# 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:
5 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:
7 years ago

Yeah, option 1is correct but explain me the associativity and explain the difference between opt 1&opt c.

Parag said:
9 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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