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Consider the logic circuit given below:

$\text{Q =}$ __________?

  1. $\overline{\text{A}} \text{C} + \text{B} \overline{\text{C}} +\text{CD}$
  2. $\text{ABC} + \overline{\text{C}} \text{D}$
  3. $\text{AB + B} \overline{\text{C}} + \text{B} \overline{\text{D}}$
  4. $\text{A} \overline{\text{B}} + \text{A} \overline{\text{C}} + \overline{\text{C}} \text{D}$
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5 Answers

4 votes
4 votes

Answer: AB + BC' + BD'

Output of  NAND gate which has input As D and C is (DC)'

Output of NAND gate which has input As  A and B is (AB)'

Output of NAND gate which has input As (DC)' and B is ((DC)' . B)'

Output of NAND gate which has input As  ((DC)' . B)' and (AB)' is (((DC)' . B)' . (AB)')'

If we solve it (((DC)' . B)' . (AB)')' then

(DC)' . B + AB 

(D' + C' )B + AB

BD' + BC' +AB

0 votes
0 votes
option C seems correct,

eq = (((cd)' b )' ( ab )' )'

      = ( cd )' b + ab

       = ( c' + d' ) b  + ab      -----------D morgans law (ab)' = a' + b'
Answer:

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