Category: Digital Circuits Objective Questions

250+ TOP MCQs on Flip Flops – 2 and Answers

Digital Electronic/Circuits online quiz on “Flip Flops”.

1. What is an ambiguous condition in a NAND based S’-R’ latch?
A. S’=0, R’=1
B. S’=1, R’=0
C. S’=1, R’=1
D. S’=0, R’=0

Answer: D
Clarification: In a NAND based S-R latch, If S’=0 & R’=0 then both the outputs (i.e. Q & Q’) goes HIGH and this condition is called an ambiguous/forbidden state. This state is also known as an Invalid state as the system goes into an unexpected situation.

2. In a NAND based S’-R’ latch, if S’=1 & R’=1 then the state of the latch is ____________
A. No change
B. Set
C. Reset
D. Forbidden

Answer: A
Clarification: In a NAND based S’-R, latch if S’=1 & R’=1 then there is no any change in the state. It remains in its prior state. This state is used for the storage of data.

3. A NAND based S’-R’ latch can be converted into S-R latch by placing ____________
A. A D latch at each of its input
B. An inverter at each of its input
C. It can never be converted
D. Both a D latch and an inverter at its input

Answer: D
Clarification: A NAND based S’-R’ latch can be converted into S-R latch by placing either a D latch or an inverter at its input as it’s operations will be complementary.

4. One major difference between a NAND based S’-R’ latch & a NOR based S-R latch is ____________
A. The inputs of NOR latch are 0 but 1 for NAND latch
B. The inputs of NOR latch are 1 but 0 for NAND latch
C. The output of NAND latch becomes set if S’=0 & R’=1 and vice versa for NOR latch
D. The output of NOR latch is 1 but 0 for NAND latch

Answer: A
Clarification: Due to inverted input of NAND based S’-R’ latch, the inputs of NOR latch are 0 but 1 for NAND latch.

5. The characteristic equation of S-R latch is ____________
A. Q(n+1) = (S + Q(n))R’
B. Q(n+1) = SR + Q(n)R
C. Q(n+1) = S’R + Q(n)R
D. Q(n+1) = S’R + Q'(n)R

Answer: A
Clarification: A characteristic equation is needed when a specific gate requires a specific output in order to satisfy the truth table. The characteristic equation of S-R latch is Q(n+1) = (S + Q(n))R’.

6. The difference between a flip-flop & latch is ____________
A. Both are same
B. Flip-flop consist of an extra output
C. Latches has one input but flip-flop has two
D. Latch has two inputs but flip-flop has one

Answer: C
Clarification: Flip-flop is a modified version of latch. To determine the changes in states, an additional control input is provided to the latch.

7. How many types of flip-flops are?
A. 2
B. 3
C. 4
D. 5

Answer: C
Clarification: There are 4 types of flip-flops, viz., S-R, J-K, D, and T. D flip-flop is an advanced version of S-R flip-flop, while T flip-flop is an advanced version of J-K flip-flop.

8. The S-R flip flop consist of ____________
A. 4 AND gates
B. Two additional AND gates
C. An additional clock input
D. 3 AND gates

Answer: B
Clarification: The S-R flip flop consists of two additional AND gates at the S and R inputs of S-R latch.

9. What is one disadvantage of an S-R flip-flop?
A. It has no Enable input
B. It has a RACE condition
C. It has no clock input
D. Invalid State

Answer: D
Clarification: The main drawback of s-r flip flop is invalid output when both the inputs are high, which is referred to as Invalid State.

10. One example of the use of an S-R flip-flop is as ____________
A. Racer
B. Stable oscillator
C. Binary storage register
D. Transition pulse generator

Answer: C
Clarification: S-R refers to set-reset. So, it is used to store two values 0 and 1. Hence, it is referred to as binary storage element. It functions as memory storage during the No Change State.

11. When is a flip-flop said to be transparent?
A. When the Q output is opposite the input
B. When the Q output follows the input
C. When you can see through the IC packaging
D. When the Q output is complementary of the input

Answer: B
Clarification: Flip-flop have the property of responding immediately to the changes in its inputs. This property is called transparency.

12. On a positive edge-triggered S-R flip-flop, the outputs reflect the input condition when ________
A. The clock pulse is LOW
B. The clock pulse is HIGH
C. The clock pulse transitions from LOW to HIGH
D. The clock pulse transitions from HIGH to LOW

Answer: C
Clarification: Edge triggered device will follow when there is transition. It is a positive edge triggered when transition takes place from low to high, while, it is negative edge triggered when the transition takes place from high to low.

13. What is the hold condition of a flip-flop?
A. Both S and R inputs activated
B. No active S or R input
C. Only S is active
D. Only R is active

Answer: B
Clarification: The hold condition in a flip-flop is obtained when both of the inputs are LOW. It is the No Change State or Memory Storage state if a flip-flop.

14. If an active-HIGH S-R latch has a 0 on the S input and a 1 on the R input and then the R input goes to 0, the latch will be ________
A. SET
B. RESET
C. Clear
D. Invalid

Answer: B
Clarification: If S=0, R=1, the flip flop is at reset condition. Then at S=0, R=0, there is no change. So, it remains in reset. If S=1, R=0, the flip flop is at the set condition.

15. The circuit that is primarily responsible for certain flip-flops to be designated as edge-triggered is the _____________
A. Edge-detection circuit
B. NOR latch
C. NAND latch
D. Pulse-steering circuit

Answer: A
Clarification: The circuit that is primarily responsible for certain flip-flops to be designated as edge-triggered is the edge-detection circuit.

250+ TOP MCQs on Shift Register Counters and Answers

Digital Electronics/Circuits Multiple Choice Questions on “Shift Register Counters”.

1. What is a recirculating register?
A. Serial out connected to serial in
B. All Q outputs connected together
C. A register that can be used over again
D. Parallel out connected to Parallel in
Answer: A
Clarification: A recirculating register is a register whose serial output is connected to the serial input in a circulated manner.

2. When is it important to use a three-state buffer?
A. When two or more outputs are connected to the same input
B. When all outputs are normally HIGH
C. When all outputs are normally LOW
D. When two or more outputs are connected to two or more inputs
Answer: A
Clarification: When two or more outputs are connected to the same input, in such situation we use of tristate buffer always because it has the capability to take upto three inputs. A buffer is a circuit where the output follows the input.

3. A bidirectional 4-bit shift register is storing the nibble 1110. Its input is LOW. The nibble 0111 is waiting to be entered on the serial data-input line. After two clock pulses, the shift register is storing ________
A. 1110
B. 0111
C. 1000
D. 1001
Answer: D
Clarification: Given,

Stored nibble | waiting nibble
         0111 | 1110, Initially  
          111 | 1100, 1st pulse
           11 | 1001, 2nd pulse.

4. In a parallel in/parallel out shift register, D0 = 1, D1 = 1, D2 = 1, and D3 = 0. After three clock pulses, the data outputs are ________
A. 1110
B. 0001
C. 1100
D. 1000
Answer: B
Clarification: Parallel in parallel out gives the same output as input. Thus, after three clock pulses, the data outputs are 0001.

5. The group of bits 10110111 is serially shifted (right-most bit first) into an 8-bit parallel output shift register with an initial state 11110000. After two clock pulses, the register contains ______________
A. 10111000
B. 10110111
C. 11110000
D. 11111100
Answer: D
Clarification: After first clock pulse, the register contains 11111000. After second clock pulse, the register would contain 11111100. Since the bits are shifted to the right at every clock pulse.

6. By adding recirculating lines to a 4-bit parallel-in serial-out shift register, it becomes a ________ ________ and ________ out register.
A. Parallel-in, serial, parallel
B. Serial-in, parallel, serial
C. Series-parallel-in, series, parallel
D. Bidirectional in, parallel, series
Answer: A
Clarification: One bit shifting takes place just after the output obtained on every register. Hence, by adding recirculating lines to a 4-bit parallel-in serial-out shift register, it becomes a Parallel-in, Serial, and Parallel-out register. Since, the bots can be inputted all at the same time, while the data can be outputted either one at a time or simultaneously.

7. What type of register would have a complete binary number shifted in one bit at a time and have all the stored bits shifted out one at a time?
A. Parallel-in Parallel-out
B. Parallel-in Serial-out
C. Serial-in Serial-out
D. Serial-in Parallel-out
Answer: C
Clarification: Serial-in Serial-out register would have a complete binary number shifted in one bit at a time and have all the stored bits shifted out one at a time. Since in serial transmission, bits are transmitted or received one at a time and not simultaneously.

8. In a 4-bit Johnson counter sequence, there are a total of how many states or bit patterns?
A. 1
B. 3
C. 4
D. 8
Answer: D
Clarification: In johnson counter, total number of states are determined by 2N = 2*4 = 16
Total Number of Used states = 2N = 2*4 = 8
Total Number of Unused states = 16 – 8 = 8.

9. If a 10-bit ring counter has an initial state 1101000000, what is the state after the second clock pulse?
A. 1101000000
B. 0011010000
C. 1100000000
D. 0000000000
Answer: B
Clarification: After shifting 2-bit we get the output as 0011010000 (Since two zeros are at 1st position and 2nd position which came from the last two bits). As in a ring counter, the bits rotate in clockwise direction.

10. How much storage capacity does each stage in a shift register represent?
A. One bit
B. Two bits
C. Four bits
D. Eight bits
Answer: A
Clarification: A register is made of flip-flops. And each flip-flop stores 1 bit of data. Thus, a shift register has the capability to store one bit and if another bit is to store, in such a situation it deletes the previous data and stores them.

250+ TOP MCQs on Random Access Memory – 4 and Answers

Basic Digital Electronic/Circuits Interview questions and answers on “Random Access Memory-4”.

1. DRAM is fabricated by using IC __________
A. 2114
B. 7489
C. 4116
D. 2776
Answer: C
Clarification: DRAM is Dynamic RAM which takes more access time compared to SRAM and is thus, slower in operation comparatively. Although, in general it offers high speed and is used in most computers nowadays. DRAM is fabricated by using IC 4116.

2. IC 4116 is of ______ storage memory.
A. 16 KB
B. 32 KB
C. 64 MB
D. 2 KB
Answer: A
Clarification: IC 4116 is a DRAM of 16 KB storage memory. It requires three supply voltages (+5V, -5V, and +12V) to operate the IC unit.

3. How many supply voltage IC 4116 requires to operate the IC unit?
A. 3
B. 2
C. 1
D. 4
Answer: A
Clarification: IC 4116 is a DRAM of 16 KB storage memory. It requires three supply voltages (+5V, -5V, and +12V) to operate the IC unit.

4. The full form of PSRAM is __________
A. Plugged Static RAM
B. Plugged Stored RAM
C. Pseudo Stored RAM
D. Pseudo Static RAM
Answer: D
Clarification: The full form of PSRAM is Pseudo Static RAM. It is a dynamic RAM which is implemented as a SRAM.

5. Pseudo static RAM is a __________
A. Static RAM
B. Dynamic RAM
C. Cache
D. ROM
Answer: B
Clarification: The full form of PSRAM is Pseudo Static RAM. It is a dynamic RAM having built-in fresh logic, which is implemented as an SRAM.

6. When PSRAM is performing internal refresh __________
A. The read operation is performed
B. The write operation is performed
C. It can not be accessed for read or write
D. The voltage goes HIGH
Answer: C
Clarification: The full form of PSRAM is Pseudo Static RAM. It is a dynamic RAM having built-in fresh logic, which is implemented as a SRAM. So, it can not be accessed for read or write during the refresh operation.

7. RAMs are utilized in the computer as __________
A. Scratch-pad
B. Buffer
C. Main memory
D. All of the Mentioned
Answer: D
Clarification: RAMs are utilized in the computer as a scratch-pad, buffer and main memories. These are the applications of RAMs. Mostly, these RAMs are DRAMs as they provide high speed.

8. The advantages of RAMs are __________
A. Non destructive read out
B. Fast operating speed
C. Low power dissipation
D. All of the Mentioned
Answer: D
Clarification: The advantages of RAM are Non-destructive read out, Fast operating speed and Low power dissipation.

9. Which one is more economical?
A. ROM
B. RAM
C. EROM
D. PROM
Answer: B
Clarification: RAM is more economical than ROM because MOS memories are more economical than the magnetic core for small and medium sized systems.

10. Which one is self-compatible?
A. ROM
B. RAM
C. EROM
D. PROM
Answer: B
Clarification: As semiconductor memories enjoy common interface and technology between sensing and decoding circuitry and the storage element itself, so RAMs are self-compatible. Also, they provide high speed and fast operation.

11. The memory which is used for storing programs and data currently being processed by the CPU is called __________
A. PROM
B. Main Memory
C. Non-volatile memory
D. Mass memory
Answer: A
Clarification: PROM has the capability to store the data due to the presence of MOSFET which is processed by the CPU. It is one-time programmable by the user.

12. CD-ROM is a __________
A. Memory register
B. Magnetic memory
C. Semiconductor memory
D. Non-volatile memory
Answer: D
Clarification: CD-ROM is a non-volatile memory. Once a program is uploaded in it then it can’t be erasable. Thus, it stores the data permanently.

13. A place which is used as storage location in a computer __________
A. A bit
B. A record
C. An address
D. A byte
Answer: C
Clarification: A storage location of a computer is an address/memory location, used to store instructions and data.

14. Which of the following is not a primary storage device?
A. Optical disk
B. Magnetic tape
C. Magnetic disk
D. RAM
Answer: D
Clarification: RAM (i.e. Random Access Memory) is not a primary storage device.

basic questions on all areas of Digital Electronic Circuits,

250+ TOP MCQs on Number System and Answers

Digital Electronics/Circuits Multiple Choice Questions on “Number System”.

1. The given hexadecimal number (1E.53)16 is equivalent to ____________
A. (35.684)8
B. (36.246)8
C. (34.340)8
D. (35.599)8

Answer: B
Clarification: First, the hexadecimal number is converted to it’s equivalent binary form, by writing the binary equivalent of each digit in form of 4 bits. Then, the binary equivalent bits are grouped in terms of 3 bits and then for each of the 3-bits, the respective digit is written. Thus, the octal equivalent is obtained.
(1E.53)16 = (0001 1110.0101 0011)2
= (00011110.01010011)2
= (011110.010100110)2
= (011 110.010 100 110)2
= (36.246)8.

2. The octal number (651.124)8 is equivalent to ______
A. (1A9.2A.16
B. (1B0.10)16
C. (1A8.A3)16
D. (1B0.B0)16

Answer: A
Clarification: First, the octal number is converted to it’s equivalent binary form, by writing the binary equivalent of each digit in form of 3 bits. Then, the binary equivalent bits are grouped in terms of 4 bits and then for each of the 4-bits, the respective digit is written. Thus, the hexadecimal equivalent is obtained.
(651.124)8 = (110 101 001.001 010 100)2
= (110101001.001010100)2
= (0001 1010 1001.0010 1010)2
= (1A9.2A.16.

3. The octal equivalent of the decimal number (417)10 is _____
A. (641)8
B. (619)8
C. (640)8
D. (598)8

Answer: A
Clarification: Octal equivalent of decimal number is obtained by dividing the number by 8 and collecting the remainders in reverse order.
8 | 417
8 | 52 — 1
8 | 6 – 4
So, (417)10 = (641)8.

4. Convert the hexadecimal number (1E2)16 to decimal.
A. 480
B. 483
C. 482
D. 484

Answer: C
Clarification: Hexadecimal to Decimal conversion is obtained by multiplying 16 to the power of base index along with the value at that index position.
(1E2)16 = 1 * 162 + 14 * 161 + 2 * 160 (Since, E = 14)
= 256 + 224 + 2 = (482)10.

5. (170)10 is equivalent to ____________
A. (FD.16
B. (DF)16
C. (AA.16
D. (AF)16

Answer: C
Clarification: Hexadecimal equivalent of decimal number is obtained by dividing the number by 16 and collecting the remainders in reverse order.
16 | 170
16 | 10 – 10
Hence, (170)10 = (AA.16.

6. Convert (214)8 into decimal.
A. (140)10
B. (141)10
C. (142)10
D. (130)10

Answer: A
Clarification: Octal to Decimal conversion is obtained by multiplying 8 to the power of base index along with the value at that index position.
(214)8 = 2 * 8v + 1 * 81 + 4 * 80
= 128 + 8 + 4 = (140)10.

7. Convert (0.345)10 into an octal number.
A. (0.16050)8
B. (0.26050)8
C. (0.19450)8
D. (0.24040)8

Answer: B
Clarification: Converting decimal fraction into octal number is achieved by multiplying the fraction part by 8 everytime and collecting the integer part of the result, unless the result is 1.
0.345*8 = 2.76 2
0.760*8 = 6.08 6
00.08*8 = 0.64 0
0.640*8 = 5.12 5
0.120*8 = 0.96 0
So, (0.345)10 = (0.26050)8.

8. Convert the binary number (01011.1011)2 into decimal.
A. (11.6875)10
B. (11.5874)10
C. (10.9876)10
D. (10.7893)10

Answer: A
Clarification: Binary to Decimal conversion is obtained by multiplying 2 to the power of base index along with the value at that index position.
(01011)2 = 0 * 24 + 1 * 23 + 0 * 22 + 1 * 21 + 1 * 20 = 11
(1011)2 = 1 * 2-1 + 0 * 2-2 + 1 * 2-3 + 1 * 2-4 = 0.6875
So, (01011.1011)2 = (11.6875)10.

9. Octal to binary conversion: (24)8 =?
A. (111101)2
B. (010100)2
C. (111100)2
D. (101010)2

Answer: B
Clarification: Each digit of the octal number is expressed in terms of group of 3 bits. Thus, the binary equivalent of the octal number is obtained.
(24)8 = (010100)2.

10. Convert binary to octal: (110110001010)2 =?
A. (5512)8
B. (6612)8
C. (4532)8
D. (6745)8

Answer: B
Clarification: The binary equivalent is segregated into groups of 3 bits, starting from left. And then for each group, the respective digit is written. Thus, the octal equivalent is obtained.
(110110001010)2 = (6612)8.

250+ TOP MCQs on Transistor Transistor Logic(TTL) and Answers

Digital Electronics/Circuits Multiple Choice Questions on “250+ TOP MCQs on Transistor Transistor Logic(TTL or T2L)”.

1. Transistor–transistor logic (TTL) is a class of digital circuits built from ____________
A. JFET only
B. Bipolar junction transistors (BJT)
C. Resistors
D. Bipolar junction transistors (BJT) and resistors
Answer: D
Clarification: Transistor–transistor logic (TTL) is a class of digital circuits built from bipolar junction transistors (BJT) and resistors. However, resistors have a small role to play and both logic gating and amplifying functions are performed by the transistors.

2. TTL is called transistor–transistor logic because both the logic gating function and the amplifying function are performed by ____________
A. Resistors
B. Bipolar junction transistors
C. One transistor
D. Resistors and transistors respectively
Answer: B
Clarification: TTL is called transistor–transistor logic because both the logic gating function and the amplifying function are performed by bipolar junction transistors (BJTs).

3. TTL was invented in 1961 by ____________
A. Baker clamp
B. James L. Buie
C. Chris Brown
D. Frank Wanlass
Answer: B
Clarification: TTL was invented in 1961 by James L Buie.

4. The full form of TCTL is ____________
A. Transistor-coupled transistor logic
B. Transistor-capacitor transistor logic
C. Transistor-complemented transistor logic
D. Transistor-complementary transistor logic
Answer: A
Clarification: The full form of TCTL is transistor-coupled transistor logic.

5. The _______ ancestor to the first personal computers.
A. PARAM 1
B. SATYAM 1
C. KENBAK 1
D. MITS Altair
Answer: C
Clarification: The KENBAK 1, ancestor to the first personal computers.

6. TTL inputs are the emitters of a ____________
A. Transistor-transistor logic
B. Multiple-emitter transistor
C. Resistor-transistor logic
D. Diode-transistor logic
Answer: B
Clarification: TTL inputs are the emitters of a multiple-emitter transistor.

7. TTL is a ____________
A. Current sinking
B. Current sourcing
C. Voltage sinking
D. Voltage sourcing
Answer: A
Clarification: Like DTL, TTL is a current-sinking logic since a current must be drawn from inputs to bring them to a logic 0 level. Current Sink means it accepts current coming out from a source.

8. Standard TTL circuits operate with a __ volt power supply.
A. 2
B. 4
C. 5
D. 3
Answer: C
Clarification: Standard TTL circuits operate with a 5-volt power supply.

9. TTL devices consume substantially ______ power than equivalent CMOS devices at rest.
A. Less
B. More
C. Equal
D. Very High
Answer: B
Clarification: TTL devices consume substantially more power than equivalent CMOS devices at rest. Thus, CMOS devices are faster than TTL devices.

10. A TTL gate may operate inadvertently as an ____________
A. Digital amplifier
B. Analog amplifier
C. Inverter
D. Regulator
Answer: B
Clarification: A TTL gate may operate inadvertently as an analog amplifier if the input is connected to a slowly changing input signal that traverses the unspecified region from 0.7V to 3.3V.

11. The speed of ______ circuits is limited by the tendency of common emitter circuits to go into saturation.
A. TTL
B. ECL
C. RTL
D. DTL
Answer: A
Clarification: The speed of TTL circuits is limited by the tendency of common emitter circuits to go into saturation due to the injection of minority carriers into the collector region. Hence, it functions slowly compared to CMOS devices.

250+ TOP MCQs on Fast Adder & Serial Adder – 2 and Answers

Digital Electronic/Circuits Quiz on “Fast Adder & Serial Adder – 2”.

1. A serial subtractor can be obtained by converting the serial adder by using the _____________
A. 1’s complement system
B. 2’s complement system
C. 10’s complement
D. 9’s complement
Answer: B
Clarification: A serial subtractor can be obtained by converting the serial adder by using the 2’s complement system. 9’s complement and 10’s complement are used for decimal numbers while adders deal with binary numbers.

2. The hexadecimal number (4B.16 is transmitted as an 8-bit word in parallel. What is the time required for this transmission if the clock frequency is 2.25 MHz?
A. 444 ns
B. 444 s
C. 3.55 s
D. 3.55 ms
Answer: A
Clarification: Because the clock pulse of 4-bit transmits the data of 8-bit word in parallel mode and this transmission is done at 2.25 MHz frequency. We know that: f=1/t and we can find the time required for this transmission by the clock pulse.
Therefore, time = (1/2.25) = 0.4444 us = 444.44 ns ~ 444ns.

3. Internally, a computer’s binary data are always transmitted on parallel channels which is commonly referred to as the __________
A. Parallel bus
B. Serial bus
C. Data bus
D. Memory bus
Answer: C
Clarification: A process consists of 3 types of buses: Control Bus, Data Bus and Address Bus. A computer’s data is always in the binary form which is stored in the bus that transmits the data on any channels. It doesn’t matter that it’s in parallel or serial.

4. What is the frequency of a clock waveform if the period of that waveform is 1.25sec?
A. 8 kHz
B. 0.8 kHz
C. 0.8 MHz
D. 8 MHz
Answer: C
Clarification: By using the formula of frequency, we can find the frequency of clock waveform. Time period(t) of the waveform is = 1.25microseconds
f=1/t
Where ‘t’ is the time taken by the clock waveform;
f=(1/1.25)
so, f=0.8 MHz.

5. Why is parallel data transmission preferred over serial data transmission for most applications?
A. It is much slower
B. It is cheaper
C. More people use it
D. It is much faster
Answer: D
Clarification: Parallel data transmission preferred over serial data transmission for most applications because it is much faster as bits are transmitted simultaneously, whereas in serial data transmission, bits are transmitted one by one.

6. With surface-mount technology (SMT), the devices should __________
A. Utilize transistor outline connections
B. Mount directly
C. Have parallel connecting pins
D. Require holes and pads
Answer: B
Clarification: Surface-mount technology (SMT) is a method for producing electronic circuits in which the components are mounted or placed directly onto the surface of printed circuit boards (PCBs). An electronic device so made is called a surface-mount device (SMD.. In the industry, it has largely replaced the through-hole technology construction method of fitting components with wire leads into holes in the circuit board. Both technologies can be used on the same board for components not suited to surface mounting such as large transformers and heat-sinked power semiconductors.

7. In most applications, transistor switches used in place of relays?
A. They consume less power
B. They are faster
C. They are quieter and smaller
D. All of the Mentioned
Answer: D
Clarification: Transistors are of less consuming power, faster, quieter, smaller and its implementation is too easy. So, in most applications transistor switches are more preferred. And also, transistors can be current-controlled or voltage-controlled depending on our choice.

8. What can a relay provide between the triggering source and the output that semiconductor switching devices cannot?
A. Total isolation
B. Faster
C. Higher current rating
D. Total isolation and higher current rating
Answer: D
Clarification: A relay provides total isolation and higher current rating between the triggering source and the output that semiconductor switching devices cannot provide. This is why relays are used to drive high watt appliances at offices or other buildings.

9. The serial format for transmitting binary information uses __________
A. A single conductor
B. Multiple conductors
C. Infrared technology
D. Fiber-optic
Answer: A
Clarification: A conductor accepts the whole data and arranges it in a serial manner, which is transmitted as binary information. In serial transmission, data is transmitted bit by bit while in parallel transmission data is transmitted simultaneously.

10. Serial communication can be sped up by __________
A. Using silver or gold conductors instead of copper
B. Using high-speed clock signals
C. Adjusting the duty cycle of the binary information
D. Using silver or gold conductors instead of copper and high-speed clock signals
Answer: B
Clarification: For any serial data transmission there is required of continuously data supply and if the input supply (i.e. high speed clock signals) in a high amount the speed of serial communication can be increased. In serial communication, data is transmitted bit by bit. So the use of high speed clock pulses would make the process faster.

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