Question 128 of 209

The output $Y$ for the given inputs $A$ and $B$ to the circuit is:

ABCD

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Answer: (B)

From the circuit:
Top AND gate output: $A \cdot B$
Bottom AND gate output: $\bar{A} \cdot B$ (the $A$ input passes through a NOT gate)
The OR gate combines these:
$Y = (A \cdot B) + (\bar{A} \cdot B)$

Factoring out $B$:
$Y = B \cdot (A + \bar{A})$

Using $A + \bar{A} = 1$:
$Y = B \cdot 1 = B$

So $Y$ is independent of $A$ and follows $B$ exactly.

Reading the input waveform of $B$:
From $t = 0$ to $t = 1\text{ s}$: $B = 1$
From $t = 1$ to $t = 2\text{ s}$: $B = 1$
From $t = 2$ to $t = 3\text{ s}$: $B = 0$

Hence, $Y$ is $1$ from $0$ to $2\text{ s}$ and $0$ from $2$ to $3\text{ s}$.

This matches the waveform shown in option $(2)$.

Hence, the correct option is $(2)$.

JEE Main Physics practice question on Semiconductors (Logic gates), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.