|Truth table for system of four NAND gates as shown in figure is :
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
A | B | Y |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 1 |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 0 |
1 | 1 | 0 |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Text Solution
Verified by ExpertsThe correct answer is:
A
y = 
= 
= 
+ 
+ 
+ 
y = 0 + 
+ 
+ 0
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
For a common emitter configuration, if and have their usual meanings, the incorrect relationship …
In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the i…
In a P-N junction diode which is not connected to any circuit-
In a......biased P-N junction the net flow holes is from N -region to the P -region-
The part of a transistor which is most heavily doped to produce large number of majority charge car…
When P-N junction diode is forward biased, then-