(a) Find the energy stored in the inductor when the current reaches its ma 115.38 x Apply the expression for the energy stored in an inductor in terms of its in

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
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Chapter33: Inductors And Ac Circuits
Section: Chapter Questions
Problem 11PQ
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Question
A 24-V battery is connected in series with a resistor and an inductor, with R = 5.2 2 and L = 5.0 H, respectively.
(a) Find the energy stored in the inductor when the current reaches its maximum value.
115.38
X
Apply the expression for the energy stored in an inductor in terms of its inductance and the current through it. J
(b) Find the energy stored in the inductor one time constant after the switch is closed.
182.34
X
Find what fraction of current has been attained in one time constant after the circuit is closed and use it to find the energy stored in the inductor at that point. J
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Transcribed Image Text:A 24-V battery is connected in series with a resistor and an inductor, with R = 5.2 2 and L = 5.0 H, respectively. (a) Find the energy stored in the inductor when the current reaches its maximum value. 115.38 X Apply the expression for the energy stored in an inductor in terms of its inductance and the current through it. J (b) Find the energy stored in the inductor one time constant after the switch is closed. 182.34 X Find what fraction of current has been attained in one time constant after the circuit is closed and use it to find the energy stored in the inductor at that point. J Need Help? Mentor
Question
24 V
Inductor
a-
Given
R= 5.22
L=5.06.2
V=24
a)
24
5.2
from HW
batter is
4.615384 A
1 x 62 x 4.6.15847²
5-0
2
:£ x 6.0
2
=
4.615384 (1-e-¹)
= 2.91748
• 1 * x 9.6.15384 = 457.82784
5-0
458.00
Ch 20
Connected in
2
4 attempt
Series with
- 1/2 X 5.0 x 4.615 384
11.53846
I
115.38 A
115.38 (1-e¹)
72.93
-4.615791(16²¹)
11.53846 (1-e-')
7.2937
1x50x7.2937
/18.23987
23
182.342
Transcribed Image Text:Question 24 V Inductor a- Given R= 5.22 L=5.06.2 V=24 a) 24 5.2 from HW batter is 4.615384 A 1 x 62 x 4.6.15847² 5-0 2 :£ x 6.0 2 = 4.615384 (1-e-¹) = 2.91748 • 1 * x 9.6.15384 = 457.82784 5-0 458.00 Ch 20 Connected in 2 4 attempt Series with - 1/2 X 5.0 x 4.615 384 11.53846 I 115.38 A 115.38 (1-e¹) 72.93 -4.615791(16²¹) 11.53846 (1-e-') 7.2937 1x50x7.2937 /18.23987 23 182.342
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