UGC NET CSE | June 2010 | Part 2 | Question: 6

Question

The logic expression for the output of the circuit shown in the figure is

• A. $\bar{A} \bar{C}+ \bar{B}\bar{C}+CD$
• B. $A\bar{C}+B\bar{C}+\bar{C}D$
• C. $ABC + \bar{C}\bar{D}$
• D. $\bar{A}\bar{B}+\bar{B}\bar{C}+\bar{C}\bar{D}$

Comments

My answer also not matching .

Answer 1

draw the k map of expression and chek the result

no answer is matching with k map

Answer 2

Final simplified expression   A' B' D + C

No option matching...

Answer 3

The logical expression can be written as:

$\left[\overline{(A+B)}+C\right].\left[(C+D)\right]$

Applying De-morgans' law:

$\left[(\bar A.\bar B)+C\right].(C+D)$

$\bar A\bar B C+\bar A\bar BD+C+CD$

taking $C$ as common here;

$C(\bar A\bar B+1+D)+\bar A\bar BD$

$\because (1+x=1)$

$\therefore \bar A\bar BD+C$

Note: none of the options are correct here.