Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 4)
4.
Derive the Boolean expression for the logic circuit shown below:
Discussion:
9 comments Page 1 of 1.
DAKSHU said:
3 years ago
Thank you all.
Vishnu said:
4 years ago
1+1 = 2.
The binary value of 2 is 10 hence 1+0=1.
Hence proved.
The binary value of 2 is 10 hence 1+0=1.
Hence proved.
Orwa dalmas said:
5 years ago
Thanks all for explaining.
Veyant said:
6 years ago
Yeah, option 1is correct but explain me the associativity and explain the difference between opt 1&opt c.
Parag said:
8 years ago
1+1=1.
In Boolean algebra its a rule, 1+1=1.
In Boolean algebra its a rule, 1+1=1.
Nakowa said:
1 decade ago
Thanks for you explaination. And any one here please can prove 1+1=1 using boolean algebra expration? Thanks again.
Aruna said:
1 decade ago
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}
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}
Vinay sharma(shoolini university-73) said:
1 decade ago
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.
Anji said:
1 decade ago
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
(((A+B)C)D)E and finally by simplifying it we ll get as C(A+B)DE
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