The truth table for the circuit given in the fig. is:

Question

The truth table for the circuit given in the fig. is: (A) |A&B&Y 0&0&1 0&1&1 1&0&0 1&1&0| (B) |A&B&Y 0&0&0 0&1&0 1&0&1 1&1&1| (C) |A&B&Y 0&0&1 0&1&0 1&0&0 1&1&0| (D) |A&B&Y 0&0&1 0&1&1 1&0&1 1&1&1|

Tardigrade Admin · Accepted Answer

C = A + B and y = overlineA.C beginarrayc|c A B C= (A + B) A.C. &y = overlineA.C hline 0 &0 &0&0&1 0 &1&1&0&1 1&0&1&1&0 1&1&1&1&0 endarray

Q. The truth table for the circuit given in the fig. is:

$∣ ∣ A 0011 B 0101 Y 1100 ∣ ∣$

$∣ ∣ A 0011 B 0101 Y 0011 ∣ ∣$

$∣ ∣ A 0011 B 0101 Y 1000 ∣ ∣$

$∣ ∣ A 0011 B 0101 Y 1111 ∣ ∣$

$C = A + B$
and $y = \overline{A . C}$
$A 0011 B 0101 C = (A + B) 0111 A . C . 0011 y = \overline{A . C} 1100$

Q. The truth table for the circuit given in the fig. is:

∣∣​A0011​B0101​Y1100​∣∣​

∣∣​A0011​B0101​Y0011​∣∣​

∣∣​A0011​B0101​Y1000​∣∣​

∣∣​A0011​B0101​Y1111​∣∣​

C=A+B and y=A.C A0011​B0101​C=(A+B)0111​A.C.0011​y=A.C1100​​

$∣ ∣ A 0011 B 0101 Y 1100 ∣ ∣$

$∣ ∣ A 0011 B 0101 Y 0011 ∣ ∣$

$∣ ∣ A 0011 B 0101 Y 1000 ∣ ∣$

$∣ ∣ A 0011 B 0101 Y 1111 ∣ ∣$

$C = A + B$
and $y = \overline{A . C}$
$A 0011 B 0101 C = (A + B) 0111 A . C . 0011 y = \overline{A . C} 1100$