Question

The diagram of a logic gate circuit is given below. The output $Y$ of the circuit is represented by

• A. $A\cdot (B+C)$
• B. $A\cdot (B\cdot C)$
• C. $A+(B\cdot C)$
• D. $A+(B+C)$

Answer 1

The correct option is C $A+(B\cdot C)$

Output of upper OR gate $=A+B$
Output of lower OR gate $=A+C$
Net output $Y=(A+B)(A+C)$
$=AA+AC+BA+BC$
$=A(1+C)+BA+BC[\because AA=A]$
$=A+BA+BC[\because 1+C=1]$
$=A(1+B)+BC[\because 1+B=1]$
$=A+BC=A+(B\cdot C)$.