Category: Digital Circuits Objective Questions

Digital Electronic/Circuits Puzzles on “Introduction of Memory Devices-3”.

1. Memories are classified into _____ categories.
A. 3
B. 4
C. 5
D. 6
Answer: C
Clarification: Memory is typically classified of 2 types: Primary and Secondary. These are further classified into 5 types of memories and these are Secondary, RAM, Dynamic/Static, Volatile/Non-volatile, Magnetic/Semiconductor Memory.

2. Secondary memory is also known as ___________
A. Registers
B. Main Memory
C. RAM
D. Both registers and main memory
Answer: D
Clarification: Secondary memory is also known as Registers/Main Memory. In secondary memory, data is usually stored for a long-term.

3. In a computer, registers are present __________
A. Within control unit
B. Within RAM
C. Within ROM
D. Within CPU
Answer: D
Clarification: In a computer, registers are present within the CPU to store data temporarily during arithmetic and logical operations and during the functioning of the ALU.

4. Which of the following has the lowest access time?
A. RAM
B. ROM
C. Registers
D. Flag
Answer: C
Clarification: Registers has the lowest access time, as they are available inside the CPU. Registers are present within the CPU to store data temporarily during arithmetic and logical operations and during the functioning of the ALU.

5. Main memories of a computer, usually made up of __________
A. Registers
B. Semiconductors
C. Counters
D. PLDs
Answer: B
Clarification: Main memories of a computer, usually made up of semiconductors which are available external to the CPU to store program and data during execution of a program. Registers are present within the CPU to store data temporarily during arithmetic and logical operations and during the functioning of the ALU.

6. As the storage capacity of the main memory is inadequate, which memory is used to enhance it?
A. Secondary Memory
B. Auxiliary Memory
C. Static Memory
D. Both Secondary Memory and Auxiliary Memory
Answer: D
Clarification: As the storage capacity of the main memory is inadequate, Secondary memory is used to enhance it and it is also known as auxiliary memory. Secondary memory is also known as Registers/Main Memory. In secondary memory, data is usually stored for a long-term.

7. Which memories are if magnetic memory type?
A. Main Memory
B. Secondary Memory
C. Static Memory
D. Volatile Memory
Answer: B
Clarification: Usually, secondary memories are of magnetic memory type that are used to store large type quantities of data. In secondary memory, data is usually stored for a long-term.

8. Which of the following comes under secondary memory/ies?
A. Floppy disk
B. Magnetic drum
C. Hard disk
D. All of the Mentioned
Answer: D
Clarification: All of the mentioned equipments are of external storage which is known as secondary memories. In secondary memory, data is usually stored for a long-term.

9. Based on method of access, memory devices are classified into ____________ categories.
A. 2
B. 3
C. 4
D. 5
Answer: A
Clarification: Based on the method of access, memory devices are classified into two categories and these are sequential access memory and RAM. A sequential access memory is one in which a particular memory location is accessed sequentially.

10. A sequential access memory is one in which __________
A. A particular memory location is accessed rapidly
B. A particular memory location is accessed sequentially
C. A particular memory location is accessed serially
D. A particular memory location is accessed parallely
Answer: B
Clarification: A sequential access memory is one in which A particular memory location is accessed sequentially (i.e. the ith memory location is accessed only after sequencing through previous (i-1) memory locations).

11. An example of sequential access memory is __________
A. Floppy disk
B. Hard disk
C. Magnetic tape memory
D. RAM
Answer: C
Clarification: A sequential access memory is one in which a particular memory location is accessed sequentially. In magnetic tape memory, data is accessed sequentially.

12. A Random Access Memory is one in which __________
A. Any location can be accessed sequentially
B. Any location can be accessed randomly
C. Any location can be accessed serially
D. Any location can be accessed parallely
Answer: B
Clarification: A Random Access Memory is one in which any location can be accessed randomly.

13. An example of RAM is __________
A. Floppy disk
B. Hard disk
C. Magnetic tape memory
D. Semiconductor RAM
Answer: D
Clarification: A Random Access Memory is one in which any location can be accessed randomly. A semiconductor RAM is too much fast and can occupy any space in the memory location.

14. A static memory is one in which __________
A. Content changes with time
B. Content doesn’t changes with time
C. Memory is static always
D. Memory is dynamic always
Answer: D
Clarification: A static memory is one in which content doesn’t changes with time (i.e. stable). Dynamic memory is one in which content changes with time (i.e. unstable).

15. A dynamic memory is one in which __________
A. Content changes with time
B. Content doesn’t changes with time
C. Memory is static always
D. Memory is dynamic always
Answer: D
Clarification: A static memory is one in which content doesn’t change with time (i.e. stable). Dynamic memory is one in which content changes with time (i.e. unstable).

puzzles on all areas of Digital Electronic Circuits,

Digital Electronics/Circuits Multiple Choice Questions on “Binary Coded Decimal(BCD.”.

1. Binary coded decimal is a combination of __________
A. Two binary digits
B. Three binary digits
C. Four binary digits
D. Five binary digits
Answer: C
Clarification: Binary coded decimal is a combination of 4 binary digits. For example-8421.

2. The decimal number 10 is represented in its BCD form as __________
A. 10100000
B. 01010111
C. 00010000
D. 00101011
Answer: C
Clarification: The decimal number 10 is represented in its BCD form as 0001 0000, in accordance to 8421 for each of the two digits.

3. Add the two BCD numbers: 1001 + 0100 = ?
A. 10101111
B. 01010000
C. 00010011
D. 00101011
Answer: C
Clarification: Firstly, Add the 1001 and 0100. We get 1101 as output but it’s not in BCD form. So, we add 0110 (i.e. 6) with 1101. As a result we get 10011 and it’s BCD form is 0001 0011.

4. Carry out BCD subtraction for (68) – (61) using 10’s complement method.
A. 00000111
B. 01110000
C. 100000111
D. 011111000
Answer: A
Clarification: First the two numbers are converted into their respective BCD form using 8421 sequence. Then binary subtraction is carried out.

5. Code is a symbolic representation of __________ information.
A. Continuous
B. Discrete
C. Analog
D. Both continuous and discrete
Answer: B
Clarification: Code is a symbolic representation of discrete information, which may be present in the form of numbers, letters or physical quantities. Mostly, it is represented using a particular number system like decimal or binary and such like.

6. When numbers, letters or words are represented by a special group of symbols, this process is called __________
A. Decoding
B. Encoding
C. Digitizing
D. Inverting
Answer: B
Clarification: When numbers, letters or words are represented by a special group of symbols, this process is called encoding. Encoding in the sense of fetching the codes or words in a computer. It is done to secure the transmission of information.

7. A three digit decimal number requires ________ for representation in the conventional BCD format.
A. 3 bits
B. 6 bits
C. 12 bits
D. 24 bits
Answer: C
Clarification: The number of bits needed to represent a given decimal number is always greater than the number of bits required for a straight binary encoding of the same. Hence, a three digit decimal number requires 12 bits for representation in BCD format.

8. How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes?
A. 12
B. 14
C. 16
D. 18
Answer: B
Clarification: Total number of decimals to be represented = 10000 = 10⁴ = 2ⁿ (where n is the number of bits requireD. = 2^13.29. Therefore, the number of bits required for straight binary encoding = 14.

9. The excess-3 code for 597 is given by __________
A. 100011001010
B. 100010100111
C. 010110010111
D. 010110101101
Answer: A
Clarification: The addition of ‘3’ to each digit yields the three new digits ‘8’, ’12’ and ’10’. Hence, the corresponding four-bit binary equivalents are 100011001010, in accordance to 8421 format.

10. The decimal equivalent of the excess-3 number 110010100011.01110101 is _____________
A. 970.42
B. 1253.75
C. 861.75
D. 1132.87
Answer: A
Clarification: The conversion of binary numbers into digits ‘1100’, ‘1010’, ‘0011’, ‘0111’ and ‘0101’ gives ’12’, ‘5’, ‘3’, ‘7’ and ‘5’ respectively. Hence, the decimal number is 970.42.

Digital Electronics/Circuits Multiple Choice Questions on “Characteristics of CMOS”.

1. The full form of CMOS is ____________
A. Capacitive metal oxide semiconductor
B. Capacitive metallic oxide semiconductor
C. Complementary metal oxide semiconductor
D. Complemented metal oxide semiconductor
Answer: C
Clarification: The full form of CMOS is complementary metal oxide semiconductor. In this type of device, both n-type and p-type transistors are used in a complementary way.

2. The full form of COS-MOS is ____________
A. Complementary symmetry metal oxide semiconductor
B. Complementary systematic metal oxide semiconductor
C. Capacitive symmetry metal oxide semiconductor
D. Complemented systematic metal oxide semiconductor
Answer: A
Clarification: The full form of COS-MOS is complementary systematic metal oxide semiconductor. In this type of device, both n-type and p-type transistors are used in a complementary way. Usually, the transistors used are MOSFETs.

3. CMOS is also sometimes referred to as ____________
A. Capacitive metal oxide semiconductor
B. Capacitive symmetry metal oxide semiconductor
C. Complementary symmetry metal oxide semiconductor
D. Complemented symmetry metal oxide semiconductor
Answer: C
Clarification: CMOS is also sometimes referred to as complementary systematic metal oxide–semiconductor (COS-MOS). In this type of device, both n-type and p-type transistors are used in a complementary way. Usually, the transistors used are MOSFETs.

4. CMOS technology is used in ____________
A. Inverter
B. Microprocessor
C. Digital logic
D. Both microprocessor and digital logic
Answer: D
Clarification: CMOS technology is used in Microprocessor, Microcontroller, static RAM and other digital logic circuits. CMOS technology is also used for several analog circuits such as image sensors (CMOS sensor), data converters and highly integrated transceivers for many types of communication.

5. Two important characteristics of CMOS devices are ____________
A. High noise immunity
B. Low static power consumption
C. High resistivity
D. Both high noise immunity and low static power consumption
Answer: D
Clarification: Two important characteristics of CMOS devices are high noise immunity and low static power consumption. Since one transistor of the pair is always off and the series combination draws significant power only momentarily during switching between on and off states. Also, the performance of CMOS is not altered with the presence of noise and thus it has high noise immunity.

6. CMOS behaves as a/an ____________
A. Adder
B. Subtractor
C. Inverter
D. Comparator
Answer: C
Clarification: Since, the outputs of the PMOS and NMOS transistors are complementary such that when the input is low, the output is high and when the input is high, the output is low. Because of this behaviour of input and output, the CMOS circuit’s output is the inverse of the input. Whereas, adders and subtractors are combinational circuits.

7. An important characteristic of a CMOS circuit is the ____________
A. Noise immunity
B. Duality
C. Symmetricity
D. Noise Margin
Answer: B
Clarification: An important characteristic of a CMOS circuit is the duality that exists between its PMOS transistors and NMOS transistors. Due to the presence of two different types of transistors, the device has a complementary function.

8. CMOS logic dissipates _______ power than NMOS logic circuits.
A. More
B. Less
C. Equal
D. Very High
Answer: B
Clarification: CMOS logic dissipates less power than NMOS logic circuits because CMOS dissipates power only when switching (“dynamic power”). Thus, CMOS has less power consumption and is more efficient.

9. Semiconductors are made of ____________
A. Ge and Si
B. Si and Pb
C. Ge and Pb
D. Pb and Au
Answer: A
Clarification: Semiconductors are made of Silicon (Si) and Germanium (Ge). Semiconductors are devices having conductivity between conductors and insulators.

10. Which chip were the first RTC and CMOS RAM chip to be used in early IBM computers, capable of storing a total of 64 bytes?
A. The Samsung 146818
B. The Samsung 146819
C. The Motorola 146818
D. The Motorola 146819
Answer: C
Clarification: The Motorola 146818 was the first RTC and CMOS RAM chip to be used in early IBM computers; capable of storing a total of 64 bytes.

Digital Electronic/Circuits Questions & Answers for entrance exams on “Realisation of one Flip-flop using other Flip-flops”.

1. To realise one flip-flop using another flip-flop along with a combinational circuit, known as ____________
A. PREVIOUS state decoder
B. NEXT state decoder
C. MIDDLE state decoder
D. PRESENT state decoder
Answer: B
Clarification: To realise one flip-flop using another flip-flop along with a combinational circuit, known as NEXT state decoder which acts as like a flip-flop.

2. For realisation of JK flip-flop from SR flip-flop, the input J and K will be given as ___________
A. External inputs to S and R
B. Internal inputs to S and R
C. External inputs to combinational circuit
D. Internal inputs to combinational circuit
Answer: A
Clarification: If a JK Flip Flop is required, the inputs are given to the combinational circuit and the output of the combinational circuit is connected to the inputs of the actual flip flop. So, J and K will be given as external inputs to S and R. As SR flip-flop have invalid state and JK flip-flop don’t.

3. For realisation of JK flip-flop from SR flip-flop, if J=0 & K=0 then the input is ___________
A. S=0, R=0
B. S=0, R=X
C. S=X, R=0
D. S=X, R=X
Answer: B
Clarification: If J=0 & K=0, the output will be as: Q(n)=0, Q(n+1)=0 and it is fed into both the AND gates which results as S=0 & R=X(i.e. don’t care).