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JEE Main Physics Class 12 Electronic Devices Part 1 Questions
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© examsnet.com
Question : 67
Total: 99
A system of four gates is set up as shown. The ‘truth table’ corresponding to this system is :
[Online April 23, 2013]
A
B
Y
0
0
1
0
1
0
1
0
0
1
1
1
A
B
Y
0
0
0
0
1
0
1
0
1
1
1
0
A
B
Y
0
0
1
0
1
0
1
0
1
1
1
0
A
B
Y
0
0
1
0
1
1
1
0
0
1
1
0
Validate
Solution:
In the given system all four gate is NOR gate
Truth Table
A
B
(
y
′
=
A
+
B
)
(
y
′
′
=
A
+
y
′
)
y
′
′
′
=
(
A
+
y
′
′
)
y
=
y
′
′
+
y
′
′
′
0
0
1
0
0
1
0
1
0
1
0
0
1
0
0
0
1
0
1
1
0
0
0
1
i.e.,
A
B
Y
0
0
1
0
1
0
1
0
0
1
1
1
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