The 5.15 A current through a 1.50 H inductor is dissipated by a 2.15 Q resistor in a circuit like that in the figure below with the switch in position 2. 0.632/ C A L (a) 0.368/ 0+ 0 = L/R 2T 3r 4 (b) (a) What is the initial energy (in J) in the inductor? 0 t = L/R 2t (c) Эт 4t 19.89 ] (b) How long will it take (in s) the current to decline to 5.00% of its initial value? 2.09 S (c) Calculate the average power (in W) dissipated, and compare it with the initial power dissipated by the resistor. 28.5 1.96 x W X (ratio of initial power to average power)
The 5.15 A current through a 1.50 H inductor is dissipated by a 2.15 Q resistor in a circuit like that in the figure below with the switch in position 2. 0.632/ C A L (a) 0.368/ 0+ 0 = L/R 2T 3r 4 (b) (a) What is the initial energy (in J) in the inductor? 0 t = L/R 2t (c) Эт 4t 19.89 ] (b) How long will it take (in s) the current to decline to 5.00% of its initial value? 2.09 S (c) Calculate the average power (in W) dissipated, and compare it with the initial power dissipated by the resistor. 28.5 1.96 x W X (ratio of initial power to average power)
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter23: Faraday’s Law And Inductance
Section: Chapter Questions
Problem 8OQ
Related questions
Question
![The 5.15 A current through a 1.50 H inductor is dissipated by a 2.15 Q resistor in a circuit like that in the figure below with the switch in position 2.
0.632/
C A L
(a)
0.368/
0+
0 = L/R 2T
3r 4
(b)
(a) What is the initial energy (in J) in the inductor?
0 t = L/R 2t
(c)
Эт 4t
19.89
]
(b) How long will it take (in s) the current to decline to 5.00% of its initial value?
2.09
S
(c) Calculate the average power (in W) dissipated, and compare it with the initial power dissipated by the resistor.
28.5
1.96
x W
X (ratio of initial power to average power)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49444d66-96b7-45b8-992f-0f6c51b0e4d0%2Fa85a87eb-4b76-4633-8bd3-d30d3fa28bc5%2Fpr1zmea_processed.png&w=3840&q=75)
Transcribed Image Text:The 5.15 A current through a 1.50 H inductor is dissipated by a 2.15 Q resistor in a circuit like that in the figure below with the switch in position 2.
0.632/
C A L
(a)
0.368/
0+
0 = L/R 2T
3r 4
(b)
(a) What is the initial energy (in J) in the inductor?
0 t = L/R 2t
(c)
Эт 4t
19.89
]
(b) How long will it take (in s) the current to decline to 5.00% of its initial value?
2.09
S
(c) Calculate the average power (in W) dissipated, and compare it with the initial power dissipated by the resistor.
28.5
1.96
x W
X (ratio of initial power to average power)
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