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Digital Electronics and Logic Design MCQs | Page - 57

Dear candidates you will find MCQ questions of Digital Electronics and Logic Design here. Learn these questions and prepare yourself for coming examinations and interviews. You can check the right answer of any question by clicking on any option or by clicking view answer button.

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Q. 561) In Q output of the last flip-flop of the shift register is connected to the data input of the first
flip-flop of the shift register.

(A) moore machine
(B) meally machine
(C) johnson counter
(D) ring counter
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Q. 562) 5-BIT JOHNSON COUNTER SEQUENCES THROUGH STATES

(A) 7
(B) 10
(C) 32
(D) 25
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Q. 563) AT T0 THE VALUE STORED IN A 4-BIT LEFT SHIFT WAS “1”. WHAT WILL BE THE VALUE OF REGISTER AFTER THREE CLOCK PULSES?

(A) 2
(B) 4
(C) 6
(D) 8
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Mr. Dubey • 51.17K Points
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Q. 564) The alternate solution for a multiplexer and a register circuit is                 

(A) parallel in / serial out shift register
(B) serial in / parallel out shift register
(C) parallel in / parallel out shift register
(D) serial in / serial out shift register
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Q. 565) A synchronous decade counter will have flip-flops

(A) 3
(B) 4
(C) 7
(D) 10
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Q. 566) In outputs depend only on the current state.

(A) mealy machine
(B) moore machine
(C) state reduction table
(D) state assignmen t table
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Q. 567) Given the state diagram of an up/down counter, we can find                

(A) the next state of a given present state
(B) the previous state of a given present state
(C) both the next and previous states of a given state
(D) the state diagram shows only the inputs/out puts of a given states
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Q. 568) THE HOURS COUNTER IS IMPLEMENTED USING                   

(A) only a single mod- 12 counter is required
(B) mod-10 and mod-6 counters
(C) mod-10 and mod-2 counters
(D) a single decade counter and a flip-flop
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Q. 569) The design and implementation of synchronous counters start from

(A) truth table
(B) k-map
(C) state table
(D) state diagram
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Q. 570) Q2 :=Q1 OR X OR Q3 The above ABEL expression will be

(A) q2:= q1 $ x $ q3
(B) q2:= q1 # x # q3
(C) q2:= q1 & x & q3
(D) q2:= q1 ! x ! q3
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