DIGITAL ELECTRONICS MCQS PART 02

DIGITAL ELECTRONICS MCQS PART 02

1.The Gray code for decimal number 6 is equivalent to
(A) 1100    
(B) 1001
(C) 0101    
(D) 0110
Ans: C

2. The digital logic family which has minimum power dissipation is
(A) TTL    
(B) RTL    
(C) DTL    
(D) CMOS
Ans: D
Explanation: The digital logic family which has minimum power dissipation is CMOS.
(CMOS being an unipolar logic family, occupy a very small fraction of silicon Chip area)

3. The output of a logic gate is 1 when all its inputs are at logic 0. the gate is either
(A) a NAND or an EX-OR    
(B) an OR or an EX-NOR
(C) an AND or an EX-OR    
(D) a NOR or an EX-NOR
Ans: D

4. Data can be changed from special code to temporal code by using
(A) Shift registers  
(B) counters 
(C) Combinational circuits
(D) A/D converters.
Ans: A
Explanation: Data can be changed from special code to temporal code by using Shift Registers. (A Register in which data gets shifted towards left or right when clock pulses are applied is known as a Shift Register.

5. A ring counter consisting of five Flip-Flops will have
(A)5 states        
(B) 10 states
(C) 32 states    
(D) Infinite states.
Ans: A
Explanation: A ring counter consisting of Five Flip-Flops will have 5 states.

6. The 2’s complement of the number 1101101 is
(A) 0101110       
(B) 0111110   
(C) 0110010    
(D) 0010011
Ans: D
Explanation: The 2’s complement of the number 1101101 is 0010011
(1’s complement of the number 1101101 is 0010010
2’s complement of the number 1101101is 0010010 + 1 =0010011)

7. The correction to be applied in decimal adder to the generated sum is
(A) 00101    
(B) 00110    
(C) 01101    
(D) 01010
Ans: B
Explanation: The correction to be applied in decimal adder to the generated sum is 00110.
When the four bit sum is more than 9 then the sum is invalid. In such  cases, add +6(i.e.  0110) to the four bit sum to skip the six invalid states. If a carry is generated when adding 6, add the carry to the next four bit group .

8. When simplified with Boolean Algebra (x + y)(x + z) simplifies to
(A)x       
(B) x + x(y + z)    
(C) x(1 + yz)    
(D) x + yz
Ans: D

9.The excess 3 code of decimal number 26 is
(A) 0100 1001    
(B) 01011001
(C) 1000 1001    
(D) 01001101
 Ans: B 
(26)10 in BCD is  ( 00100110 ) BCD
Add 011 to each BCD 01011001 for excess – 3

10. The code where all successive numbers differ from their preceding number by single bit is
(A)Binary code.
(B) BCD.
(C) Excess – 3.
(D) Gray.
Ans: D
The code where all successive numbers differ from their preceding number by single bit is Gray Code.
(It is an unweighted code. The most important characteristic of this code is that only a single bit change occurs when going from one code number to next.)