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.
Rashmi said:
1 decade ago
For nor gate the basic truth table s A B Y
0 0 1
0 1 0
1 0 0
1 1 0
and in the same way truth tablefor 3input norgate s
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 here only one output s high.hence option A
0 0 1
0 1 0
1 0 0
1 1 0
and in the same way truth tablefor 3input norgate s
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 here only one output s high.hence option A
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)
Devendra babu said:
1 decade 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 s
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 s 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 s
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 s high.hence option A is the correct option to be elected which was given in the choices
Rashmi said:
1 decade ago
Truth table of NOR
Now in 8 connection only 1 can have 0 0 0 i/p so that 1 connectn can remain high.
i/pa i/pb o/p
0 0 1
0 1 0
1 0 0
1 1 0
Now in 8 connection only 1 can have 0 0 0 i/p so that 1 connectn can remain high.
Js dhaker said:
1 decade ago
A B C = OR NOR
0 0 0 0 1
0 0 1 1 0
0 1 0 1 0
0 1 1 1 0
1 0 0 1 0
1 0 1 1 0
1 1 0 1 0
1 1 1 1 0
0 0 0 0 1
0 0 1 1 0
0 1 0 1 0
0 1 1 1 0
1 0 0 1 0
1 0 1 1 0
1 1 0 1 0
1 1 1 1 0
Aarti said:
1 decade ago
NOR is OR+INVERT so OR will be 1 when any one of inputs is '1' but in NOR output will be inverted OR. So when all inputs are '0' OR output will be '0' and its inverted means NOR output will be '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)
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)
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)
Al-imran masum said:
1 decade ago
Only, . A=0, B=0, C=0 is equal output is 1. That means high. But the another possibilities output is also 0. That means low. So our answer is 1.
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