Boolean Algebra MCQ Quiz in தமிழ் - Objective Question with Answer for Boolean Algebra - இலவச PDF ஐப் பதிவிறக்கவும்

Concept-

• The branch of algebra which deals with the values of the variables in the form of the truth values true and false is called Boolean algebra.
• True and false are usually denoted by 1 and 0 respectively.

Explanation-

If we take complement, we get negation of the variable but if we again take the complement of complemented variable, we get the same variable.

$F\left(A\right)=\bar{\overline{A}}=A$

Laws of Boolean Algebra:

Application:

f(A, B) = ∑ m(0, 1, 2, 3)

= A̅ B̅ + A̅ B + A B̅ + AB

= A̅ ( B + B̅) + A (B̅ + B)

= A̅ + A = 1

F(A, B, C) = C + AB’C’

F(A, B, C) = (C + C’)(C + AB’)

The given circuit gives the output:

Z(a, b, c) = $a+\stackrel{\leftharpoonup }{b}c$ 

Expanding it into canonical form to obtain the minterms

Z(a, b, c) = $a\left(b+\overline{b}\right)\left(c+\overline{c}\right)+\left(a+\overline{a}\right)\overline{b}c$ 

= $abc+ab\overline{c}+a\overline{b}c+a\overline{b}\overline{c}+\overline{a}\overline{b}c$

After rearranging the canonical terms, this corresponds to min-terms: ∑ (1,4, 5, 6, 7)

Alternate solution:

The output of the circuit is $a+\stackrel{\leftharpoonup }{b}c$ 

K Map for this Boolean expression

The above K Map corresponds to min-terms: ∑ (1,4, 5, 6, 7)

From truth table:

F(X, Y) = XY̅ + XY

= X[Y̅ + Y]

$f\left(X,Y,Z,W\right)=X\overline{Y}Z\overline{W}+\overline{X}\overline{Y}\overline{Z}\overline{W}+X\overline{Y}\overline{Z}\overline{W}+\overline{X}\overline{Y}Z\overline{W}+\overline{X}Y\overline{Z}W+\overline{X}YZW+XY\overline{Z}W+XYZW$

$f\left(X,Y,Z,W\right)=\overline{X}\overline{Y}\overline{Z}+\overline{X}\overline{Y}Z\overline{W}+\overline{X}Y\overline{Z}W+\overline{X}YZW+X\overline{Y}\overline{Z}\overline{W}+X\overline{Y}Z\overline{W}+XY\overline{Z}W+XYZW$

$f\left(X,Y,Z,W\right)=\overline{X}\overline{Y}\overline{W}\left(Z+\overline{Z}\right)+\overline{X}YW\left(Z+\overline{Z}\right)+X\overline{Y}\overline{W}\left(Z+\overline{Z}\right)+XYW\left(Z+\overline{Z}\right)$

$f\left(X,Y,Z,W\right)=\overline{X}\overline{Y}\overline{W}+X\overline{Y}\overline{W}+\overline{X}YW+XYW$

$f\left(X,Y,Z,W\right)=\left(\overline{X}+X\right)\left(\overline{Y}\overline{W}\right)+\left(\overline{X}+X\right)\left(YW\right)$

$f\left(X,Y,Z,W\right)=\overline{Y}\overline{W}+YW$

$f\left(X,Y,Z,W\right)=Y\odot W$

XNOR gate with NAND gates:

Hence 5 NAND gates is needed

Tips and Tricks:

The above Boolean function can also be solved using K-map:

X = ABC + A̅B + ABC̅

Rearranging:

X = ABC + ABC̅ + A̅B

X = AB(C + C̅) + A̅B

X = AB + A̅B   [C + C̅ = 1]

X = (A + A̅)B  

X = 1.B   [A + A̅ = 1]

Concept-

• The branch of algebra which deals with the values of the variables in the form of the truth values true and false is called Boolean algebra.
• True and false are usually denoted by 1 and 0 respectively

Calculation:

Let F(A, B) = A + A̅ .B + A.B̅

F(A, B) = A (1 + B̅) + A̅.B

F(A, B) = A + A̅.B

F(A, B) = (A + A̅).(A + B)

F(A, B) = A + B 

A $ B = A̅ + B (1)

W = A + B̅ (2)

Y = W̅ $ A̅

Y = W + A̅ (From 1)

Y = A + B̅ + A̅ (From 2)

The correct answer is "option 1".

EXPLANATION:

Option 1: TRUE

The truth table X is:

F is equal to X.

Option 2: FALSE

The truth table X+Y is:

F is not equal to X+Y.

Option 3: FALSE

The truth table X’Y + XY' is:

F is not equal to X’.Y +X.Y'.

Option 4: FALSE

The truth table Y is :

 F is not equal to Y.

Hence, the correct answer is "option 1".

From truth table, the sum of minterms

F(X, Y) = X.Y̅ + X.Y = X(Y̅ + Y) = X