The inputs A, B and C to be given in order to get an output Y = 1 from the following circuit are
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In logic-gate problems, solving backward from the final output is often the fastest method.
For an AND gate output to be \(1\), every input entering the gate must already be \(1\).
Concept:
Logic gates operate on binary values \(0\) and \(1\) according to the rules of Boolean algebra.
To determine the required inputs, we analyze the circuit gate by gate.
The circuit contains:
• A NOR gate with inputs \(A\) and \(B\)
• An AND gate receiving the NOR output and input \(C\)
The Boolean expressions are:
\[
X = \overline{A+B}
\]
for the NOR gate, and
\[
Y = X \cdot C
\]
for the AND gate.
Important Gate Rules: • NOR Gate:
Output becomes \(1\) only when all inputs are \(0\).
\[
\overline{0+0}=1
\]
• AND Gate:
Output becomes \(1\) only when all inputs are \(1\).
\[
1 \cdot 1 = 1
\]
Step 1: Analyze the final AND gate.
The final output is
\[
Y = X \cdot C
\]
We need
\[
Y = 1
\]
For an AND gate to produce output \(1\), both inputs must be \(1\).
Therefore,
\[
X = 1
\quad \text{and} \quad
C = 1
\]
Step 2: Analyze the NOR gate output.
The first gate is a NOR gate:
\[
X = \overline{A+B}
\]
From Step 1, we already know that
\[
X = 1
\]
A NOR gate gives output \(1\) only when both inputs are \(0\).
Hence,
\[
A = 0
\quad \text{and} \quad
B = 0
\]
Step 3: Determine all required input values.
Combining the results obtained:
\[
A=0,\qquad B=0,\qquad C=1
\]
Step 4: Verification using Boolean expression.
First gate output:
\[
X = \overline{A+B}
\]
Substituting \(A=0\) and \(B=0\),
\[
X = \overline{0+0}
\]
\[
X = \overline{0}
\]
\[
X = 1
\]
Now evaluate the final AND gate:
\[
Y = X \cdot C
\]
Substituting \(X=1\) and \(C=1\),
\[
Y = 1 \cdot 1
\]
\[
Y = 1
\]
Thus, the condition is satisfied.
Final Answer:
\[
\boxed{A=0,\; B=0,\; C=1}
\]
Hence, the correct option is:
\[
\boxed{\text{Option (E)}}
\]
