Semiconductors

JEE Main level, single correct, medium

159 of 209

Question 159 of 209

For the following circuit and given inputs AA and B,B, choose the correct option for output ‘Y’‘Y’

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Answer: (A)

The signal AA is given to a NOR gate and BB is given to an OR gate. The output of these two signals is given to a NAND gate. Using the laws of Boolean algebra,
Y=A⋅Bˉ=A+BˉY=\bar{A\cdot B}=A+\bar{B}.
Analysing from the input graphs, one can see,
t1−t2:A=0,Bˉ=0,Y=0t2−t3:A=1,Bˉ=0,Y=1t3−t4:A=0,Bˉ=1,Y=1t4−t5:A=1,Bˉ=0,Y=1t5−t6:A=1,Bˉ=1,Y=1{t}_{1}-{t}_{2}:A=0,\bar{B}=0,Y=0 {t}_{2}-{t}_{3}:A=1,\bar{B}=0,Y=1 {t}_{3}-{t}_{4}:A=0,\bar{B}=1,Y=1 {t}_{4}-{t}_{5}:A=1,\bar{B}=0,Y=1 {t}_{5}-{t}_{6}:A=1,\bar{B}=1,Y=1

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