Applaing lap lace trangtom to given Circuit at + >0 Switch is closed and ci r cuit Be comes R LS US) I6) v (s) I ls) R+LS + る 20 given VIt) = do ê at -) v.!s) = L: 1H =) LS : S C: IMf -) 106 I (s) = S+a aos 10 +s+ l06 (sta) (s+10s+106) A BS + C St2 - 5 S + 20 124998 + St2 184998 (Sts)4 999975 5/124998) (Sts), 833325 (s+5)4999975 -s J1R4998 41666 (S+s)²4 99995 S+2 CH) = inverse lap lace d Ils) -5 ;(+) = -at st e"cos/ 999975 t)+ 55555J3 L04.999 194.900

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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Can you explain the Bs+C part ?

Applaing
lap lace trangfom to given Circuit
at + >0
Switch is closed and ci r cuit
Be comes
R
LS
I6)
v (s)
I ls)
R+LS +
20
given VIt) = do ê at -)
v.!s) =
=)
LS : S
C : IMf -)
106
CS
I (s) =
S+a
aos
10 +s+ l06
(sta) (st+105+106)
A
BS + C
St2
- 5
S + 20
124998
+
St2
184998
(Sts)4 999975
-s 124998
G
5/124998) (S+s), 833325
(s+5)4199975
41666 (S+s)
99995
S+2
CH) = inverse
laplace g Ils)
-5
;(+) =
-at
st
55555J3
41666 J1333
5t sin(J999975+
e"cos J99975 t)+
124-998
124998
Transcribed Image Text:Applaing lap lace trangfom to given Circuit at + >0 Switch is closed and ci r cuit Be comes R LS I6) v (s) I ls) R+LS + 20 given VIt) = do ê at -) v.!s) = =) LS : S C : IMf -) 106 CS I (s) = S+a aos 10 +s+ l06 (sta) (st+105+106) A BS + C St2 - 5 S + 20 124998 + St2 184998 (Sts)4 999975 -s 124998 G 5/124998) (S+s), 833325 (s+5)4199975 41666 (S+s) 99995 S+2 CH) = inverse laplace g Ils) -5 ;(+) = -at st 55555J3 41666 J1333 5t sin(J999975+ e"cos J99975 t)+ 124-998 124998
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