To get output '1' at R, for the given logic g | vedclass

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(C) From the circuit diagram, the inputs to the final $NOR$ gate are $P$ and $Q$.$P = \overline{X} + Y$$Q = \overline{X \cdot \overline{Y}}$The output

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$X = 1, Y = 0$

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(C) From the circuit diagram, the inputs to the final $NOR$ gate are $P$ and $Q$.$P = \overline{X} + Y$$Q = \overline{X \cdot \overline{Y}}$The output $R$ is given by the $NOR$ operation: $R = \overline{P + Q}$.Substituting the expressions for $P$ and $Q$:$R = \overline{(\overline{X} + Y) + (\overline{X \cdot \overline{Y}})}$Using De Morgan's Law, $\overline{A + B} = \overline{A} \cdot \overline{B}$:$R = (\overline{\overline{X} + Y}) \cdot (\overline{\overline{X \cdot \overline{Y}}})$$R = (X \cdot \overline{Y}) \cdot (X \cdot \overline{Y})$$R = X \cdot \overline{Y}$For $R = 1$, we must have $X = 1$ and $\overline{Y} = 1$, which implies $Y = 0$.Therefore, the input values must be $X = 1$ and $Y = 0$.

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

To get output '$1$' at $R$, for the given logic gate circuit the input values must be

Similar Questions

$A$ system of four gates is set up as shown. The truth table corresponding to this system is

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The combination of gates shown below yields:

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