Question

The following arrangement performs the logic function of _____ gate. All the gates present below are NAND gates.

A. AND
B. OR
C. NAND
D. EXOR

Answer 1

Hint: In this solution first we have to find out the expression for the first two gates A and B and then we can find out the expression for Y.
So to solve this question we have to first know about logic gates-
A logic gate is a physical electronic device executing a Boolean function is a logic gate.
There are many types of logic gates- The most important ones are- NAND gate and NOR gate.
Both NAND and NOR gates are known as universal logic gates because any logic gate can be formed from these two gates.

Complete step by step solution:
NAND gate is actually formed by the combination of NOR and AND gates.
The truth table for NAND gate is-

A	B	Output
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

As we can see from the above truth table, a NAND gate gives the output zero if both the inputs have value $1$ otherwise the outputs are $1$.
Here all the gates in the figure are NAND gates.
So, the output of left side gates will be $\overline A $ and $\overline B $ (since the inputs are $A$ and $B$)
So, the output of right side gate Y will be $\overline {\overline A .\overline B } $

Now we have to apply the De Morgan’s theorem to the above expression.
So, Y= $\overline {\overline A .\overline B } = \overline {\overline A } + \overline {\overline B } = A + B$
Truth table for OR gate-

A	B	Output
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

From the above truth table we can see that the expression of Y $ = A + B$ is the expression for OR gate also when the inputs are $A$ and $B$.
Thus the given arrangement performs the logic function of OR gate.
Hence option B is correct.

Note: For solving this question we have to remember De Morgan’s law and the truth table for all the gates. According to De Morgan’s law a NAND gate is equal to a bubbled OR gate and an OR gate is equal to a bubbled AND gate.