Digital Electronics - Logic Gates - Discussion
Discussion Forum : Logic Gates - General Questions (Q.No. 2)
2.
If a 3-input NOR gate has eight input possibilities, how many of those possibilities will result in a HIGH output?
Discussion:
38 comments Page 1 of 4.
Sarasti said:
2 years ago
Option A is the correct answer.
(1)
Subash said:
4 years ago
For a NOR gate, whenever there is a 1 in Input, the output will remain LOW (0). So, count the number of possibilities to have both inputs as zero. In our case it's+ only 1 possibility.
(3)
Ali said:
4 years ago
If a 3-input NOR gate has eight input possibilities how many of these possibilities will result in a low output? Please anyone tell me.
(3)
MUHAMMAD SHAIGAN said:
5 years ago
I am not getting. Can anyone explain this? please.
(2)
SHIVAM said:
5 years ago
Because in nor gate if any of the input is high the output is low.
So, we are left with only one case when output is high i.e. A=0, B=0 and C=0.
So, we are left with only one case when output is high i.e. A=0, B=0 and C=0.
(1)
Nitin Gawande said:
6 years ago
In nor gate if any input is the high answer will be 0.
(1)
Avadhesh Kumar soni said:
6 years ago
NOR gate will only produce a logic 1 output when all inputs are simultaneously at logic 0. Any other input combination will produce a logic 0 output.
(1)
Vijaya said:
7 years ago
Thank you all.
Ashutosh Jaiswal said:
8 years ago
Actually the truth table for NOR Gate is given by
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
And similarly truth table for 3input nor gate is;
A B C Y
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 0
So, only one output is high. Hence option A is the correct option to be elected which was given in the choices.
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
And similarly truth table for 3input nor gate is;
A B C Y
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 0
So, only one output is high. Hence option A is the correct option to be elected which was given in the choices.
(7)
JAMES said:
8 years ago
I have a question.
Prove this for me AB+(barB)C+AB = AB+(barB)C.
Prove this for me AB+(barB)C+AB = AB+(barB)C.
(1)
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