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
• • • • • •
I!08 09) V5FE<<6;;`3 =D>3 DF2 QD=5H ;6F #9 =572DF E5FE154 67=K a9 MD335Q2 45@2 57QDF5D74 E5FE1543 !9 767(=572DF E5FE1543 67=K :9 A64< =572DF D7H 767(=572DF E5FE1543 09L #== 4<2 F23534D7E23 57 ;5?1F2 3<6>7 A2=6> DF2 ) b 2DE<9 %<2 QD=12 6; E1FF274 57 #@M2F2 4
#9 )c)T a9 Lc)T !9 Sc)T :9 Xc)T 09O %>6 F235346F3 ') D7H 'L ?5Q27 E6@A572H F23534D7E2 6; S9T b ><27 57 32F523 D7H ) b ><27 57 MDFD==2=d 4<2 F23534D7E2 DF2 #9 )9T b D7H O b a9 L b D7H L9T b !9 ) b D7H O9T b :9 S b D7H ^9T b 09S +7 ?5Q27 ;5?1F2d 4<2 QD=12 6; F23534D7E2 ' 57 b 53
#9 )^ a9 L^ !9 O^ :9 S^ 09T %<2 1754 6; 57H1E4D7E2 53 "27FK9 +4 ED7 D=36 A2 F2MF232742H D3 #9 Nc/2E a9 N ( /2E !9 Nc# :9 N ( /2Ec# 09U %<2 272F?K 346F2H 57 D7 57H1E46F 6; 57H1E4D7E2 C "27FK 53 F2MF232742H D3d #9 5LC a9 5CL !9 [)cL]C95L :9 CLc5 09W %<2 Q6=4D?2 57H1E2H 57 D7 57H1E46F 6; C "27FK 53 F2MF232742H D3d #9 C5 a9 C H5cH4
!9 :9
Cc5 CL5
09X # MDFD==2= M=D42 EDMDE546F 227 4<2@ 53 57?3 53 4<2 2JMF233567 ;6F 272F?K 346F2H 57 D EDMDE546F #9 !Q a9 ! HQcH4 !9 !cQ :9 [)cL]!QL 09)^ # AFD7E< 6; D 724>6Fe 53 3D5H 46 A2 DE45Q2 ><27 54 E6735343 6; 6 72 #9 F235346F a9 Q6=4D?2 361FE2 !9 57H1E46F :9 EDMDE546F
!"#$%&'(L !+'!,+% #.#Cf/+/ %&!".+0,&/ 0123456738
1. Consider the circuit shown in Figure 1. Reduce the portion of the circuit to the left of terminals a–b to (a) a Thévenin equivalent and (b) a Norton equivalent. Find the current through R=16, and comment on whether resistance matching is accomplished for maximum power transfer. 2. Consider the circuit of Figure 2, including a dependent source. Obtain the Thévenin equivalent at terminals a–b. 3. By means of nodal analysis, find the current delivered by the 10-V source and the voltage across the 10-resistance in the circuit shown in Figure-3 4. For the network shown in Figure 4, find the current delivered by the 10-V source and the voltage across the 3-resistor by means of mesh-current analysis. 5. Determine the voltage across the 20-resistor in the following circuit of Figure-5 with the application of superposition. 6. Use delta–wye transformation for network reduction and determine the current through the 12-ohm resistor in the circuit of Figure-6 7. Determine the Thévenin and Norton equivalent circuits as viewed by the load resistance Rin the network of Figure-7 (i) Find the value of R if the power dissipated by Ris to be a maximum. (ii) Obtain the value of the power in part (b). 8. Reduce the circuit of Figure-8 to a Thévenin and a Norton equivalent circuit. 9. Find the Thévenin and Norton equivalent circuitsfor configuration of Figure-9 as viewed from terminals a–b.
the
10. Obtain the Thévenin and Norton equivalent circuits for the portion of the circuit to the left of terminals a–b in Figure 10, and find the current in the 200-ohm resistance. 11. In the circuit given in Figure 11, determine the currentIthrough the 2-ohm resistor by (a) the nodalvoltage method, and (b) meshcurrent analysis.
12. Find the voltage across the 8-A current source in the circuit of Figure 12 with the use of nodal analysis & Determine the current in the 0.5-resistor of the circuit by mesh analysis.
Figure.1
Figure.2
Figure.3
Figure.4
Figure.5
Figure.6
Figure.7
Figure.8
Figure.9
Figure.10
Figure.11
Figure.12
I!08
1. In an electrical network to neglect a current source the current source is: a) b) c) d)
Open Circuited. Short Circuited. Replaced by a capacitor. Replaced by an Inductor.
2. Which of the following theorem is applicable for both linear and nonlinear circuits? a) b) c) d)
Superposition theorem. Thevenin's theorem. Norton's theorem. none of these.
3. In an electrical network to neglect a voltage source the voltage source is: a) b) c) d)
Open Circuited. Short Circuited. Replaced by a capacitor. Replaced by an Inductor.
4. In figure, the total power consumed is a) b) c) d)
10 12 16 20
W W W W
5. What is maximum power transfer condition? a) b) c) d)
Rs = R L Rs < R L Rs > R L None of the above.
6. In Thevenin’s Theorem which term to be determined? a) b) c) d)
Vth, Isc Vth,Rth Rth, Isc None of the above
7. In Norton’s Theorem which term to be determined? a) b) c) d)
Vth, Isc Vth,Rth Rth, Isc None of the above
8. In Thevenin’s equivalent circuit Rth is Parallel connection with Vth? a) b) c) d)
True False May be None of the above
9. In Norton’s equivalent circuit Rth is Parallel connection with Isc ? a) b) c) d)
True False May be None of the above
10. Star network is also known as a) b) c) d)
T-network X- network Y-network Z-network
!"#$%&'(O #.#C-_ a,+C:+._ aC-!V/ #.: -$&'#%+-.#C #I$C+G+&'/ 0123456738 )9 L9 O9 S9 T9 U9 W9
&JM=D57 +H2D= 6M2FD4567D= D@M=5;52F D7H 543 E
X9
:2;5728 5] -$&.(C--$ N-C%#_& _#+. # 55] +.$,% '&/+/%#.!& '5 555] -,%$,% '&/+/%#.!& '6 5Q] !6@@67(I6H2 '2h2E4567 'D456 [!I''] Q] +.$,% -GG/&% N-C%#_& N -/ Q5] +.$,% a+#/ !,''&.% + a Q55] +.$,% -GG/&% !,''&.% + -/ Q555] $6>2F(/1MM=K '2h2E4567 'D456 [$/''] 5J] /=2> 'D42 Z9 &JM=D57 6M2FD4567 6; D7D=6? E6@M142F39 )^9 &JM=D57 MFDE45ED= DMM=5ED4567 6; 6M(D@M8 #7 D146@645Q2 M6>2F(D335342H 3422F57? 3K342@9 I!08 )9 9 =
789:8 $# ";"-< :5;;2F2745D= D@M=5;52F '2E45;52F
! >
-3E5==D46F .672 6; 4<2 DA6Q2
L9 9 =
?,(*"@) @"$- ,A "- $B)"( ,CD"EC $#< +7;57542 C6>
! >
N2FK <5?< N2FK =6>
O9 9 =
!"-BF$B*G ,A "- $B)"( ,CD"EC $#< +7;57542 C6>
! >
N2FK <5?< N2FK =6>
S9 9 =
75*C5* $EC)B"-%) ,A "- $B)"( ,CD"EC $#< +7;57542 ! N2FK <5?< C6> > g2F6
T9 9 =
=:HH ,A "- $B)"( ,CD"EC $#< +7;57542 C6>
! >
N2FK <5?< N2FK =6>
U9 9 =
I()F +"*) ,A "- $B)"( ,CD"EC $#< +7;57542 C6>
W9 9
7CD"EC $# "< N6=4D?2(E674F6==2H Q6=4D?2 361FE2 [N!N/] !1FF274(E674F6==2H Q6=4D?2 361FE2 [!!N/]
=
! >
N2FK <5?< N2FK =6>
!
N6=4D?2(E674F6==2H E1FF274 361FE2 [N!!/] !1FF274(E674F6==2H E1FF274 361FE2 [!!!/]
>
X9 9 =
7CD9EC 5#)#< -7=K iQ2 N6=4D?2 :1D= 31MM=Kd 5929d k NEE
Z9 9 =
JG) @"$- "-B =:HH %"- "(#, K) )LC+)##)B $- MMMMMN .2>467B3 ! :2E5A2=3 N6=4cD@M2F23 > .672 6; 4<2 DA6Q2
! >
-7=K jQ2 Q6=4D?2 .672 6; 4<2 DA6Q2
)^9 JG) *OC$%"( ,C)- (,,C @"$- ,A "- P=QR0 789:8 $# MMN 9 L J )^T ! L J )^U = +7;57542 > g2F6 ))9 9 =
I()F *"+) ,A P= QR0 789:8 $# MMMMMMMN ^9O lNcN ! ^9T @NcN >
^9T Ncl3 ^9O eNcN
!"#$%&'(S :+_+%#C a,+C:+._ aC-!V/ #.: !-I$,%&' /f/%&I/ 0123456738
1. Obtain the truth table for the logic block shown in Figure
2. For the NOR and inverter realizations shown in Figure, find the truth table, the type of gate realized, and the expression for the logic output,in each case.
3. Draw the logic diagram for the following Boolean expressions (without any simplification).
4.
5.
6. Draw the logic diagram of an SR latch using only NAND gates, and obtain the truth table for that implementation. 7. J and K are the external inputs to the JKFF shown in Figure. Note that gates 1 and 2 are enabled only when the clock pulse is high. Consider the four cases of operation and explain what happens.
8. Draw a block diagram for a 2-to-4 decoder. Obtain the truth table, and develop a logic diagram. 9. Show a block diagram of a 4-bit, parallel-input shift-right register and briefly explain its operation. 10. What is the basic difference between the weighted resistor and the R–2R ladder D/A converters? I!08 1.
Convert hexadecimal value 16 to decimal.
A.
2210
B.
1610
C.
1010
D.
2010
2. Convert the following decimal number to 8-bit binary. 187 A.
10111011 2
B.
11011101 2
C.
10111101 2
D.
10111100 2
3.The output of an AND gate with three inputs, A, B, and C, is HIGH hen ________. A.
A = 1, B = 1, C = 0
B.
A = 0, B = 0, C = 0
C.
A = 1, B = 1, C = 1
D.
A = 1, B = 0, C = 1
4. The output of an exclusive-NOR gate is HIGH if ________. A.
the inputs are equal
B.
one input is HIGH, and the other input is LOW
C.
the inputs are unequal
D.
none of the above
5.Determine the values of A, B, C, and D that make the sum term
equal to zero.
A.
A = 1, B = 0, C = 0, D = 0
B.
A = 1, B = 0, C = 1, D = 0
C.
A = 0, B = 1, C = 0, D = 0
D.
A = 1, B = 0, C = 1, D = 1
6.Which of the following expressions is in the sum-of-products (SOP) form? A.
(A +
B.
(A )B (CD )
C.
AB (CD )
D.
AB + CD
7. How is a
B )(C + D )
J-K flip-flop
A.
J =
0,
K =
0
B.
J =
1,
K =
0
C.
J =
0,
K =
1
D.
J =
1,
K =
1
made to toggle?
8.A positive edge-triggered D flip-flop will store a 1 when ________. A.
the D input is HIGH and the clock transitions from HIGH to LOW
B.
the D input is HIGH and the clock transitions from LOW to HIGH
C.
the D input is HIGH and the clock is LOW
D.
the D input is HIGH and the clock is HIGH
9.If both inputs of an goes high?
S-R flip-flop
are LOW, what will happen when the clock
A.
No change will occur in the output.
B.
An invalid state will exist.
C.
The output will toggle.
D.
The output will reset.
10. Implementation of the Boolean expression ________. A.
three AND gates, one OR gate
B.
three AND gates, one NOT gate, one OR gate
C.
three AND gates, one NOT gate, three OR gates
D.
three AND gates, three OR gates
results in
!"#$%&'(T /+_.#C $'-!&//+._ 0123456738 )9 L9 O9 S9 T9 U9
W9 X9 Z9 )^9
:FD> D7H 2JM=D57 4<2 A=6Ee H5D?FD@ 6; D745(76532 3K342@ 46 31MMF233 4<2 76532 57 D7 D5FEFD;49 :5;;2F2745D42 A24>227 E674571613 D7H H53EF242 35?7D=39 &JM=D57 <6> D AD35E 35?7D= MF6E23357? 3K342@ >6Fe39 Y57? >DQ238 [D] '2E4D7?1=DF M1=32 4FD57 [A] %F5D7?1=DF >DQ2 [E] /D>4664< >DQ2 [H] /P1DF2 >DQ2 [2] "D=;(F2E45;52H 3572 >DQ2 &JM=D57 4<2 ;F2P127EK 4FD73=D4567 MF6E233 >54< <2=M 6; MF6H1E4 @6H1=D46F9 &JM=D57 3D@M=57? >54< <2=M 6; 3>54E<57? 3D@M=2F9 &JM=D57 4<2 76532 57 4<2 E57? MF6M2F4523 6; 35?7D=3 >54< 2JD@M=2 6; 2DE<8 [D] $2F56H5E54K D7H !D13D=54K [A] &Q277233 D7H -HH7233 [E] &72F?K(4KM2 D7H M6>2F 4KM29
I!08 09) !54< D] 3D@M=57? 45@2 A] 45@2 E674571613=K E] 3D@M=57? ;F2P127EK H] 7672 6; 4<232 09O :53EF242 35?7D= 53 F2MF232742H AK D] E6H57? A] @6H1=D4567 E] H2@6H1=D4567 H] 32P127E2 6; 3D@M=23 09S !DFF52F 35?7D=
D] E6734D74 @D?7541H2 A] E6734D74 ;F2P127EK E] QDF5DA=2 ;F2P127EK H] QDF5DA=2 45@2 09T I6H1=D457? 35?7D= ;F2P127EK A] =6> @6H1=D4567 E] <5?< ;F2P127EK H] 7672 6; 4<232 09U +7 6FH2F 46 F2H1E2 5742F;2F27E2d 4<2 35?7D= 3<61=H A2 D] D@M=5;52H A] @1=45M=52H E] H2@6H1=D42H H] @6H1=D42H 09W I233D?23 4FDQ2= ;F6@ 4FD73@5442F 46 F2E25Q2F >54< <2=M 6; D] %FD73@5442F A] '2E25Q2F E] EDQ2 A] 5@M1=32 35?7D= E] 3D@M=2 35?7D= H] .672 6; 4<232 09Z .67 2=2E4F5E 35?7D= 53 E67Q2F42H 5746 2=2E4F5ED= 35?7D= AK D] 4FD73@5442F A] F2E25Q2F E] =572 H] 7672 6; 4<232 09)^ +7 D@M=541H2 @6H1=D4567d D@M=541H2 53 D] E6734D74 A] m2F6 E] QDF5DA=2 H] 7672 6; 4<232
!"#$%&'(U !-II,.+!#%+-. /f/%&I/ 0123456738 09)
09L
09O 09S 09T 09U 09W 09X 09Z
&JM=D57 4<2 @6H2 6; MF6MD?D4567 6; 2=2E4F6@D?7245E >DQ2 57 ;F22 3MDE2 D7H D4@63M<2F29 #=36 H2;572 /5?7D= @1=45MD42F _D57 [2]a2D@ >5H4< [;]$6=DF5mD4567 [?] fD?5 ,HD #74277D Y54< 3154DA=2 >DQ2;6F@d D=36 H2;572 @6H1=D4567 +7H2J9 &JM=D57 4<2 ?272FD4567 6; :/a /! #I9 Y227 #I D7H GI9 :FD> D7H &JM=D57 4<2 A=6Ee H5D?FD@ 6; 31M2F<24F6HK72 F2E25Q2F9 /4D42 /D@M=57? %<26F2@9 :2;572 !6@MD7H57?9 Y
09)^ &JM=D57 @6H2 6; MF6MD?D4567 6; 4FD73@533567 =5729 I!08 09) 09L
09O 09S 09T 09U
09W
09X
09Z
!6DJ5D= 4FD73@533567 =572 E6@@67=K 6M2FD423 57 nnn [D] %& @6H2 [A] %I @6H2 [!] %&I @6H2 [H]7672 N!- 53 nnnnnn [D] N6=4D?2 E674F6= 63E5==D46F [A]N5H26 E674F6= 63E5==D46F [!] N6=4D?2 E6@@67 63E5==D46F [H] N5H26 !6@@67 63E5==D46F Y
09)^
'G /4D7H3 ;6F [D] 'DH56 GF2P127EK [A] '2E25Q2F GF2P127EK [!] '2h2E4567 GF2P127EK [H] '2M2D42F ;F2P127EK
!"#$%&'(W a#/+! !-.%'-C /f/%&I/ 0123456738 )9
YF542 76423 67 6M27 =66M E674F6= 3K342@3 D7H E=632H =66M E674F6= 3K342@3 >54< 4<2 <2=M 6; 72D4 A=6Ee H5D?FD@39 _5Q2 36@2 3154DA=2 F2D= =5;2 2JD@M=23 6; A64< 4KM23 6; 3K342@39 L9 !6@MDF2 -M27 C66M Q2F313 !=632H C66M E674F6= 3K342@39 O9 !=D335;K 4<2 ;22HADEe E674F6= 3K342@3 AK M1FM6329 S9 :2;572 \ 2JM=D57 4<2 4FD73;2F ;17E4567 1357? a=6Ee H5D?FD@ 6; D7 2=2@274DFK ;22HADEe E674F6= 3K342@9 T9 YF542 D4 =2D34 35J A=6Ee H5D?FD@ F2H1E4567 42E<75P123 >54< A=6Ee H5D?FD@39 U9 &JM=D57 4<2 2;;2E4 6; ;22HADEe 67 /273545Q54Kd /4DA5=54Kd :K7D@5E F23M6732 D7H aD7H>5H4<9 W9 &JM=D57 HK7D@5E F23M6732 6; E674F6= 3K342@ >54< 4<2 <2=M 6; 72D4 A=6Ee H5D?FD@9 X9 :2;572 \ &JM=D57 342DHK 34D42 2FF6F9 Z9 _5Q2 4<2 E=D335;5ED4567 6; G22HADEe !674F6= /K342@3 AK !674F6= #E45679 )^9 &JM=D57 :5?54D= !674F6= /K342@ >54< 4<2 <2=M 6; 2JD@M=29
I!08 0N 9 =
/N
P- "- ,C)- (,,C %,-*+,( #O#*)E -14M14 53 57H2M27H274 6; E674F6= 57M14 -7=K 3K342@ MDFD@242F3
! >
-14M14 53 H2M27H274 67 E674F6= 57M14 .672 6; 4<2 DA6Q2
9 =
9 %,-*+,( #O#*)E $- FG$%G *G) %,-*+,( "%*$,- $# #,E)G,F B)C)-B)-* ,- *G) ,5*C5* $# 6-,F!=632H =66M 3K342@ ! /2@5(E=632H =66M 3K342@ -M27 3K342@ > .672 6; 4<2 DA6Q2
SN 9 =
9- "5*,E"*$% *,"#*)+ $# " MMMMMM (,,C %,-*+,( #O#*)EN 6M27 ! E=632H MDF45D==K E=632H > D7K 6; 4<2 DA6Q2
RN 9 =
JG) $-$*$"( +)#C,-#) FG)- *G) ,5*C5* $# -,* )T5"( *, $-C5* $# %"(()B %FD735274 F23M6732 ! &FF6F F23M6732 :K7D@5E F23M6732 > &54<2F 6; 4<2 DA6Q2
UN 9 =
MMMMMMM $# " %(,#)B (,,C #O#*)EN #146(M5=64 ;6F D7 D5FEFD;4 !DF 34DF42F
2N
JG) *+"-#A)+ A5-%*$,- $# "CC($%"K() *, FG$%G ,A *G) A,((,F$-@V
! >
:5F2E4 E1FF274 ?272FD46F &=2E4F5E 3>54E<
9
C572DF D7H 45@2(57 QDF5D74 3K342@3
!
=
C572DF 3K342@3
>
C572DF D7H 45@2(QDF5D74 3K342@3 .67(=572DF 3K342@3
QN 9 =
W$*G A))BK"%6 MMMMM $-%+)"#)#N 3K342@ 34DA5=54K ?D57
! >
3273545Q54K 2;;2E43 6; H5341FA57? 35?7D=3
XN 9 =
I*)"BO I*"*) '++,+ A,+ JOC)D1 #O#*)E A,+ 5-$*) #*)C $-C5* $# MMMMMMMMMMN G57542 ! +7;57542 g2F6 > #== 6; DA6Q2
YN 9 =
I*)"BO I*"*) '++,+ A,+ JOC)D/ #O#*)E A,+ 5-$*) +"EC $-C5* $# MMMMMMMMMMN G57542 ! +7;57542 g2F6 > #== 6; DA6Q2
01N I*)"BO I*"*) '++,+ A,+ JOC)D0 #O#*)E A,+ 5-$*) "%%)()+"*$,- $-C5* $# MMMMMMMMMMN 9 G57542 ! +7;57542 = g2F6 > #== 6; DA6Q2
