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The Boolean expression $P+\overline{P}Q$ , where P and Q are the inputs of the logic circuit, represents
(A) AND gate
(B) NAND gate
(C) NOT gate
(D) OR gate

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Answer
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Hint:
In the given question, we have been provided with a Boolean expression and have been asked to tell us which logic gate the output represents. In all questions regarding Boolean expression and logic gates, our approach should be to draw the truth table of the given expression and then compare the value obtained with the truth table of known logic gates. Let’s see the detailed step by step solution.

Complete step by step solution:
We have been given two inputs P and Q and the expression has been formed using a combination of the sum and products of the inputs or their complements. The truth table for the given Boolean expression can be drawn as given below.
$P$ $Q$ $\overline{P}$ $\overline{P}Q$ $P+\overline{P}Q$
00100
01111
10001
11001

Now that we have drawn the truth table for the given Boolean expression, we have to compare the values in the truth table against the values of the output given by the logic gates mentioned in the options. For that, we will draw another truth table that will show us the results of the application of AND, NAND and OR gate respectively.
We need not draw the truth table for option (C) because the NOT gate is only applicable for a single input and hence has no use in case of two inputs. We can therefore leave NOT gate out of the options under consideration.
The truth table for the other logic gates is drawn below. Note that NAND gate is the complement of the AND logic gate.
$P$ $Q$ $P.Q$ (AND gate) $\overline{P.Q}$ (NAND gate) $P+Q$ (OR gate)
00010
01011
10011
11101

Comparing the two truth tables, we can say that the logic gate that best resembles our given Boolean expression is the OR gate.
Therefore, we can say that option (D) is the correct answer.

Note:
It is very important to have the basic knowledge about some of the very common logic gates as it helps us to solve questions. For example, if we did not know that OR gate produces a false output only if both the input values are false and in all the other cases, it produces a true output, our work would not have been this easy. We should also know how many inputs a particular logic gate requires as it helped us to eliminate option (C).