Laws of Boolean Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Laws of Boolean Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

De Morgan’s First Theorem:

• According to De Morgan’s first theorem, a NAND gate is equivalent to a Bubbled OR gate.
• The Boolean expressions for the bubbled OR gate can be expressed by the equation shown below.

\(\overline {A.B} = \bar A + \bar B\)

De Morgan’s second theorem:

• According to De Morgan’s first theorem, a NOR gate is equivalent to a Bubbled AND gate.
• The Boolean expressions for the bubbled AND gate can be expressed by the equation shown below.

\(\overline {A + B} = \bar A.\bar B\)

{A + A̅ ·B} = (A + A̅) (A + B) 

= 1 . (A + B) 

= (A + B)

Concept:

Some of the important boolean identities are as follows:

• 1 + A = 1
• \(A \space +\space \overline{A}\space = \space 1 \)
• 0 + A = A
• \(A \space\space \overline{A}\space = \space 0 \)

Calculation:

Y = (B + BC) (B + B̅C) (B + D)

Y = { B (1 + C) } { (B + B̅) (B + C) } (B + D)

Y = { B } { (B + C) } (B + D)

Y = BBB + BBD + BBC + BCD

Y = B + BD + BC + BCD

Y = B (1 + D) + BC (1 + D)

Y = B + BC

Y = B (1 + C)

Y = B 

Laws of Boolean Algebra: 

Application:

f(X, Y, Z) = ∑ m(2, 3, 4, 5)

= X̅YZ̅ + X̅YZ + XY̅Z̅ + XY̅Z

= X̅Y(Z + Z̅) + XY̅(Z + Z̅)

= X̅Y + XY̅

The correct option is 4

Concept:

De Morgan’s law states that:

\(\overline {\left( {{A_1}.{A_2} \ldots {A_n}} \right)} = \left( {\overline {{A_1}} + \overline {{A_2}} + \ldots + \overline {{A_n}} } \right)\)

\(\overline {\left( {{A_1} + {A_2} + \ldots + {A_n}} \right)} = \left( {\overline {{A_1}} \;.\;\overline {{A_2}} \;.\;..\;\overline {{A_n}} } \right)\)

∴ De-Morgan's expression  will be:

\(\overline {\left( {{A} + {B} + {C}} \right)} = \left( {\overline {{A}} \;.\;\overline {{B}} \;.\overline {{C}} } \right)\)

The correct answer is 'option 4'

Solution:

From the figure,

At 1 the combination is AB

At2 the combination is C

At 3 combination is \(\overline {(AB).C}\)

At 4 combination is \(\overline{(D+E)}\)

\(\implies Y=\overline{\overline{(ABC)}.\overline{(D+E)}}\)

Using De Morgan's law,

\(\implies Y=ABC+D+E\)

Concept:

Some of the important boolean expressions are:

1.) \(A + \bar{A} = 1\)

2.) \(A \bar{A} = 0\)

3.) \(1+A=1\)

4.) \(0+A= A\)

Calculation:

Given, \(f = A \overline B + AB + AC \overline D\)

\(f = A (\overline B + B) + AC \overline D\)

\(f = A + AC \overline D\)

\(f = A ( 1+ C \overline D)\)   : Use Domination Law 1 + A = 1 

\(f = A \)

\(\begin{array}{l} \left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\\ = \left( {\bar A + \bar B} \right)\left[ {\bar A + \left( {\overline {B + C} } \right)} \right] + A\left( {\bar B + \bar C} \right) \end{array}\)

\(\begin{array}{l} = \left( {\bar A + \bar B} \right)\left[ {\bar A + \bar B\bar C} \right] + A\bar B + A\bar C\\ = \bar A + \bar A\bar B\bar C + \bar A\bar B + \bar B\bar C + A\bar B + A\bar C \end{array}\)

\(\begin{array}{l} = \bar A\left( {1 + \bar B\bar C + \bar A\bar B} \right) + \bar B\bar C + A\bar B + A\bar C\\ = \bar A + \bar B\bar C + A\bar B + A\bar C \end{array}\)

= A̅ + A.B̅ + B̅.C̅ + A.C̅

= A̅ + B̅   + B̅.C̅ + A.C̅

= A̅ + B̅ (1 + C̅) + A.C̅

= A̅ + B̅ + A.C̅

=A̅ + C̅ + B̅ 

\(\begin{array}{l} = \bar A + \bar B + \bar C\\ = \overline {ABC} \end{array}\)

Concept:

Calculation:

Given,

F = AB + ABC + AB (D + E)

or, F = AB (1 + C) + AB (D + E) [∵ 1 + C = 1]

or, F = AB + AB (D + E)

or, F = AB (1 + D + E) [∵ 1 + D+ E = 1]

or, F = AB

Concept:

Analysis:

Given:

Y = (1+A+B+C)

By using Null Law;

Y = 1

Hence, option 4 is correct.