Question 172 of 209

In the following logic circuit the sequence of the inputs $A,B$ are $(0,0),(0,1),(1,0)$ and $(1,1).$ The output $Y$ for this sequence will be :

A$0,1,0,1$B$0,0,1,1$C$1,1,1,0$D$1,0,1,0$

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Answer: (C) $1,1,1,0$

Look at the above circuit, the value of $P$ can be found by the idea of AND gate, $P=A\cdot B$ and the value of $Q$, can be calculated by the idea of OR gate, $Q=A+B$. Then expression for $Y$ can be taken by the idea of NAND gate, $\Rightarrow Y=\bar{P\cdot Q}$ by the idea of Demorgan's theorem, $Y=\bar{P}+\bar{Q}=\bar{A}+\bar{B}+\bar{A}\cdot \bar{B}$


$\begin{matrix}A & B & Y=\bar{A}+\bar{B}+\bar{A}\cdot \bar{B} \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{matrix}$

JEE Main Physics practice question on Semiconductors (Logic gates), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.