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Assuming that flip-flops are in reset condition initially, the count sequence observed at QA in the circuit shown is
- a)0010111
- b)0001011
- c)0101111
- d)0110100
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Assuming that flip-flops are in reset condition initially, the count s...
Initially, QA = QB = QC = 0
DA = QB ⊕ QC = 0, DB = QA = 0
DC = QB = 0
After one clock pulse,
QA = 1, QB = 0, QC = 0
DA = QB⊕QC = 0
DB = QA = 1, DC = QB = 0
After two clock pulse,
QA = 1, QB = 1, QC = 0
DA = QB⊕QC = 1
DB = QA = 1, DC = QB = 1
After three clock pulses,
QA = 0, QB = 1, QC = 1
DA = QB⊕QC = 0
DB = QA = 0, DC = QB = 1
After four clock pulse,
QA = 1, QB = 0, QC = 1
DA = QB⊕QC = 1
DB = QA = 1, DC = QB = 0
After five clock pulse,
QA = 0, QB = 1, QC = 0
DA = QB⊕QC = 1
DB = QA = 0, DC = QB = 1
After six clock pulse,
QA = 0, QB = 0, QC = 1
Therefore, the count sequence observed at QA is 0010111.....
Most Upvoted Answer
Assuming that flip-flops are in reset condition initially, the count s...
Initially, QA = QB = QC = 0
DA = QB ⊕ QC = 0, DB = QA = 0
DC = QB = 0
After one clock pulse,
QA = 1, QB = 0, QC = 0
DA = QB⊕QC = 0
DB = QA = 1, DC = QB = 0
After two clock pulse,
QA = 1, QB = 1, QC = 0
DA = QB⊕QC = 1
DB = QA = 1, DC = QB = 1
After three clock pulses,
QA = 0, QB = 1, QC = 1
DA = QB⊕QC = 0
DB = QA = 0, DC = QB = 1
After four clock pulse,
QA = 1, QB = 0, QC = 1
DA = QB⊕QC = 1
DB = QA = 1, DC = QB = 0
After five clock pulse,
QA = 0, QB = 1, QC = 0
DA = QB⊕QC = 1
DB = QA = 0, DC = QB = 1
After six clock pulse,
QA = 0, QB = 0, QC = 1
Therefore, the count sequence observed at QA is 0010111.....
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Assuming that flip-flops are in reset condition initially, the count sequence observed at QA in the circuit shown isa)0010111b)0001011c)0101111d)0110100Correct answer is option 'A'. Can you explain this answer?
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