Question

The output Y of the following digital logic circuit will be '1' (one) for the inputs

• A. \(A = 1\) , \(B = 0\) , \(C = 0\)
• B. \(A = 1\) , \(B = 0\) , \(C = 1\)
• C. \(A = 1\) , \(B = 1\) , \(C = 0\)
• D. \(A = 1\) , \(B = 1\) , \(C = 1\)

Hint:

Y = (NOT A) OR (B AND C).

Answer 1

The Correct Option is D

Step 1: Understanding the Figure:
Input A passes through a NOT gate. Inputs B and C go into an AND gate. The outputs of the NOT gate and the AND gate feed an OR gate, which gives Y.

Step 2: Boolean expression:
\[ Y = \overline{A} + B\cdot C \]

Step 3: Check each option:
(A) \(A = 1, B = 0, C = 0\): \(\overline A = 0\), \(BC = 0\), so \(Y = 0\).
(B) \(A = 1, B = 0, C = 1\): \(\overline A = 0\), \(BC = 0\), so \(Y = 0\).
(C) \(A = 1, B = 1, C = 0\): \(\overline A = 0\), \(BC = 0\), so \(Y = 0\).
(D) \(A = 1, B = 1, C = 1\): \(\overline A = 0\), \(BC = 1\), so \(Y = 0 + 1 = 1\).

Step 4: Conclusion:
With \(A = 1\), the NOT gate gives 0, so \(Y\) is 1 only when both \(B\) and \(C\) are 1.

Final Answer:
Y is 1 for \(A = B = C = 1\), option (D). \[ \boxed{A=1,\ B=1,\ C=1} \]