Electronics - Logic Circuit Simplification - Discussion

9. 

One of DeMorgan's theorems states that mcq19_1007_1.gif. Simply stated, this means that logically there is no difference between:

[A]. a NAND gate and an AND gate with a bubbled output
[B]. a NOR gate and an AND gate with a bubbled output
[C]. a NOR gate and a NAND gate with a bubbled output
[D]. a NAND gate and an OR gate with a bubbled output

Answer: Option B

Explanation:

No answer description available for this question.

Sooryashibu said: (Feb 25, 2014)  
NAND = NOT of AND.

Lakshmi said: (Apr 21, 2015)  
Can you please elaborate?

Abedelkareem said: (Jul 11, 2015)  
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.

Scaneu said: (Aug 31, 2016)  
I think it should be: OR gate with bubble on each input is equivalent to a NAND gate.

Mohan said: (Jun 11, 2017)  
Option B is Correct.

A NOR gate and a bubbled input of AND gate.

Madarasi said: (Jun 11, 2017)  
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.

Affan said: (Mar 17, 2018)  
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.

Sneha said: (Jan 15, 2019)  
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).

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