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+3 votes
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Consider the logic circuit given below:

Q=__________?

  1. $\bar{A} C + B \bar{C} +CD$
  2. $ABC + \bar{C} D$
  3. $AB + B \bar{C} + B \bar{D}$
  4. $A \bar{B} + A \bar{C} + \bar{C} D$
in Digital Logic by Veteran (105k points) | 2.2k views
+1
write the output terms after the gate(and dont evaluate for each gate) evaluate only the final output
0
replace last nand with bubbled or.
0
whenever 2 sequential NAND gates or NOR gates are there then we can replace those NAND / NOR gates by AND - OR / OR - AND gates respectively and do simplification after replacement.

5 Answers

+12 votes
Best answer

Option C

Q =((D'+C')B+AB)

  =AB+BC'+BD'

Therefore option C

by Boss (13.8k points)
selected by
+1
Thank you for such a nice explanation..
+3 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

by Boss (45.3k points)
0 votes
option C seems correct,

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

      = ( cd )' b + ab

       = ( c' + d' ) b  + ab      -----------D morgans law (ab)' = a' + b'
by Active (3.5k points)
0 votes

Hence,Option(C)AB+BC'+BD' is the correct choice.

by Boss (41k points)
0 votes
option (c) is the right answer
by (131 points)
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