# Logic Gates

 Question 1
In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a
 A NAND gate B NOR gate C XOR gate D XNOR gate
GATE EE 2019   Digital Electronics
Question 1 Explanation:
The Boolean expression for the output of the digital circuit is shown below

The above expression is of XOR gate.
 Question 2
In the logic circuit shown in the figure, Y is given by
 A Y = ABCD B Y = (A + B)(C + D) C Y = A + B + C + D D Y = AB + CD
GATE EE 2018   Digital Electronics
Question 2 Explanation:

 Question 3
For a 3-input logic circuit shown below, the output Z can be expressed as
 A $Q+\bar{R}$ B $P\bar{Q}+R$ C $\bar{Q}+R$ D $P+\bar{Q}+R$
GATE EE 2017-SET-2   Digital Electronics
Question 3 Explanation:

$Z=\overline{\overline{P\bar{Q}}\cdot Q\cdot \overline{Q\cdot R} }$
$\;\;=P\bar{Q}+\bar{Q}+Q R$
$\;\;=\bar{Q}(P+1)+QR$
$\;\;=\bar{Q}+QR$
$\;\;=(\bar{Q}+Q)(\bar{Q}+R)$
$\;\;=\bar{Q}+R$
 Question 4
Which of the following logic circuits is a realization of the function F whose Karnaugh map is shown in figure

 A A B B C C D D
GATE EE 2014-SET-1   Digital Electronics
Question 4 Explanation:

$F=\bar{A}\bar{C}+BC$
 Question 5
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles
 A and AND gate B an OR gate C an XOR gate D a NAND gate
GATE EE 2013   Digital Electronics
Question 5 Explanation:
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.
 Question 6
The output Y of the logic circuit given below is
 A 1 B 0 C X D $\bar{X}$
GATE EE 2011   Digital Electronics
Question 6 Explanation:

$Y=x\bar{\bar{x}}+\bar{x}\bar{x}=1$
 Question 7
The complete set of only those Logic Gates designated as Universal Gates is
 A NOT, OR and AND Gates B XNOR, NOR and NAND Gates C NOR and NAND Gates D XOR, NOR and NAND Gates
GATE EE 2009   Digital Electronics
Question 7 Explanation:
NOR and NAND are designated as universal logic gates because using any one of them, we can omplement all the logic gates.
 Question 8
A,B,C and D are input, and Y is the output bit in the XOR gate circuit of the figure below. Which of the following statements about the sum S of A,B,C,D and Y is correct ?
 A S is always with zero or odd B S is always either zero or even C S = 1 only if the sum of A,B,C and D is even D S = 1 only if the sum of A,B,C and D is odd
GATE EE 2007   Digital Electronics
Question 8 Explanation:
$Y=A\oplus B\oplus C\oplus D$ 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)
 Question 9
For the circuit shown in figure, the Boolean expression for the output Y in terms of inputs P, Q, R and S is
 A $\bar{P}+\bar{Q}+\bar{R}+\bar{S}$ B $P+Q+R+S$ C $\left ( \bar{P}+\bar{Q} \right )+\left ( \bar{R}+\bar{S} \right )$ D $\left ( P+Q \right )\left ( R+S \right )$
GATE EE 2002   Digital Electronics
Question 9 Explanation:
$Y=\overline{(\bar{P}\cdot \bar{Q})\cdot (\bar{R}\cdot \bar{S})}$ $\because \;\; \overline{\bar{A}\cdot \bar{B}}=(A+B)$
$\therefore \;Y=\overline{\bar{P}\cdot \bar{Q}}+\overline{\bar{R}\cdot \bar{S}}$
$\Rightarrow \;Y=(P+Q+R+S)$
 Question 10
The output of a logic gate is "1" when all its inputs are at logic "0". The gate is either
 A a NAND or an EX-OR gate B a NOR or an EX-OR gate C an AND or an EX-NOR gate D a NOR or an EX-NOR gate
GATE EE 2001   Digital Electronics
Question 10 Explanation:

$Y=\bar{A}\bar{B}=\overline{A+B}\rightarrow$ NOR GATE
$Y=AB+\bar{A}\bar{B}\rightarrow$ EX-NOR GATE
There are 10 questions to complete.