Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

4. 

Derive the Boolean expression for the logic circuit shown below:

[A].
[B].
[C].
[D].

Answer: Option A

Explanation:

No answer description available for this question.

Anji said: (Feb 16, 2011)  
OR gate o/p is A+B and 1st AND gate o/p is (A+B)C and the 2nd AND gate o/p is ((A+B)C)D and the last AND gate o/p is

(((A+B)C)D)E and finally by simplifying it we ll get as C(A+B)DE

Vinay Sharma(Shoolini University-73) said: (Oct 4, 2011)  
B and D are not corret because there is bar on E, and in option D it should be A+B because there is or gate for this input. So option A is correct.

Aruna said: (Jan 23, 2012)  
OR GATE OP=(A+B)
1ST AND OP=(A+B)C
2ND AND OP=[(A+B)C]D
3RD AND OP=[(A+B)C]D)E
ANS: [(A+B)C]D)E ....{AFTER SIMPLIFICATION}

Nakowa said: (Mar 22, 2012)  
Thanks for you explaination. And any one here please can prove 1+1=1 using boolean algebra expration? Thanks again.

Parag said: (Dec 18, 2015)  
1+1=1.

In Boolean algebra its a rule, 1+1=1.

Veyant said: (Sep 3, 2017)  
Yeah, option 1is correct but explain me the associativity and explain the difference between opt 1&opt c.

Orwa Dalmas said: (Apr 20, 2019)  
Thanks all for explaining.

Vishnu said: (Sep 22, 2019)  
1+1 = 2.

The binary value of 2 is 10 hence 1+0=1.

Hence proved.

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