Logic Gates Practice Questions

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$Z=\overline{\overline{P\bar{Q}}\cdot Q\cdot \overline{Q\cdot R} }$ $\;\;=P\bar{Q}+\bar{Q}+Q R$ $\;\;=\bar{Q}(P+1)+QR$ $\;\;=\bar{Q}+QR$ $\;\;=(\bar{Q}+Q)(\bar{Q}+R)$ $\;\;=\bar{Q}+R$

$F=\bar{A}\bar{C}+BC$

Truth table of XOR gate is, So, from the XOR gate truth table, it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switched irrespective of the state of the other switch.

$Y=x\bar{\bar{x}}+\bar{x}\bar{x}=1$

NOR and NAND are designated as universal logic gates because using any one of them, we can omplement all the logic gates.

$Y=A\oplus B\oplus C\oplus D$ from the given diagram we know that sum of anu no. of bits is XOR of all bits. So, $S=A\oplus B\oplus C\oplus D\oplus Y$ $S=Y\oplus Y$ $S=$ either zero or even because LSB is zero (always)

$Y=\overline{(\bar{P}\cdot \bar{Q})\cdot (\bar{R}\cdot \bar{S})}$ $\because \;\; \overline{\bar{A}\cdot \bar{B}}=(A+B)$ $\therefore \;Y=\overline{\bar{P}\cdot \bar{Q}}+\overline{\bar{R}\cdot \bar{S}}$ $\Rightarrow \;Y=(P+Q+R+S)$

$Y=\bar{A}\bar{B}=\overline{A+B}\rightarrow$ NOR GATE $Y=AB+\bar{A}\bar{B}\rightarrow$ EX-NOR GATE