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Consider the circuit shown below. IN a certain steady state, ‘Y’ is at logical ‘l’. What are the possible values of A, B, C?

  1. A=0, B=0, C=1
  2. A=0, B=C=1
  3. A=1, B=C=0
  4. A= B=1, C=1
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assuming f as Y in order to  y=1 (steady) state the last AND gate must have both input as 1

so   C=1 (so choice C is out) and for other input to be 1  either B=0 (which ensures output of 2nd NAND gate as 1)  

or A=1 as it ensure the output of second  NAND gate as 0 if other input is also 1 , now which is f which may be 0 or 1 so to be safe side leave this

i.e         B=0  and  A=1 or  0     C=1    choice A)satisfy and also choice D)

hence ans should be A and also D
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The answer is A . C can`t be 0 so option(c) is canceled and if you put option (b) values then Y= 0 (i think 'f' is Y here ). So (b) is canceled . Remaining A and D both satisfied the circuit (Y =1 ) but if you see they say the "STEADY state" , so before turning on the circuit the value of Y (f) may be 0 , now 

option (d)    then A(1) nand 0(f) = 1 ------>  B(1) nand 1  =  0 -----> C(1) and 0 = 0 (f) 

option (a)    then A(0) nand 0(f) = 1 ------>  B(0) nand 1  =  1 -----> C(1) and 1 = 1 (f) satisfied the circuit .

Answer:

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