solutions_chapter21

1 50.0 v 2 1 n 1 n 2 2 2 5 1 50.0 v 21 10 2 2 5 5000

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Unformatted text preview: 1 50.0 V 2 1 N 1 N 2 2 2 5 1 50.0 V 21 10 2 2 5 5000 V . R 5 V 2 P 5 1 120 V 2 2 2.88 W 5 5000 V . P 5 V 2 R P 5 VI 5 1 12.0 V 21 0.240 A 2 5 2.88 W I 2 5 V 2 R 5 12.0 V 50.0 V 5 0.240 A N 1 N 2 5 V 1 V 2 5 120 V 12.0 V 5 10 P 5 VI 5 V 2 R . V 2 V 1 5 N 2 N 1 . I 2 V 2 5 1 13.3 A 21 120 V 2 5 1600 W. R eff 5 9.00 V 1 1 / 2 2 2 5 36.0 V . R 5 V 2 P 5 1 120 V 2 2 1600 W 5 9.00 V . I 5 P V 5 1600 W 240 V 5 6.67 A. P 5 VI , N 2 / N 1 5 1 2 . V 2 5 120 V, V 1 5 240 V V 1 I 1 5 V 2 I 2 . P 1 5 P 2 P 5 V 2 R . R eff 5 R 1 N 2 / N 1 2 2 . V 2 V 1 5 N 2 N 1 . R eff 5 R 1 N 2 / N 1 2 2 5 125 V 1 834 / 275 2 2 5 13.6 V V 2 5 V 1 1 N 2 / N 1 2 5 1 25.0 V 21 834 / 275 2 5 75.8 V V 1 5 25.0 V. N 2 5 275 R eff 5 R 1 N 2 / N 1 2 2 . V 2 V 1 5 N 2 N 1 . L 5 1 4 p 3 10 2 7 T # m / A 2 p 1 7.50 3 10 2 4 m 2 2 1 50 2 2 5.00 3 10 2 2 m 5 1.11 3 10 2 7 H 5 0.111 m H L 5 m N 2 A l . L 5 1 4 p 3 10 2 7 Wb / m # A 21 1800 2 2 1 0.480 3 10 2 4 m 2 2 2 p 1 0.120 m 2 5 2.59 3 10 2 4 H 5 0.259 mH L 5 m N 2 A 2 p r . 21-10 Chapter 21 21.44. Set Up: and Solve: so 21.45. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is The energy stored in an inductor is Solve: and Reflect: The stored energy is proportional to the square of the number of turns even though the magnetic field within the solenoid is directly proportional to the number of turns. We will see in Section 21.10 that the magnetic field energy depends on the square of the magnetic field. 21.46. Set Up: Solve: (a) (b) This is not a reasonable current in a circuit; it is too large. 21.47. Set Up: The self-inductance of a solenoid is found in Problem 21.37 to be The length l of the solenoid is the number of turns divided by the turns per unit length. Solve: (a) (b) If is the number of turns per unit length, then and For this coil This is not a practical length for laboratory use. Reflect: The number of turns is The length of wire in the solenoid is the circumference C of one turn times the number of turns. The length of wire is This length of wire will have a large resist- ance and electrical energy loses will be very large. 21.48. Set Up: The energy density is The energy U is u (volume). Solve: 21.49. Set Up: Solve: (a) (b) 21.50. Set Up: and Solve: I 2 5 I 1 1 U 2 U 1 2 1 / 2 5 I 1 9.0 mJ 3.0 mJ 2 1 / 2 5 I " 3 . U 1 I 1 2 5 U 2 I 2 2 . U I 2 5 1 2 L 5 constant U 5 1 2 LI 2 . L 5 2 U I 2 5 2 1 1.73 3 10 7 J 2 1 80.0 A 2 2 5 5.41 3 10 3 H Energy 5 1 200 W 21 24 h 21 3600 s / h 2 5 1.73 3 10 7 J U 5 1 2 LI 2 . Energy 5 Pt . U 5 B 2 2 m 1 volume 2 5 1 0.560 T 2 2 2 1 4 p 3 10 2 7 T # m / A 2 1 0.0290 m 3 2 5 3.62 3 10 3 J u 5 B 2 2 m . I 2 R 1 0.126 m 21 5.63 3 10 5 2 5 7.1 3 10 4 m 5 71 km. C 5 p d 5 p 1 4.00 3 10 2 2 m 2 5 0.126 m....
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1 50.0 V 2 1 N 1 N 2 2 2 5 1 50.0 V 21 10 2 2 5 5000 V R 5...

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