Q192 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
The truth table for the given circuit is
- A
A B E 0 0 1 0 1 0 1 0 1 1 1 0 - B
A B E 0 0 1 0 1 0 1 0 0 1 1 0 - Correct answerC
A B E 0 0 0 0 1 1 1 0 0 1 1 1 - D
A B E 0 0 0 0 1 1 1 0 1 1 1 0
Solution
From the given logic circuit, the boolean expressions for the outputs of the two AND gates are:
$C = A \cdot B$
$D = \bar{A} \cdot B$
The final output $E$ from the OR gate is:
$E = C + D$
$E = A \cdot B + \bar{A} \cdot B$
Taking $B$ common, we get:
$E = B(A + \bar{A})$
Since $A + \bar{A} = 1$,
$E = B \cdot 1$
$E = B$
Therefore, the output $E$ is identical to the input $B$.
The corresponding truth table is:
$A = 0, B = 0 \Rightarrow E = 0$
$A = 0, B = 1 \Rightarrow E = 1$
$A = 1, B = 0 \Rightarrow E = 0$
$A = 1, B = 1 \Rightarrow E = 1$
Answer:
$C = A \cdot B$
$D = \bar{A} \cdot B$
The final output $E$ from the OR gate is:
$E = C + D$
$E = A \cdot B + \bar{A} \cdot B$
Taking $B$ common, we get:
$E = B(A + \bar{A})$
Since $A + \bar{A} = 1$,
$E = B \cdot 1$
$E = B$
Therefore, the output $E$ is identical to the input $B$.
The corresponding truth table is:
$A = 0, B = 0 \Rightarrow E = 0$
$A = 0, B = 1 \Rightarrow E = 1$
$A = 1, B = 0 \Rightarrow E = 0$
$A = 1, B = 1 \Rightarrow E = 1$
Answer:
| A | B | E |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Hence, the correct answer is option: 'C'.