16) 205 (12 → C
     192
     13 → D
So, (205)₁₀ = (CD)₁₆ = (x)₁₆
Therefore, x = CD.

Correct Option: A

16) 205 (12 → C
     192
     13 → D
So, (205)₁₀ = (CD)₁₆ = (x)₁₆
Therefore, x = CD.

Binary to Gray code converter is shown below:

From above figure w e conclude that B₃ does not require any gate. Hence alternative (B) is the correct answer.

Correct Option: B

Binary to Gray code converter is shown below:

From above figure w e conclude that B₃ does not require any gate. Hence alternative (B) is the correct answer.

Y = AB + AB
This means, half adder requires 4 NAND gates. Hence alternative (C) is the correct answer.

Correct Option: C

Y = AB + AB
This means, half adder requires 4 NAND gates. Hence alternative (C) is the correct answer.

Here switching delay means after 1 μ sec.,the switch is moved between the point A and B.

Hence, Time period = 2μ sec
∴

Correct Option: B

Here switching delay means after 1 μ sec.,the switch is moved between the point A and B.

Hence, Time period = 2μ sec
∴

F1 = (AB)′
F2 = (BA)′
Y = F1F2 = [( AB)′ ( BA)′]′
= [(A + B′) ( B + A)] ′ = AB + BA
Alternative Method:
NAND-NAND circuit can be replaced by AND-OR Hence,

Now, F1 = AB,
F2 = BA
Y=F1 + F2 = AB + BA (this is shortcut method).

Correct Option: D

F1 = (AB)′
F2 = (BA)′
Y = F1F2 = [( AB)′ ( BA)′]′
= [(A + B′) ( B + A)] ′ = AB + BA
Alternative Method:
NAND-NAND circuit can be replaced by AND-OR Hence,

Now, F1 = AB,
F2 = BA
Y=F1 + F2 = AB + BA (this is shortcut method).