question_answer

To get in output 1 from the circuit shown in the figure, the input must be :

A)  \[A=0,\,B=1,\,C=1\]    

B)  \[A=1,\,B=0,\,C=0\]

C)  \[A=1,\,B=0,\,C=1\]    

D)  \[A=1,\,B=1,\,C=0\]

Correct Answer: C

Solution :

We can release the gate circuit as below Output of gate 1, \[{{Y}_{1}}=A+B\] Output of gate 2, \[{{Y}_{2}}={{Y}_{1}}\,.C=(A+B).C\] [a] if A = 0, B = 1, C = O then \[{{Y}_{2}}=(1+0)\,.0=1.0=0\] [b] if A = 1, B = 0, C = O then \[{{Y}_{2}}=(1+0)\,.0=1.0=0\] [c] if A = 1, B = 0, C = 1 then \[{{Y}_{2}}=(1+0)\,.1=1.1=1\] [d] if A = 1, B = 1, C = O then \[{{Y}_{2}}=(1+0)\,.0=1.0=0\]