Name the gate which represents the Boolean expression, \[Y=\overline{A\text{ }.\text{ }B}\]
A.NAND
B.AND
C.NOT
D.NOR
Answer
Verified
466.2k+ views
Hint: The expressions in Boolean Algebra can be reduced to simple terms very easily using identities. The reduced expressions can be checked with the binary digits 0 and 1 to determine whether the given expression is a NAND, AND, NOT or NOR equivalent.
Complete answer:
There are mainly three types of gates in Boolean algebra. They are:
AND: It is a binary operation or gate which gives output only if both the statements are TRUE. The table of AND operation looks like:
OR: It is a binary operation which gives the output if any of the variables is TRUE (1). The table of OR (+) operation looks like:
NOT: It is a unary operation which inverts the variable. The table of NOT operation looks like:
Now let us consider the given expression,
\[Y=\overline{A\text{ }.\text{ }B}\]
We can use the similar tables to find the equivalent expression.
When we look at the basic gates, we can release that the table of \[Y=\overline{A\text{ }.\text{ }B}\] is the NOT operation on the AND operation.
The table of NAND operation can be derived from AND table as:
Therefore, the equivalent gate for \[Y=\overline{A\text{ }.\text{ }B}\] is a NAND gate.
The correct solution is option A.
Note:
Using the de Morgan’s theorem for Boolean algebra, \[Y=\overline{A\text{ }.\text{ }B}\] can be written as \[\overline{A\text{ }.\text{ }B}=\overline{A}+\overline{B}\]. We can find the solution to the question by using this theorem also. We need to find the complement values of A and B separately and use the OR operator.
All the operations in Boolean algebra can be derived from each other.
Complete answer:
There are mainly three types of gates in Boolean algebra. They are:
AND: It is a binary operation or gate which gives output only if both the statements are TRUE. The table of AND operation looks like:
OR: It is a binary operation which gives the output if any of the variables is TRUE (1). The table of OR (+) operation looks like:
NOT: It is a unary operation which inverts the variable. The table of NOT operation looks like:
Now let us consider the given expression,
\[Y=\overline{A\text{ }.\text{ }B}\]
We can use the similar tables to find the equivalent expression.
When we look at the basic gates, we can release that the table of \[Y=\overline{A\text{ }.\text{ }B}\] is the NOT operation on the AND operation.
The table of NAND operation can be derived from AND table as:
Therefore, the equivalent gate for \[Y=\overline{A\text{ }.\text{ }B}\] is a NAND gate.
The correct solution is option A.
Note:
Using the de Morgan’s theorem for Boolean algebra, \[Y=\overline{A\text{ }.\text{ }B}\] can be written as \[\overline{A\text{ }.\text{ }B}=\overline{A}+\overline{B}\]. We can find the solution to the question by using this theorem also. We need to find the complement values of A and B separately and use the OR operator.
All the operations in Boolean algebra can be derived from each other.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Class 12 Question and Answer - Your Ultimate Solutions Guide
Master Class 12 Economics: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
What are the major means of transport Explain each class 12 social science CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
What is a transformer Explain the principle construction class 12 physics CBSE
Explain sex determination in humans with the help of class 12 biology CBSE