The correct output for the given logic circui | vedclass

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(D) The logic circuit consists of a $NOR$ gate followed by an $AND$ gate. The inputs to the $NOR$ gate are $A$ and $B$. The output of the $NOR$ gate i

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(D) The logic circuit consists of a $NOR$ gate followed by an $AND$ gate. The inputs to the $NOR$ gate are $A$ and $B$. The output of the $NOR$ gate is $\overline{A+B}$.This output is then fed into an $AND$ gate along with input $B$. Thus,the final output $C$ is given by:$C = (\overline{A+B}) \cdot B$Let's analyze the truth table for this expression:If $A=0, B=0$,then $C = (\overline{0+0}) \cdot 0 = 1 \cdot 0 = 0$.If $A=1, B=0$,then $C = (\overline{1+0}) \cdot 0 = 0 \cdot 0 = 0$.If $A=0, B=1$,then $C = (\overline{0+1}) \cdot 1 = 0 \cdot 1 = 0$.If $A=1, B=1$,then $C = (\overline{1+1}) \cdot 1 = 0 \cdot 1 = 0$.Wait,let's re-examine the circuit diagram. The input $B$ is connected to both the $NOR$ gate and the $AND$ gate. The output of the $NOR$ gate is $\overline{A+B}$. The $AND$ gate takes this and $B$ as inputs. So $C = (\overline{A+B}) \cdot B$. Since $(\overline{A+B}) \cdot B = (\overline{A} \cdot \overline{B}) \cdot B = \overline{A} \cdot (\overline{B} \cdot B) = \overline{A} \cdot 0 = 0$,the output is always $0$.

The correct output for the given logic circuit and inputs is:

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