# Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

### Discussion :: Boolean Algebra and Logic Simplification - General Questions (Q.No.4)

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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