Truth table for the given circuit (Fig. 14.6) is
A.
A | B | C |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
B.
A | B | C |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
C.
A | B | C |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 0 |
1 | 1 | 1 |
D.
A | B | C |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
The correct answer is the option c.
Explanation:-
(c) In this problem the input C of OR gets and when which is an output of AND gate. So, "C equals A AND B" or C=A.B and "D equals NOT A AND B" or D=.B
and "E equals C AND D "or E=C+D=(A.B)+(A.B).
Now we can generate the truth table of this arrangement of gates can be given :
A | B | C=A.B | d=.B | E=(C+D) | |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 1 |