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(ELEC101)[2008](f)midterm~kytangab^_10259.pdf
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(a) Using Boolean algebra, show that
AB + (A +B) =AB
F = AB + (A + B) = (A + B) + AB = (A + B) + AB
= A + B(1+ A) = A + B = AB
(b) A logic circuit has 4 inputs ABCD and 1 output E. E is 1 when ABCD are 0000, 0010, 0110, 1000, 1001 and is 0 for 0001, 0011, 0100, 0101, 0111. 1010 to 1111 are dont cares. Design the logic circuit using ONLY NAND gates.
AB
CD
E = A + BD + CD
A
B
E
D
A
B
E
D
Q1n-Q1n-
n
Q
(a) Write the truth table for the following JKFF. Hence sketch the waveform of Q. Q is 0 initially.
nQ
Q 1n
-
J K Q
0 0 NO CHANGE
0 1 RESET
1 0 SET
1 1 TOGGLE
CHANGE
2A
-2V 2. -6V
1A + I
2.
-10V
0V
4V
Apply KVL
Io*2.+4V+-10V+(1A+Io)*2.= 0 .4V+Io*4.= 0 Io=1A
-10V
1A
4V
0V
0A
Power for 1A = 1A*4V = 4W
consumed
3. In the circuit , find Io . Find also the power consumed or delivered by the 1A source .
2.
Io
-10V
0V
a
V2
14mA 1k.
V2 = 7V
R = 1k.
ba
a
V2
R between ab = 1k
Hence RN = 1k I1 V1
I1 V1
When R = 0, I3 = 14mA
Hence Isc = 14mA 4. When R = 0., I3 = 14mA. When R = 1k., find V2.
I1, V1 and I2 are unknowns.
I2
Kill 20V
3k.
20mA
Io
6k.
Io= 20mA* =12mA
6k.+4k.
20V
Kill 20mA
Io
20V
Io== 2mA
6k.+4k.
Io=12mA+2mA=14mA
24mA 20V
56V 20mA 14mA
5. Use the Superposition Method to find Io .
20V
Io
20mA
6 Va 2k. 12V
4k.
12V
0V
2k.
0V
Va . 0 Va Va-12V
2mA = ++
6k. 3k. 2k. 12V = Va + 2Va + 3Va -36V Va = 8V 2k. 8V
Vo = Va* =
6k. 3
4k. 8mA/6 8V 2k. 2mA 12V 12V
2k. 0V
2mA/3
6. Use nodal analysis to find Vo .
2k. 4k.
12V 2k.
3k.
-6V
1mA
Vo
4k.
2mA
2k.
2V
Vo
4k.
2k. 2k. 4V
Vo
4k.
2V + 4V
Vo = * 4k.= 2V
12k.
1 mA
3k.
1V -2V
-6V 4V
1mA
2V
4k. 2V0V 0.5mA
7. Use source transformation to find Vo .
3k. -6V
1mA
Vo
4k.
12. 2. 4VO
VO
4.
4.
4Vo
4Vo . Vo Vo
= 5A +
2. 4. 6Vo = 20V + Vo Vo = 4V
V1 = -56V
4V
2.
12. 16V
-56V
4A
5A
4.
4.
1A
4Vo
0V
8. In the circuit, find V1. The dependent source (voltage
controlled voltage source) is equal to 4Vo and is in volt. .
12. Vo 2.
V1
4.
4.
5A
4Vo
20V
Va
20.
9.5A/3
20.
110V/3 20V 170V/3
5.5A/3
80V
20.
3.5A/3
0V 80V
5.5A/3
20.
20.
9. Use node voltage method to find Va.
0V 80V
Apply KCL to 20V, Iin = Iout
Va Va+ 20V .70V Va+ 20V .80V
5A =+5A ++ 20.
20. 20. 20. 0 = Va+ (Va.50V) + (Va+ 20V) .80V 110V
Va =
80V
3
10 18V
IX I1
IO
12k.
18V .IO *12k..(IO .6mA)*12k.= 0 90V .IO *24k.= 0 15
IO = 4 mA
45V 18V
3m/4
3m/4 9m/2
3m/2 15m/4 9m/4
12k.
27V
12k.
10. Use loop current method to find Io .
18V
12k.
IO
12k.
12k.
6A
11
R