250+ TOP MCQs on Digital Electronics and Logic Gates | Class12 Physics

Physics Multiple Choice Questions on “Digital Electronics and Logic Gates”.

1. Which of the following is correct about logic gates?
a) Logic gates have one or more input signals and only one output signal
b) Logic gates have only one input and output signal
c) Logic gates are analogous circuits
d) Logic gates have only one input and many output signals
Answer: a
Clarification: Logic gates are the basic building blocks of any digital system. A digital circuit with one or more input signals but only one output signal is known as a logic gate. All the other statements are not valid.

2. How many basic logic gates are there?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The logic gates are considered to be the building blocks of a digital system. Each logic gate follows a certain logical relationship between input and output voltage. There are 3 basic logic gates, namely, OR gate, AND gate, and NOT gate.

3. Which of the following gates can have only one input?
a) OR gate
b) NOT gate
c) AND gate
d) NAND gate
Answer: b
Clarification: The NOT gate is the simplest of all logic gates. It has only one input and one output signal. NOT gate is also called as the inverter because t inverts the input signal. All the other gates can have one or more input signals.

4. The truth table shows both the input and output signals.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. The truth table is a table that shows all possible input combinations and the corresponding output combinations for a particular logic gate. These are used in connection with Boolean algebra, Boolean functions, and propositional calculus.

5. Which gate will a NAND gate be equivalent to when two inputs of NAND gates are shorted?
a) AND gate
b) OR gate
c) NAND gate
d) NOT gate
Answer: d
Clarification: When two inputs of a NAND gate are shorted, then the Boolean expression for it becomes:
Y=not(A.A)
Y=not(A)
Therefore, the NAND gate will be equivalent to the NOT gate.

6. What should be done to obtain an OR gate from a NAND gate?
a) We need only 3 NAND gates
b) We need two NOT gates obtained from NAND gates and one NAND gate
c) We need 3 NOT gates obtained from NAND gates and 3 NAND gates
d) We need 2 NAND gates and 4 AND gates obtained from NAND gates.
Answer: b
Clarification: We need OR gate (Y = A + B) from NAND gate (Y = not (A.B))
Y = not (not (A.B))
Y = not (not (A)) + not (not (B))        [Using the Boolean identify ➔ not (A.B) = not (A) + not (B)]
So, Y = A + B [Since not (not (A)) = A and not (not (B)) = B]
Therefore, to obtain an OR gate from a NAND gate, we need two NOT gates obtained from NAND gates and one NAND gate.

7. In Boolean algebra, what will not (A + not (B)).C) be equal to?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: not (A + not (B)).C) = not (A + not (B)) + not (C)     [From De Morgan’s theorem]
not (A + not (B)).C) = not (A).not (B) + not(C)     [From De Morgan’s theorem]
not (A + not (B)).C) = not (A).B + not (C)