In the following combinations of logic gates, | vedclass

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(B) For combination $(i)$:The inputs to the $NAND$ gate are $1$ and $1$,so its output is $\overline{1 \cdot 1} = 0$. This $0$ passes through a $NOT$ g

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$0, 1, 0$

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(B) For combination $(i)$:The inputs to the $NAND$ gate are $1$ and $1$,so its output is $\overline{1 \cdot 1} = 0$. This $0$ passes through a $NOT$ gate,becoming $1$. The inputs to the $NOR$ gate are $1$ and $0$. The output of a $NOR$ gate is $1$ only if both inputs are $0$. Since one input is $1$,the output $A = 0$.For combination $(ii)$:The inputs $0$ and $1$ pass through $NOT$ gates,becoming $1$ and $0$ respectively. These are the inputs to a $NAND$ gate. The output is $\overline{1 \cdot 0} = \overline{0} = 1$. Thus,$B = 1$.For combination $(iii)$:The inputs $1$ and $1$ pass through a $NOR$ gate,giving an output of $\overline{1 + 1} = 0$. This $0$ and the input $1$ are fed into an $AND$ gate. The output is $0 \cdot 1 = 0$. Thus,$C = 0$.Therefore,the outputs are $A = 0, B = 1, C = 0$.

In the following combinations of logic gates,the outputs of $A, B$ and $C$ are respectively:

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