The battery terminal voltage in the figure below is E = 6.90 V and the current I reaches half its maximum value of 4.00 A at t = 0.160 s after the switch is closed.HINTRw(a) Calculate the time constant 7 (in s).0.16X S(b) What is the potential difference (in V) across the inductor at t = 0.160 s?-6.9X V(c) What is the potential difference (in V) across the inductor in the instant after the switch is closed at t = 0?6.90✓ V 4-O- given &9=6.90A=400t=0·160a)еI -Io (1-e² )0.160t0-160en (s)=b)6.90 xc) E. eIIn(*(501X 00.1)= 0.1600.16- (1-05)afhe=>690 xe6-0.166.90 x 1 x 20 t700.160T6.9Zenel=1=/0.160essaa =P&-01 x ²246 = A

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Answered: The battery terminal voltage in the…

The battery terminal voltage in the figure below is &= 6.90 V and the current I reaches half its maximum value of 4.00 A at t= 0.160 s after the switch is closed.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

The battery terminal voltage in the figure below is E = 6.90 V and the current I reaches half its maximum value of 4.00 A at t = 0.160 s after the switch is closed.
HINT
R
w
(a) Calculate the time constant 7 (in s).
0.16
X S
(b) What is the potential difference (in V) across the inductor at t = 0.160 s?
-6.9
X V
(c) What is the potential difference (in V) across the inductor in the instant after the switch is closed at t = 0?
6.90
✓ V 4
-O-

given &
9=6.90
A=400
t=0·160
a)
е
I -
Io (1-e² )
0.160
t
0-160
en (s)
=
b)
6.90 x
c) E. e
I
In
(*(501X 00.1)
= 0.160
0.16
- (1-05)
afhe
=>
690 xe6
-0.16
6.90 x 1 x 20 t
70
0.160
T
6.9
Zenel=1
=/0.160
es
saa =
P
&-01 x ²246 = A

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Step 3: (b) The expression for potential difference across the inductor

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Step 4: (c) Potential difference at the inductor

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