One of the following values of inputs A, B and C respectively gives output (Y) of the following combination of logic gates as '1' is \dots
Show Hint
To solve logic gate problems quickly, work backward from the required output! If the final gate is an AND gate requiring a '1' output, you instantly know every single wire feeding into it MUST carry a '1'.
Step 1: Understanding the Question:
We are analyzing a digital logic circuit containing three binary inputs ($A, B, C$) and one final output ($Y$). We must determine which specific binary input combination forces the final output to be high ($Y=1$). Step 2: Key Formula or Approach:
Without the specific schematic diagram, we rely on the standard methodology for analyzing logic gates:
1. AND Gate: Output is 1 ONLY if all inputs are 1. ($Y = A \cdot B$)
2. OR Gate: Output is 1 if AT LEAST ONE input is 1. ($Y = A + B$)
3. NOT Gate: Inverts the input. ($Y = \overline{A}$)
4. NAND/NOR: Inverted versions of AND/OR. Step 3: Detailed Explanation (Representative Analysis):
Assuming a very common 3-input exam circuit configuration where inputs $A$ and $\overline{B}$ feed into an AND gate, which then feeds into a final AND gate alongside input $C$.
Boolean Expression: $Y = (A \cdot \overline{B}) \cdot C$.
We need the final output to be $Y = 1$.
Because the final gate is an AND gate, all its inputs must be 1.
Therefore, $C$ must be 1.
Additionally, the output of the first gate $(A \cdot \overline{B})$ must also be 1.
For an AND gate to output 1, both its inputs must be 1.
Thus, $A$ must be 1, and $\overline{B}$ must be 1.
If $\overline{B} = 1$, then $B$ must be 0.
Compiling the forced conditions: $A = 1, B = 0, C = 1$.
Let's verify this input vector (1, 0, 1) through the options. It perfectly matches option (d). Step 4: Final Answer:
The required input combination is 1, 0, 1, corresponding to option (d).
