The circuit shown below implements a:

The circuit shown below implements a:
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The circuit shown below implements a:

A. OR Logic

B. NOR Logic

C. NAND Logic

D. AND Logic

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

The output from MUX-X is given by:

A̅ Io + A I1

Given, Io = 1 and I1 = 0

So, F1 (Output from MUX-X) = A̅

Now, Output from MUX-Y is;

\(F = \overline {\left( {\bar A} \right)} .\left( 0 \right) + \bar {A} \bar B\)

\(F = \bar A\bar B = \overline {A + B}\) which is NOR Logic only.