Question

In the logic circuit diagram, when all the four inputs, A, B, C, D are 'one' the outputs \(Y_1\), \(Y_2\), \(Y_3\) are respectively (1, 1, 0). When the inputs A and C are changed to zero and B and D are still 'one', then the outputs \(Y_1\), \(Y_2\), \(Y_3\) are respectively change to

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

Hint:

Y1 = NOT(NOR(A, B)) = A OR B, Y2 = C AND D, Y3 = NOR(Y1, Y2).

Answer 1

The Correct Option is B

Step 1: Reading the circuit
The figure shows a NOR gate on inputs A and B feeding a NOT gate whose output is \(Y_1\). Inputs C and D go to an AND gate with output \(Y_2\). A final NOR gate takes \(Y_1\) and \(Y_2\) and gives \(Y_3\).

Step 2: Simplify the gates
\(Y_1 = \overline{\overline{A + B}} = A + B\) (OR). \(Y_2 = C\cdot D\). \(Y_3 = \overline{Y_1 + Y_2}\).
Check the given case: all inputs 1 gives \(Y_1 = 1\), \(Y_2 = 1\), \(Y_3 = 0\), matching (1, 1, 0).

Step 3: New inputs
Now \(A = 0\), \(B = 1\), \(C = 0\), \(D = 1\).
\(Y_1 = 0 + 1 = 1\)
\(Y_2 = 0 \cdot 1 = 0\)
\(Y_3 = \overline{1 + 0} = 0\)
So the outputs are (1, 0, 0), option (B).

Final Answer:
The outputs are \(Y_1, Y_2, Y_3 = 1, 0, 0\), option (B). \[ \boxed{(1, 0, 0)} \]