Question

The output of the given logic circuit is

• A. $A\cdot \overline{B}$
• B. $\overline{A}\cdot B$
• C. $A\cdot B+\overline{A\cdot B}$
• D. $A\cdot \overline{B}+\overline{A}\cdot B$

Answer 1

The correct option is A $A\cdot \overline{B}$

The output from the first gate $Y_1$ is,

$Y_1=\overline{A.B}$

Now, the inputs to the second gate $Y_2$ are $A\text{and}{Y}_1$.

$\Rightarrow Y_2=\overline{A.Y_1}=\overline{A.\overline{(A.B)}}$

Using DeMorgan's Theorem,

$Y_2=\overline{A}+\overline{\overline{(A.B)}}$

$\Rightarrow Y_2=\overline{A}+(A.B)$

Again for third gate,

$Y_3=B+Y_1=B+\overline{(A.B)}$

$\Rightarrow Y_3=B+\overline{A}+\overline{B}$

As $B+\overline{B}=1$

$\Rightarrow Y_3=\overline{A}+1=1$

Now the final output $Y$ can be expressed as,

$Y=\overline{Y_2.Y_3}$

$\Rightarrow Y=\overline{(\overline{A}+(A.B\left)\right).1}$

Using DeMorgan's theorem for converting multiplication to addition,

$\Rightarrow Y=\overline{\overline{A}+A.B}+0$

$\Rightarrow Y=\overline{\overline{A}}.\overline{(A.B)}$

$\Rightarrow Y=A.(\overline{A}+\overline{B})$

$\Rightarrow Y=A.\overline{A}+A.\overline{B}$

$\because A.\overline{A}=0$

$\Rightarrow Y=0+A.\overline{B}=A.\overline{B}$

Hence, option $\left(\text{A}\right)$ is correct.

Why this question? Tip: The wire connected to an input value will transmit that input to all points connected through that wire.