The Boolean expression P + overline{P}Q,where | vedclass

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(D) Using the distributive law of Boolean algebra,the expression can be simplified as follows:$P + \overline{P}Q = (P + \overline{P}) \cdot (P + Q)$Si

Vedclass · Accepted Answer

$OR$ gate

Vedclass · Answer

(D) Using the distributive law of Boolean algebra,the expression can be simplified as follows:$P + \overline{P}Q = (P + \overline{P}) \cdot (P + Q)$Since $(P + \overline{P}) = 1$,the expression becomes:$1 \cdot (P + Q) = P + Q$The expression $P + Q$ represents the output of an $OR$ gate.Alternatively,we can construct the truth table:$P$$Q$$\overline{P}$$\overline{P}Q$$P + \overline{P}Q$$0$$0$$1$$0$$0$$1$$0$$0$$0$$1$$0$$1$$1$$1$$1$$1$$1$$0$$0$$1$The output column for $P + \overline{P}Q$ matches the truth table of an $OR$ gate. Thus,the correct option is $D$.

The Boolean expression $P + \overline{P}Q$,where $P$ and $Q$ are the inputs of the logic circuit,represents

Similar Questions

The following truth table corresponds to which logic gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$X: 0, 1, 1, 1$

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Assertion : $NAND$ or $NOR$ gates are called digital building blocks.
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In the circuit below,$A$ and $B$ represent two inputs and $C$ represents the output. The circuit represents:

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