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8/17/2019 EENG_223_Final_Exam_S08-09_soln (1)
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EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
Faculty of Engineering
ELECTIC!L !N" ELECT#NIC E NGINEEING "E$!T%ENT
EENG223 Circuit Theory I
INFE221 – Electrical Circuits
FINAL EXAM
Sprin 2!!"#!$
23 &une 2''(
"uration) 12' minutes
Instructor% #* +u,rer
-ol.e all $ro0lems
-T"ENT-
N%E N!%E
-N!%E
G#$ N#*
&ro'le( &oints
1 2'
2 2'
3 2'
4 2'
2
T#T!L 1'
8/17/2019 EENG_223_Final_Exam_S08-09_soln (1)
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i o
5 !
12 6
/ 6
+
v x
12 78 6
su9erno:e
v 1 v 2 v 3
/ v x
;
i o
+
v x
127
12 6
5!
/ 6
8 6
i 1
i 2 i 3
/ v x
;
EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
1) In the circuit of Figure 1<
=a> Fin: the current io using no:al analysis* *1! pts+ =0> Fin: the current io using mesh analysis* *1! pts+
*a+
+CL for the su9erno:e )
1 1 15 ' 1*2 7
x x
vv v v v− − = = ⇒ = −
2 1 12 13*2 7v v= − = − +CL for no:e 3 )
3 2 3 1 1 23
122 3 ' 38*4 7
8 12 x
v v v v v vv v
− − +− + + = ⇒ = = − 1 3' 3*1 !
12
v vi
−= =
*'+
+7L e?n* can 0e @ritten only for mesh 1)
1 1 3
3 2 2
3 2 2 3
1 '
12 12 8= > ' =1>
= > 2
5 ! '*2 ! 5*2 !
-u0stitute in =1> 3*1 !
x
i i i
i v i i
i i i i
i i
− + + − =
= − = − − =− = ⇒ = ⇒ =
⇒ = =
Fiure 1
i o
5 !
12 6
/ 6+
v x
12 78 6
/ v x
;
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8/17/2019 EENG_223_Final_Exam_S08-09_soln (1)
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EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
3) In the circuit in Figure 3< the o9Bam9s are i:eal
=a> Fin: the .oltage v' @hen =o9en circuit> f R = ∞ * *1! pts+
=0> Fin: the .oltage v' @hen 12 , f R = Ω * *1! pts+
=a>Ahen f R = ∞
'1
2 '1
' 2
2' '*4 8 7
1' 4 7
2'
4 7
v
v v
v v
+
+
= − × = −
⇒ = = −
∴ = = −
=0>Ahen 12 , f R = Ω < the first am9lifier may 0e consi:ere: as a summing am9lifier< @ith in9uts @hichare the '*4 7 an: v'*
'1 ' '
2 '1 '
' 2 ' '
4' 4' 1''*4 =8 > 7
2 12 3
1'
=4 > 72' 3
84 7 1* 7
3
v v v
v v v
v v v v
+
+
= − × + = − + ÷
⇒ = = − +
∴ = ⇒ = − ⇒ = −
Fiure 3
;
2 , 6
4' , 6
1' , 6
1' , 6
R f
v '
'*4 7
;
2 , 6
4' , 6
1' , 6
1' , 6
v '
v '1
v ;2
'*4 7
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EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
-* The s@itch in the circuit in Figure 4 has 0een close: for a long time* It is o9ene: at t '* Fin:the ca9acitor .oltage v=t > for t D '*
!tt
'
B
the circuit is un:er :c con:itions)
+7L for the loo9)
' '
' '
' 3''=3 > 1'' '
'*' ! =' > 1'' 7
i i
i v i−
− + + =
⇒ = ⇒ = =
Ahen the s@itch is o9ene: at t ')
=' > =' > / 7v v+ −
= =
+CL at no:e !) '3C i i= − !lso< C dv
i C dt
= < '1''
vi =
'*3 '*3
3'*1 '*3
1''
= > ='>* 7t t
dv dvv v
dt dt
v t v e e− −
⇒ = − ⇒ = −
∴ = =
# the solution coul: 0e foun: 0y :etermining the time constant as eq R C τ = < @here eq R is foun: fromthe circuit
+CL) ' '1''
3 1''3
t t t eq
t
vi i v i R
i= = ⇒ = = Ω
/ 3 /1'1'' 1''*1 s = > ='>* 7
3 3
t t v t v e e
τ
τ − −∴ = × = ⇒ = =
;
E
/' 7
Fiure -
t C '
;
v = t >
1'' 6
'*1 F
i '2 i '
3'' 6
;
E
/' 7 i '2 i '
;
v =' B
>
3 i ' 1'' 6
3'' 6
;
v = t >
1'' 6
'*1 F
i '2 i '
i C
!
1'' 6
i '2 i '
!
i t
;
v t
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EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
*The s@itch in the circuit in Figure has 0een close: for a long time* It is o9ene: at t '*
*a+ Fin: the current i=t > for t D '* *2! pts+*'+ Fin: the total energy :issi9ate: in the 12' resistor after the s@itch is o9ene:* *, pts+
=a> !t t 'B the circuit is un:er :c con:itions)
=' > 1' ! =' > ' 7i v− −= − =
Ahen the s@itch is o9ene: at t '< the resulting circuit is a sourceBfree LC circuit)
!t t '; =' > =' > 1' !i i+ −= = − *
3
' 5
'
1 11' ra:*/s
'*1 1' 1'
12'5'' / s<
2 2 '*1 H
LC
R
L
ω
α α ω
−
= = =× ×
Ω= = = ⇒
×
Therefore< the res9onse is un:erB:am9e:* The solution for the current has the form
( ) 2 2
1 2 '= > cos= > sin= > 8'' ra:*/st
d d d i t e B t B t α
ω ω ω ω α
−
= + = − =
To fin: the un,no@n coefficients< the initial con:itions are a99lie:* First the initial con:ition for the time
:eri.ati.e of current must 0e :etermine:*
( )1
+7L) = > = > ' =' > =' > =' > 1'*=' 12''> 12''' !/sdi di
L v t Ri t v Ridt dt L
+ + ++ + = ⇒ = − + = − − =
1='> 1' !i B= = −
t C '1' J F
i = t >1'' mH
2'' 6
12' 61' !
Fiure ,
12' 6
; v =' >
i =' >
1' !
2'' 6
1' J F
1'' mH12' 6
; v = t >
i = t >
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EENG223 Circuit Theory I / INFE221 Electric Circuits – Final Exam
( ) ( ) ( )1 2 1 2
1 2 2
cos= > sin= > sin= > cos= >
12''' 5''= 1'>' ='> K* !
8''
t t
d d d d d d
d
die B t B t e B t B t
dt
dit B B B
dt
α α
α ω ω ω ω ω ω
α ω
− −= − + + − +
+ −= ⇒ = − + ⇒ = =
( )5''= > 1' cos=8'' > K*/sin=8'' > !t i t e t t −= − +
=0> The energy :issi9ate: in the resistor is the total energy store: in the in:uctor an: the ca9acitor att '; <
2 2 21 1 1=' > =' > '*1 1' &
2 2 2total W Li Cv
+ += + = × × =