
In the Boolean algebra $\overline{(A+\overline{B})\cdot C}$ will be equal to.
a)$(\overline{A}\cdot B)+\overline{C}$
b) $(A\cdot \overline{B})+C$
c) $(A\cdot B)\cdot \overline{C}$
d) $(A+B)+C$
Answer
579.6k+ views
Hint: Using the Boolean algebra rules of the OR, AND and DeMorgans rules the above question can be simplified and made to a simpler form. For using logic gates, the input method takes all the possible combinations of the inputs.
Complete answer:
Formula to be used:
AND Gate rules are as follows,
$\begin{align}
& A\cdot A=A \\
& A\cdot \overline{A}=0 \\
& A\cdot 1=A \\
& A\cdot 0=0 \\
\end{align}$
OR Gate rules are as follows,
$\begin{align}
& A+A=A \\
& A+\overline{A}=1 \\
& A+1=1 \\
& A+0=A \\
\end{align}$
$A+\overline{A}B=A+B$ Adsorption rule
$\overline{A+B}=\overline{A}\centerdot \overline{B}$ DeMorgans rule for OR gate
$\overline{A\centerdot B}=\overline{A}+\overline{B}$ DeMorgans rule for AND gate
$\overline{\overline{A}}=A$ Double negation
Using the formulas we can simplify any given Boolean equation,
Let us solve the above question,
$\overline{(A+\overline{B})\cdot C}$
Using DeMorgans law for AND gate we have,
$\overline{(A+\overline{B})}+\overline{C}$
Again using DeMorgans law for OR gate the above equation reduces to,
$\overline{A}\centerdot \overline{\overline{B}}+\overline{C}$
$\because \overline{\overline{B}}=B$
From double negation, we can simplify the above Boolean expression to,
$\overline{A}\cdot B+\overline{C}$
Hence the correct answer to the question is option a.
Additional information:
In digital circuits there are only two values zeroes and ones i.e. minimum and maximum. In case of an analog circuit i.e. with battery and resistances there exists different values for current in the circuit. The digital circuits are established using logic gates. A logic gate is a digital circuit that is designed to perform a particular logical operation. As it works according to some logical relationship between the input and the output voltages, it is generally known as a logic gate.
Note:
While solving any question, first solve for AND gate and then proceed with OR gate as it becomes more convenient to proceed to the solution. The above question can also be solved by using input methods in logic gates but it will be time consuming and we would have to manipulate one option after the other. For deriving the initial results of the above formulae’s one can use the Logic gates input method.
Complete answer:
Formula to be used:
AND Gate rules are as follows,
$\begin{align}
& A\cdot A=A \\
& A\cdot \overline{A}=0 \\
& A\cdot 1=A \\
& A\cdot 0=0 \\
\end{align}$
OR Gate rules are as follows,
$\begin{align}
& A+A=A \\
& A+\overline{A}=1 \\
& A+1=1 \\
& A+0=A \\
\end{align}$
$A+\overline{A}B=A+B$ Adsorption rule
$\overline{A+B}=\overline{A}\centerdot \overline{B}$ DeMorgans rule for OR gate
$\overline{A\centerdot B}=\overline{A}+\overline{B}$ DeMorgans rule for AND gate
$\overline{\overline{A}}=A$ Double negation
Using the formulas we can simplify any given Boolean equation,
Let us solve the above question,
$\overline{(A+\overline{B})\cdot C}$
Using DeMorgans law for AND gate we have,
$\overline{(A+\overline{B})}+\overline{C}$
Again using DeMorgans law for OR gate the above equation reduces to,
$\overline{A}\centerdot \overline{\overline{B}}+\overline{C}$
$\because \overline{\overline{B}}=B$
From double negation, we can simplify the above Boolean expression to,
$\overline{A}\cdot B+\overline{C}$
Hence the correct answer to the question is option a.
Additional information:
In digital circuits there are only two values zeroes and ones i.e. minimum and maximum. In case of an analog circuit i.e. with battery and resistances there exists different values for current in the circuit. The digital circuits are established using logic gates. A logic gate is a digital circuit that is designed to perform a particular logical operation. As it works according to some logical relationship between the input and the output voltages, it is generally known as a logic gate.
Note:
While solving any question, first solve for AND gate and then proceed with OR gate as it becomes more convenient to proceed to the solution. The above question can also be solved by using input methods in logic gates but it will be time consuming and we would have to manipulate one option after the other. For deriving the initial results of the above formulae’s one can use the Logic gates input method.
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