A proper combination of 3 NOT and 1 NAND gates is shown. If A=0, B=1, C=1, then the output of this combination is

Question

A proper combination of 3 NOT and 1 NAND gates is shown. If A=0, B=1, C=1, then the output of this combination is (A) 1 (B) 0 (C) not predictable (D) None of these

Tardigrade Admin · Accepted Answer

The simplified circuit is shown in figure below.

So, output

y = \overline{\overset{ˉ}{A} \cdot \overset{ˉ}{B} \cdot \overset{ˉ}{C}}

If

A = 0, B = 1, C = 1

∴ y = \overline{\overset{ˉ}{0} \cdot \overset{ˉ}{1} \cdot \overset{ˉ}{1}} = \overline{1 \cdot 0 \cdot 0} = \overset{ˉ}{0} = 1

Q. A proper combination of 3NOT and 1NAND gates is shown. If A=0,B=1,C=1, then the output of this combination is

1

0

not predictable

None of these

The simplified circuit is shown in figure below. So, output y=Aˉ⋅Bˉ⋅Cˉ If A=0,B=1,C=1 ∴y=0ˉ⋅1ˉ⋅1ˉ=1⋅0⋅0=0ˉ=1

Q. A proper combination of $3 NOT$ and $1 N A N D$ gates is shown. If $A = 0, B = 1, C = 1$ , then the output of this combination is

The simplified circuit is shown in figure below.

So, output $y = \overline{\overset{ˉ}{A} \cdot \overset{ˉ}{B} \cdot \overset{ˉ}{C}}$
If $A = 0, B = 1, C = 1$
$∴ y = \overline{\overset{ˉ}{0} \cdot \overset{ˉ}{1} \cdot \overset{ˉ}{1}} = \overline{1 \cdot 0 \cdot 0} = \overset{ˉ}{0} = 1$