Eind LI L 2 2 I 100 A 2 E 1.00 10 6 J (b) E mcT T 5.00 C mc 1000 kg 200 J/kg C 68. Unreasonable Results A 25.0 H inductor has 100 A of current turned off in 1.00 ms. (a) What voltage is induced to oppose this? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? Solution (a) E L I LI 0.250 H 150 A t 0.500 s t 0.500 s t E 75.0 V 100 A 2.50 10 6 V I 25.0 H t 1.00 10 3 s (b) The voltage is so extremely high that arcing would occur and the current would not be reduced so rapidly. (c) It is not reasonable to shut off such a large current in such a large inductor in such an extremely short time. 23.10 RL CIRCUITS 69. If you want a characteristic RL time constant of 1.00 s, and you have a 500 resistor, what value of self-inductance is needed? Solution Using the equation L , we know L R 1.00 s500 Ω 500 H. R OpenStax College Physics Instructor Solutions Manual Chapter 23 70. Your RL circuit has a characteristic time constant of 20.0 ns, and a resistance of 5.00 M . (a) What is the inductance of the circuit? (b) What resistance would give you a 1.00 ns time constant, perhaps needed for quick response in an oscilloscope? Solution (a) 71. A large superconducting magnet, used for magnetic resonance imaging, has a 50.0 H inductance. If you want current through it to be adjustable with a 1.00 s characteristic time constant, what is the minimum resistance of system? Solution 72. Verify that after a time of 10.0 ms, the current for the situation considered in Example 23.9 will be 0.183 A as stated. Solution I I 0e 73. Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and resistors ranging from 100.0 to 1000.0 M . What is the range of characteristic RL time constants you can produce by connecting a single resistor to a single inductor? Solution L L R 5.00 10 6 Ω 2.00 10 8 s 0.100 H R L 0.100 H (b) R 1.00 10 8 Ω 100 MΩ 9 1.00 10 s L L 50.0 H R 50.0 Ω R 1.00 s max t 3 3 10.0 Ae 1.0010 s 2.5010 s 0.183 A Lmax L 10.0 H 1.00 10 -9 H 100 s; min min 1.00 10 -15 s 6 Rmin 0.100 Rmax 1.00 10 74. (a) What is the characteristic time constant of a 25.0 mH inductor that has a resistance of 4.00 ? (b) If it is connected to a 12.0 V battery, what is the current after 12.5 ms? Solution (a) L 2.50 10 -2 H 6.25 10 -3 s R 4.00 V 12.0 V (b) I 0 3.00 A, so that R 4.00 Ω I I 0 1 e 75. t 3.00 A 1 e 1.2510 2 s 6.25103 s 2.59 A What percentage of the final current I 0 flows through an inductor L in series with a resistor R , three time constants after the circuit is completed? OpenStax College Physics Instructor Solutions Manual Chapter 23 Solution t I 1 e 1 e 3 0.9502 I0 The current is 95.0% of the final current after 3 time constants. 76. The 5.00 A current through a 1.50 H inductor is dissipated by a 2.00 resistor in a circuit like that in Figure 23.44 with the switch in position 2. (a) What is the initial energy in the inductor? (b) How long will it take the current to decline to 5.00% of its initial value? (c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor. Solution (a) Eind 1 2 2 LI 0.51.50 H 5.00 A 18.75 J 18.8 J 2 t L 1.50 H I I 0 e ; 0.750 s R 2.00 t I I t e ln I0 I0 (b) I or t ln 0.750 s ln 0.0500 2.247 s 2.25 s I0 (c) E 1 LI 2 0.5150 H 0.05005.00 A 2 4.69 J f f 2 E E f 18.75 J 4.69 J Pave i 6.25 W t 2.25 s P 2 Pi I 02 R 5.00 A 2.00 Ω 50.0 W i 8.00 Pave 77. Solution (a) Use the exact exponential treatment to find how much time is required to bring the current through an 80.0 mH inductor in series with a 15.0 resistor to 99.0% of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of . (c) Discuss how significant the difference is. L 8.00 10 -2 H 5.333 10 -3 s, and R 15.0 t t I I I 0 1 e 1 e I0 (a) e t I 1 I0 I t ln 1 I0 5.333 10 -3 s ln 1 0.990 2.4561 10 -2 s 24.6 ms OpenStax College Physics Instructor Solutions Manual Chapter 23 (b) Using the result from part (a), t 4.61, t 5 55.33 10 3 s 2.667 10 2 s 26.7 ms 26.7 ms (c) 1.085 9% difference 24.6 ms So, using integer values is off by about 9% which is greater than the inherent uncertainty in the given parameters. 78. (a) Using the exact exponential treatment, find the time required for the current through a 2.00 H inductor in series with a 0.500 resistor to be reduced to 0.100% of its original value. (b) Compare your answer to the approximate treatment using integral numbers of . (c) Discuss how significant the difference is. Solution L 2.00 H (a) R 0.500 4.00 s; t I I I 0 e t ln 4.00 s ln 1.00 10 -3 27.6 s I0 t (b) Using the result from part (a), 6.91, t 7 74.00 s 28.00 s 28.0 ms 28.00 s (c) 1.013 1% difference . So, using integer values is off by about 1% which 27.63 s is larger than the inherent uncertainty in the given parameters. 23.11 REACTANCE, INDUCTIVE AND CAPACITIVE 79. At what frequency will a 30.0 mH inductor have a reactance of 100 ? Solution X L 2fL f 80. What value of inductance should be used if a 20.0 k reactance is needed at a frequency of 500 Hz? Solution X L 2 fL L 81. What capacitance should be used to produce a 2.00 M reactance at 60.0 Hz? Solution C 82. At what frequency will an 80.0 mF capacitor have a reactance of 0.250 ? XL 100 Ω 531 Hz 2L 2π 3.00 10 2 H XL 2.00 10 4 Ω 6.37 H 2f 2π 500 Hz 1 1 1.326 10 9 F 1.33 nF 6 2fX C 2π 60.0 Hz 2.00 10 Ω OpenStax College Physics Instructor Solutions Manual Chapter 23 Solution XC 83. (a) Find the current through a 0.500 H inductor connected to a 60.0 Hz, 480 V AC source. (b) What would the current be at 100 kHz? Solution 1 1 1 f 7.96 Hz 2fC 2CX C 2π 8.00 10 -2 F 0.250 2 (a) X L 2fL 2π60.0 Hz 0.500 H 1.885 10 Ω, so that V 480 V I 2.55 A X L 1.885 10 2 Ω 5 5 (b) X L 2π 1.00 10 Hz 0.500 H 3.142 10 Ω, so that V 480 V I 1.53 mA X L 3.142 10 5 Ω 84. (a) What current flows when a 60.0 Hz, 480 V AC source is connected to a 0.250 F capacitor? (b) What would the current be at 25.0 kHz? Solution (a) X C 1 1 1.061 10 4 , so that -7 2fC 2π 60.0 Hz 2.50 10 F V 480 V I 45.2 mA X C 1.061 10 4 Ω, 1 2.546 101 , so that (b) X C 4 -7 2π 2.50 10 Hz 2.50 10 F V 480 V I 18.8 A X C 2.546 101 85. A 20.0 kHz, 16.0 V source connected to an inductor produces a 2.00 A current. What is the inductance? Solution X L 2fL; I L 86. Solution V V X L 2 fL V 16.0 V 63.7 μH 2fL 2π 2.00 10 4 Hz 2.00 Hz A 20.0 Hz, 16.0 V source produces a 2.00 mA current when connected to a capacitor. What is the capacitance? XC 1 V I 2.00 10 -3 A ;I 2VfC C 0.995 μF 2fC XC 2Vf 2π 16.0 V 20.0 Hz OpenStax College Physics Instructor Solutions Manual Chapter 23 87. (a) An inductor designed to filter high-frequency noise from power supplied to a personal computer is placed in series with the computer. What minimum inductance should it have to produce a 2.00 k reactance for 15.0 kHz noise? (b) What is its reactance at 60.0 Hz? Solution (a) Using the equation X L 2fL , X 2.00 103 Ω 2.122 102 H 21.2 mH X L 2fL, or L L 2f 2π 1.50 10 4 Hz (b) Again using X L 2fL , X L 2fL 2π60.0 Hz 2.122 10-2 H 8.00 Ω 88. The capacitor in Figure 23.55(a) is designed to filter low-frequency signals, impeding their transmission between circuits. (a) What capacitance is needed to produce a 100 k reactance at a frequency of 120 Hz? (b) What would its reactance be at 1.00 MHz? (c) Discuss the implications of your answers to (a) and (b). Solution (a) X C 1 2fC 1 1 C 1.326 10 -8 F 13.3 nF 5 2fX C 2π 120 Hz 1.00 10 1 1 (b) X C 12.0 6 2fC 2π 1.00 10 Hz 1.326 10 -8 F (c) This implies that the capacitor impedes low frequencies significantly more than high frequencies, and therefore does what it is designed to do. 89. The capacitor in Figure 23.55(b) will filter high-frequency signals by shorting them to earth/ground. (a) What capacitance is needed to produce a reactance of 10.0 m for a 5.00 kHz signal? (b) What would its reactance be at 3.00 Hz? (c) Discuss the implications of your answers to (a) and (b). Solution (a) X C 1 1 1 C 3.18 mF 3 2fC 2fX C 2π 5.00 10 Hz 1.00 10 2 1 1 (b) X C 16.7 2fC 2π 3.00 Hz 3.18 10 -3 F (c) The reactance from the capacitor at high frequencies is much smaller than that at lower frequencies, so the capacitor does indeed allow the high frequencies to short to ground much better than the low frequencies, therefore working as it is designed to. OpenStax College Physics Instructor Solutions Manual Chapter 23 90. Unreasonable Results In a recording of voltages due to brain activity (an EEG), a 10.0 mV signal with a 0.500 Hz frequency is applied to a capacitor, producing a current of 100 mA. Resistance is negligible. (a) What is the capacitance? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? Solution (a) X C 1 V I 0.100 A ;I 2fCV C 3.18 F 2fC XC 2fV 2π0.500 Hz 1.00 10 2 V (b) That is an enormous capacitance. A capacitor that size would fill a large room. (c) It is unreasonable to assume such a small voltage could achieve such a current at the low frequency given. Capacitive resistance is very large at low frequencies unless the capacitance is very large. 23.12 RLC SERIES AC CIRCUITS 92. Solution An RL circuit consists of a 40.0 resistor and a 3.00 mH inductor. (a) Find its impedance Z at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those found in Example 23.12, in which there was also a capacitor. (a) X L 2fL 2π60.0 Hz 3.00 10 -3 H 1.131 Ω; X C 0 since there is no capacitor. Z R2 X L X C 40.0 2 12 2 1.131` 40.02 Ω (4 sig fig. to show difference ) 2 12 X ' L 2f ' L 2π 1.00 10 4 Hz 3.00 10 -3 H 188.5 Ω, so Z ' 40.0 Ω 188.5 Ω 193 Ω (b) At 60 Hz, with a capacitor, Z 531 , over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z 190 , about the same as without the capacitor. The capacitor has a smaller effect at high frequencies. 93. 2 2 12 An RC circuit consists of a 40.0 resistor and a 5.00 F capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those found in Example 23.12, in which there was also an inductor. OpenStax College Physics Solution (a) Instructor Solutions Manual Chapter 23 1 1 530.52 ; 2fC 2π 60.0 Hz 5.00 10 -6 F X L 0 since there is no inductor. At 60 Hz : X C 40.0 Ω Z R 2 X L X C At 10 kHz : X 'C 2 12 530.52`Ω 2 12 2 532 Ω 1 1 3.183 ; 4 2f ' C 2π 1.00 10 Hz 5.00 10 -6 F Z 40.0 3.183 40.1 Ω (b) At 60 Hz Z is about the same with or without the inductor because the capacitor dominates at low frequency. At 10.0 kHz Z is much less without the inductor because the inductor dominates at high frequencies. 94. Solution 2 12 2 An LC circuit consists of a 3.00 H inductor and a 5.00 F capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those found in Example 23.12, in which there was also a resistor. (a) At 60 Hz : X L 2fL 2π60.0 Hz 3.00 10 -3 H 1.131 Ω; XC 1 1 530.52 ; 2fC 2π 60.0 Hz 5.00 10 -6 F 1.131 530.52` Z R 2 X L X C 2 12 2 12 529 Ω At 10 kHz : X ' L 2f ' L 2π 1.00 10 4 Hz 3.00 10 -3 H 188.5 Ω; X 'C 1 1 3.183 ; 4 2f ' C 2π 1.00 10 Hz 5.00 10 -6 F Z 188.5 3.183 185 Ω (b) These values are close to those obtained in Example 23.12 because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance. 2 12 95. What is the resonant frequency of a 0.500 mH inductor connected to a 40.0 F capacitor? Solution Using f 0 96. 1 2 LC 2 5.00 10 1 -4 H 4.00 10 F -5 1.125 10 3 Hz 1.13 kHz To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00 H inductor connected to a variable capacitor. What range of capacitance is needed? OpenStax College Physics Solution f0 Cmin Cmax 97. Solution Solution 1 2 LC C 1 ; 4 Lf 02 2 1 -6 -6 4 1.00 10 H 1.650 10 Hz 1 2 6 4 1.00 10 H 5.00 10 Hz 2 5 2 2 9.304 10 -9 F 9.30 nF 1.013 10 -7 F 101.3 nF 9.30 nF 101 nF 1 1 -9 2 Lmin C min 2 1.00 10 H 1.00 10 -12 F 1 0.159 Hz 12 2 10.0 H 0.100 F f 0 max 1 f0 f 02 Solution 12 5.03 GHz What capacitance do you need to produce a resonant frequency of 1.00 GHz, when using an 8.00 nH inductor? 1 4 2 LC 1 2 LC 1 C 2 4 Lf 02 4 2 8.00 10 -9 H 1.00 10 9 Hz 99. Chapter 23 Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and capacitors ranging from 1.00 pF to 0.100 F. What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor? f 0 min 98. Instructor Solutions Manual 3.17 10 -12 F 3.17 pF 2 What inductance do you need to produce a resonant frequency of 60.0 Hz, when using a 2.00 F capacitor? f0 1 2 LC L 1 1 3.52 H 2 2 2 4 Cf 0 4 2.00 10 -6 F 60.0 Hz 2 100. The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency? Solution (a) f 0 (b) C 1 2 LC L 1 1 2 4 Cf 0 4 2 2.50 10 -12 F 8.80 10 7 Hz 2 1 1 2 4 Lf 0 4 2 1.308 10 -6 H 1.08 108 Hz 2 2 1.66 pF 2 1.31 μH OpenStax College Physics 101. Solution Instructor Solutions Manual Chapter 23 An RLC series circuit has a 2.50 resistor, a 100 H inductor, and an 80.0 F capacitor. (a) Find the circuit’s impedance at 120 Hz. (b) Find the circuit’s impedance at 5.00 kHz. (c) If the voltage source has Vrms 5.60 V , what is I rms at each frequency? (d) What is the resonant frequency of the circuit? (e) What is I rms at resonance? (a) X L 2fL 2π120 Hz 1.00 10 -4 H 7.540 10 -2 Ω 1 1 XC 16.58 2 f C 2π 120 Hz 8.00 10 -5 F Z R 2 X L X C 2 2.50 Ω 2 7.54 10 2 Ω 16.58 Ω 2 (b) X L 2π5.00 10 3 Hz 1.00 10 -4 H 3.142 Ω 1 XC 3.979 10 -1 3 -4 2π 5.00 10 Hz 8.00 10 F Z 2.50 Ω 2 3.142 Ω 3.979 10 1 Ω 2 16.7 Ω 3.71 Ω Vrms 5.60 V 0.336 A Z 16.69 Ω 5.60 V and for f 5.00 kHz , I rms 1.51 A 3.712 Ω (c) For f 120 Hz , I rms (d) f 0 1 2 LC 1 2 1.00 10 H 8.00 10 F -7 -5 5.63 10 4 Hz 56.3 kHz (e) At resonance, X L X R , so that Z R thus: I rms 102. Vrms 5.60 V 2.24 A R 2.50 Ω An RLC series circuit has a 1.00 k resistor, a 150 H inductor, and a 25.0 nF capacitor. (a) Find the circuit’s impedance at 500 Hz. (b) Find the circuit’s impedance at 7.50 kHz. (c) If the voltage source has Vrms 408 V , what is I rms at each frequency? (d) What is the resonant frequency of the circuit? (e) What is I rms at resonance? OpenStax College Physics Solution Instructor Solutions Manual Chapter 23 -4 (a) X L 2fL 2π500 Hz 1.50 10 H 0.471 Ω 1 1 XC 1.273 10 4 -8 2fC 2π 500 Hz 2.50 10 F Z R 2 X L X C 1000 Ω 2 12 2 0.4714 Ω 12,730 Ω 2 12 (b) X L 2π7500 Hz 1.50 10 H 7.069 Ω 1 XC 848.8 2π 7500 Hz 2.50 10 -8 F 12.8 kΩ -4 Z 1000 Ω 7069.2 Ω 848.8 Ω 6.30 kΩ 408 V V 408 V (c) (i) I rms rms 64.8 mA 31.9 mA ; (ii) I rms 6300 Ω Z 12,800 Ω 1 1 (d) f 0 82.2 kHz 12 -4 2 LC 2 1.50 10 H 2.50 10 8 F (e) I rms 2 2 12 Vrms 408 V 0.408 A R 1000 Ω 103. An RLC series circuit has a 2.50 resistor, a 100 H inductor, and an 80.0 F capacitor. (a) Find the power factor at f 120 Hz . (b) What is the phase angle at 120 Hz? (c) What is the average power at 120 Hz? (d) Find the average power at the circuit’s resonant frequency. Solution (a) From Exercise 23.101, we know Z : cos R 2.50 0.150 Z 16.69 R (b) cos 1 cos 1 0.1498 81.4 Z (c) Pave I rms Vrms cos 3.355 10 1 A 5.60 V 0.1498 0.281 W (d) Pave 104. 2 Vrms 5.60 V 2 12.5 W R 2.50 Ω An RLC series circuit has a 1.00 k resistor, a 150 H inductor, and a 25.0 nF capacitor. (a) Find the power factor at f 7.50 kHz . (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit’s resonant frequency. OpenStax College Physics Solution Instructor Solutions Manual (a) From Exercise 23.102, we know Z : cos Chapter 23 R 1000 0.159 Z 6300 R (b) cos 1 cos 1 0.1587 80.9 Z (c) Pave I rms Vrms cos 6.48 10 2 A 408 V 26.4 W 2 2 Vrms 408 V 166.46 W 166 W R 1000 Ω (d) Pave 105. An RLC series circuit has a 200 resistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0 . (a) What is the impedance? (b) Find the circuit’s capacitance. (c) If Vrms 408 V is applied, what is the average power supplied? Solution (a) cos R R 200 Z 283 Ω Z cos cos 45 -2 3 (b) X L 2fL 2π8000 Hz 2.50 10 H 1.26 10 Ω Z R2 X L X C 2 12 so that X L X C Z 2 R 2 2 X C X L Z 2 R2 12 X C 1.2566 10 3 Ω 282.84 Ω 200 Ω XC Solution 1.0566 10 3 Ω 1 1 1 C 18.8 nF 2fC 2fX C 2π 8000 Hz 1.0566 10 3 (c) Pave I rms Vrms 106. 2 12 2 2 Vrms 408 V 2 589 W Z 282.84 Ω Referring to Example 23.14, find the average power at 10.0 kHz. X L 2fL 2π 1.00 10 4 Hz 3.00 10 -3 H 1.89 10 2 Ω XC 1 1 3.18 4 2fC 2π 1.00 10 Hz 5.00 10 -6 F Z R2 X L X C 40.0 Ω 2 12 2 188.5 Ω 3.183 Ω 2 12 189.6 Ω, so that Vrms 120 V 0.6330 A and Z 189.6 Ω 40.0 Ω 16.0 W R I rms Vrms cos I rms Vrms 0.633 A 120 V 189.6 Ω Z I rms Pave This file is copyright 2016, Rice University. All Rights Reserved.