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'

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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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    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

    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  

    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

    + += + = × × =