Electronics - Logic Circuit Simplification - Discussion
Discussion Forum : Logic Circuit Simplification - General Questions (Q.No. 9)
9.
One of DeMorgan's theorems states that 
. Simply stated, this means that logically there is no difference between:
. Simply stated, this means that logically there is no difference between:Discussion:
8 comments Page 1 of 1.
Sooryashibu said:
1 decade ago
NAND = NOT of AND.
Lakshmi said:
1 decade ago
Can you please elaborate?
AbedElkareem said:
1 decade ago
I think that, the correct answer is D.
Because a NAND is equivalent to after equal term and OR with bubbled output is equivalent for the before equal term.
Because a NAND is equivalent to after equal term and OR with bubbled output is equivalent for the before equal term.
Scaneu said:
9 years ago
I think it should be: OR gate with bubble on each input is equivalent to a NAND gate.
Mohan said:
8 years ago
Option B is Correct.
A NOR gate and a bubbled input of AND gate.
A NOR gate and a bubbled input of AND gate.
Madarasi said:
8 years ago
I think option A is correct.
Because, NAND gate output=A.B whole bar=A bar+B bar.
The Output of AND gate is A.B = bubbled output=A.B whole bar.
Because, NAND gate output=A.B whole bar=A bar+B bar.
The Output of AND gate is A.B = bubbled output=A.B whole bar.
Affan said:
8 years ago
Option A is correct.
As ( A.B whole bar = Abar + Bbar) is the result of only NAND gate.
Therefore option A as both gates are NAND in this option only.
As ( A.B whole bar = Abar + Bbar) is the result of only NAND gate.
Therefore option A as both gates are NAND in this option only.
SNEHA said:
7 years ago
I think Option D is correct.
As (A.B whole bar = Abar + Bbar) is the result of only NAND gate and (Abar + Bbar=bubbled OR).
As (A.B whole bar = Abar + Bbar) is the result of only NAND gate and (Abar + Bbar=bubbled OR).
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