In the given circuit,the output Y becomes zer | vedclass

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Question

(A) The circuit consists of an $AND$ gate with inputs $A$ and $B$,followed by a $NAND$ gate. The input $C$ passes through a $NOT$ gate before entering

Vedclass · Accepted Answer

$A = 1, B = 1, C = 0$

Vedclass · Answer

(A) The circuit consists of an $AND$ gate with inputs $A$ and $B$,followed by a $NAND$ gate. The input $C$ passes through a $NOT$ gate before entering the $NAND$ gate.Let the output of the $AND$ gate be $X = A \cdot B$.The input $C$ passes through a $NOT$ gate,so its output is $\bar{C}$.The final output $Y$ is the output of the $NAND$ gate with inputs $X$ and $\bar{C}$,which is $Y = \overline{X \cdot \bar{C}} = \overline{(A \cdot B) \cdot \bar{C}}$.For the output $Y$ to be zero,the expression $(A \cdot B) \cdot \bar{C}$ must be $1$.This requires $A \cdot B = 1$ and $\bar{C} = 1$.$A \cdot B = 1$ implies $A = 1$ and $B = 1$.$\bar{C} = 1$ implies $C = 0$.Therefore,the inputs are $A = 1, B = 1, C = 0$.

In the given circuit,the output $Y$ becomes zero for the inputs

Similar Questions

What is the logic gate with one input and one output?

The output of the following logic circuit is:

The Boolean equation $\overline{A} \cdot \overline{B} = \dots$

The truth table shown below is for which logic gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$Y: 1, 0, 0, 1$

The real-time variation of input signals $A$ and $B$ are as shown below. If the inputs are fed into a $NAND$ gate,then select the output signal from the following.

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