Question

Consider the logic circuit with inputs A,B,C, and output Y. How many combinations of A, B and C give the output Y=0?

• A. 8
• B. 5
• C. 7
• D. 1

Answer 1

The Correct Option is C

In a logic circuit with inputs A, B, and C, each input can take values 0 or 1. Since there are three binary inputs, the total number of input combinations is:

$2^3 = 8$

We are interested in the number of combinations that result in output $Y = 0$.

From the logic of the circuit, the combinations that result in $Y = 0$ are: 

1. $A = 0$, $B = 0$, $C = 0$
2. $A = 0$, $B = 0$, $C = 1$
3. $A = 0$, $B = 1$, $C = 0$
4. $A = 1$, $B = 0$, $C = 0$
5. $A = 0$, $B = 1$, $C = 1$
6. $A = 1$, $B = 0$, $C = 1$
7. $A = 1$, $B = 1$, $C = 0$

So, the number of combinations giving output $Y = 0$ is 7.

Correct option: (C): 7

Answer 2

Given Boolean Expression:

$D = \overline{\overline{(A + B)} \cdot C}$ 

Using De Morgan’s Law: $\overline{P \cdot Q} = \overline{P} + \overline{Q}$

So, this becomes:

$D = \overline{\overline{(A + B)} + \overline{C}}$

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Case 1: $A = 0$, $B = 0$, $C = 0$

• $A + B = 0 + 0 = 0$
• $\overline{(A + B)} = \overline{0} = 1$
• $\overline{C} = \overline{0} = 1$
• Now: $D = \overline{1 + 1} = \overline{1} = 0$

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Case 2: $A = 1$, $B = 1$, $C = 0$

• $A + B = 1 + 1 = 1$
• $\overline{(A + B)} = \overline{1} = 0$
• $\overline{C} = \overline{0} = 1$
• Now: $D = \overline{0 + 1} = \overline{1} = 0$

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Case 3: $A = 0$, $B = 1$, $C = 1$

• $A + B = 0 + 1 = 1$
• $\overline{(A + B)} = \overline{1} = 0$
• $\overline{C} = \overline{1} = 0$
• Now: $D = \overline{0 + 0} = \overline{0} = 1$

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Conclusion:

The output $D$ depends on the specific values of $A$, $B$, and $C$. The expression simplifies using De Morgan's Law, and for each combination, we calculate step by step. As shown above, when the sum inside is 1, and $\overline{C}$ is also 1, the result becomes 0.