solutions_chapter21

T 5 0.313 s t 5 l r 5 2.50 h 8.00 v 5 0.313 s i 5 0 v

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Unformatted text preview: t 5 0.313 s, t 5 L R 5 2.50 H 8.00 V 5 0.313 s. i 5 0, v L 5 L D i D t 5 1 2.50 H 21 2.40 A / s 2 5 6.00 V. D i D t 5 2.40 A / s t 5 0, D i D t 5 E L 5 6.00 V 2.50 H 5 2.40 A / s. i 5 t 5 0, t S ` . i 5 i max 5 E R i 5 0. t 5 0, i 5 E R 1 1 2 e 2 1 R / L 2 t 2 . L D i D t 5 v L , E 2 iR 2 L D i D t 5 0. U max 5 1 2 LI 2 5 1 2 1 11.0 3 10 2 3 H 21 0.0800 A 2 2 5 3.52 3 10 2 5 J 5 35.2 m J i 5 1 0.0800 A 21 1 2 e 2 73.3 m s / 73.3 m s 2 5 0.0506 A I 5 E R 5 12.0 V 150.0 V 5 0.0800 A. t S ` t 5 L R 5 11.0 3 10 2 3 H 150.0 V 5 7.33 3 10 2 5 s 5 73.3 m s U 5 1 2 LI 2 . i 5 E R 1 1 2 e 2 t / t 2 . t 5 L / R . P 5 I 2 R 5 1 0.300 A 2 2 1 180 V 2 5 16.2 W 5 16.2 J / s U 5 1 2 LI 2 5 1 2 1 12.0 H 21 0.300 A 2 2 5 0.540 J P 5 I 2 R . U 5 1 2 LI 2 . 21-12 Chapter 21 21.56. Set Up: Problem 21.55 shows that initially is zero and rises to 25.0 V, that starts at 25.0 V and decreases to zero, and that i starts at zero and rises to 1.67 A. so is proportional to i. A reads the current i. Solve: The graphs are sketched in Figure 21.56. Figure 21.56 21.57. Set Up: Current decay in an R- L circuit is described by Eq. (21.28). Solve: and and Reflect: As R is decreased the time constant increases and it takes longer for the current to decay. For this circuit the time constant is After one time constant the current has decayed to about 37% of its original value. It is reasonable for it to take a little over two time constants for the current to decay to 10% of its original value. 21.58. Set Up: The current as a function of time is given by The energy stored in the inductor is U reaches its maximum value when i is times its maximum value. Solve: (a) The maximum current is gives and (b) and 21.59. Set Up: The resonant angular frequency of an L- C circuit is Solve: 21.60. Set Up: The energy stored in a capacitor is The energy stored in an inductor is Energy conservation requires that the total stored energy be constant. The current is a maximum when the charge on the capacitor is zero. Solve: (a) Initially and and The total energy stored is (b) The current is maximum when and so and i max 5 Å 2 1 6.40 3 10 2 4 J 2 3.75 3 10 2 3 H 5 0.584 A. 1 2 Li max 2 5 6.40 3 10 2 4 J U L 5 6.40 3 10 2 4 J. U C 1 U L 5 6.40 3 10 2 4 J U C 5 0. q 5 0.640 mJ. U C 5 1 2 C v 2 5 1 2 1 5.00 3 10 2 6 F 21 16.0 V 2 2 5 6.40 3 10 2 4 J. U L 5 i 5 0. v 5 16.0 V U L 5 1 2 Li 2 . U C 5 1 2 C v 2 . v 5 1 "1 5.25 3 10 2 3 H 21 12.0 3 10 2 6 F 2 5 3980 rad / s v 5 1 " LC . t 5 2 L ln 1 1 2 1 / " 2 2 R 5 2 1 1.50 3 10 2 3 H 2 ln 1 1 2 1 / " 2 2 0.750 3 10 3 V 5 2.46 m s. 1 " 2 5 1 2 e 2 1 R / L 2 t t 5 2 L ln 1 1 2 2 R 5 2 1 1.50 3 10 2 3 H 2 ln 1 1 2 2 0.750 3 10 3 V 5 1.39 m s. 2 1 R L 2 t 5 ln 1 1 2 2 . e 2 1 R / L 2 t 5 1 2 ....
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t 5 0.313 s t 5 L R 5 2.50 H 8.00 V 5 0.313 s i 5 0 v L 5 L...

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