Question

The Boolean expression for the given combination of logic gates is

• A. \(Y = (A+B)\cdot \overset{̄}{C}\)
• B. \(Y = (A\cdot B)\cdot \overset{̄}{C}\)
• C. \(Y = (A+B)+C\)
• D. \(Y = (A\cdot B)+\overset{̄}{C}\)

Hint:

Trace the signal through each gate: AND on A and B, NOT on C, then OR.

Answer 1

The Correct Option is D

Step 1: Trace
Level one: gate AND takes $A$ and $B$. Gate NOT takes $C$. Level two: gate OR joins them.

Step 2: Boolean form
AND is multiplication, OR is addition, NOT is the bar. So $Y = AB + \overline{C}$.

Step 3: Test a case
If $A = B = 0$ and $C = 1$, then $AB = 0$ and $\overline{C} = 0$, so $Y = 0$. If $C = 0$, then $Y = 1$ always.

Step 4: Match
This agrees with option (D) and disagrees with the others, which give $Y = 0$ whenever $C = 0$.

Final Answer:
The expression is AB + C-bar. This is option (D). \[ \boxed{\text{(D) }Y=(A\cdot B)+\overline{C}} \]