The behavior of the circuit is like which gat | vedclass

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(A) Let the inputs be $A$ and $B$. The circuit consists of an $AND$ gate and an $OR$ gate whose outputs are fed into a final $OR$ gate.Let the output

Vedclass · Accepted Answer

$OR$

Vedclass · Answer

(A) Let the inputs be $A$ and $B$. The circuit consists of an $AND$ gate and an $OR$ gate whose outputs are fed into a final $OR$ gate.Let the output of the $AND$ gate be $Y_1 = A \cdot B$.Let the output of the first $OR$ gate be $Y_2 = A + B$.The final output $X$ is the $OR$ operation of $Y_1$ and $Y_2$:$X = Y_1 + Y_2 = (A \cdot B) + (A + B)$.Using the Boolean identity $(A \cdot B) + A + B = A + B$,we get $X = A + B$.Truth Table:$A, B$$Y_1 = A \cdot B$$Y_2 = A + B$$X = Y_1 + Y_2$$0, 0$$0$$0$$0$$0, 1$$0$$1$$1$$1, 0$$0$$1$$1$$1, 1$$1$$1$$1$Since the output $X$ follows the truth table of an $OR$ gate,the circuit behaves like an $OR$ gate.

The behavior of the circuit is like which gate?

Similar Questions

Assertion: The logic gate $NOT$ can be built using a diode.
Reason: The output voltage and the input voltage of the diode have a $180^o$ phase difference.

For a $NAND$ gate, the inputs and outputs are given in the table below. The values taken by $C, D, E, F$ are respectively:
Input $A$ Input $B$ Output $Y$
$0$ $1$ $C$
$0$ $0$ $D$
$1$ $0$ $E$
$1$ $1$ $F$

To get an output $Y=1$ from the circuit shown,the inputs $A, B,$ and $C$ must be respectively:

The given truth table is correct for:
$P: 1, 1, 0, 0$
$Q: 1, 0, 1, 0$
$R: 1, 0, 0, 0$

The correct truth table for the given logic gate circuit is:

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