Discussion :: Boolean Algebra and Logic Simplification  General Questions (Q.No.4)
4.  Derive the Boolean expression for the logic circuit shown below: 

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 University73) 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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