Question

The Boolean expression $P+\overline{P}Q$ where P and Q are the inputs of the logic circuits, represents.
$(a)$ AND gate
$(b)$ NAND gate
$(c)$ NOT gate
$(d)$ OR gate

Answer 1

- Hint: In this question consider the basics of the truth table of OR, AND and NOT gate for basic inputs of, 1. In general 4 basic cases will arise that is 0 operation 1, 0 operation 0, 1 operation 0 and 1 operation 1. Then try and create the same truth table considering the cases for the given Boolean expression that is $P+\overline{P}Q$. Compare the results obtained with the basic gates, this will help getting the right answer.

Complete step-by-step solution -

Given Boolean expression
$P+\overline{P}Q$
Where P and Q are the inputs of the logic circuit
Now make the truth table using these inputs and find out the output using the Boolean rule which is,
OR rule
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 1
AND rule
0.0 = 0
0.1 = 0
1.0 = 0
1.1 = 1
NOT rule or Inverted rule
$\overline{1}=0$
$\overline{0}=1$

S.No	Input (P)	Input (Q)	Output ($P+\overline{P}Q$)
1	0	0	(0+$\overline{0}$.0)=(0+1.0)=(0+0)=0
2	0	1	(0+$\overline{0}$.1)=(0+1.1)=(0+1)=1
3	1	0	(1+$\overline{1}$.0)=(1+0.0)=(1+0)=1
4	1	1	(1+$\overline{1}$.1)=(1+0.1)=(1+0)=1

So as we see that the Boolean expression $P+\overline{P}Q$ where P and Q are the inputs of the logic circuit follow the OR rule.
So the logic circuit represents the OR gate.
So this is the required answer.
Hence option (D) is the correct answer.

Note – The gates are very important in terms of electronic circuits or even in power systems as they can be used to control the circuit as they take very less time in comparison to odd switches to turn off and on. NAND gate is a basic gate formed by the combination of the NOT gate and the AND gate.