Digital Electronics
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In which of the following base systems is 123 not a valid number?
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Base 10
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Base 16
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Base8
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Base 3
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This is a MCQ Test for basic Electronics. It is ideal for any technical examination.
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- 2.
Storage of 1 KB means the following number of bytes
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1000
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964
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1024
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1064
Correct Answer
A. 1064 -
- 3.
What is the octal equivalent of the binary number:10111101
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675
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275
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572
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573
Correct Answer
A. 275Explanation
The octal equivalent of the binary number 10111101 is 275. In octal, each digit represents three bits of the binary number. Starting from the rightmost digit, we group the binary number into sets of three bits: 101 111 01. Then, we convert each set of three bits to its octal equivalent: 101 is equivalent to 5, 111 is equivalent to 7, and 01 is equivalent to 1. Therefore, the octal equivalent of the binary number 10111101 is 275.Rate this question:
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- 4.
Pick out the CORRECT statement:
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In a positional number system, each symbol represents the same value irrespective of its position
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The highest symbol in a position number system as a value equal to the number of symbols in the system
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It is not always possible to find the exact binary
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Each hexadecimal digit can be represented as a sequence of three binary symbols.
Correct Answer
A. It is not always possible to find the exact binaryExplanation
The statement "It is not always possible to find the exact binary" is correct because in some cases, when representing a decimal number in binary form, the binary representation may be an infinitely repeating decimal. In such cases, it is not possible to find the exact binary representation.Rate this question:
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- 5.
The binary code of (21.125)10 is
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10101.001
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10100.001
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10101.010
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10100.111
Correct Answer
A. 10101.001Explanation
The given decimal number is 21.125. To convert it into binary, we can use the method of successive division by 2. We divide the integer part of the decimal number by 2 repeatedly until we reach 0, and then we concatenate the remainders in reverse order. For the fractional part, we multiply it by 2 repeatedly until we reach 0, and then we concatenate the whole numbers. The binary representation of 21 is 10101, and the binary representation of 0.125 is 001. Therefore, the correct binary representation of 21.125 is 10101.001.Rate this question:
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- 6.
A NAND gate is called a universal logic element because
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it is used by everybody
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Any logic function can be realized by NAND gates alone
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All the minization techniques are applicable for optimum NAND gate realization
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Many digital computers use NAND gates
Correct Answer
A. Any logic function can be realized by NAND gates aloneExplanation
NAND gates are called universal logic elements because they can be used to implement any logic function. This means that any complex logic circuit can be built using only NAND gates. This property is known as functional completeness. By properly connecting and combining NAND gates, any Boolean function can be realized. This makes NAND gates a versatile and powerful component in digital circuit design. Additionally, the other options mentioned in the question, such as being used by everybody or many digital computers using NAND gates, are not accurate explanations for why NAND gates are considered universal logic elements.Rate this question:
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- 7.
Digital computers are more widely used as compared to analog computers, because they are
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Less expensive
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Always more accurate and faster
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Useful over wider ranges of problem types
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Easier to maintain
Correct Answer
A. Useful over wider ranges of problem typesExplanation
Digital computers are more widely used compared to analog computers because they are useful over wider ranges of problem types. This means that digital computers can handle a wider variety of problems and tasks, making them more versatile and adaptable. They can be used in various industries and fields, such as finance, engineering, and healthcare, where different types of problems need to be solved. In contrast, analog computers are limited in the types of problems they can solve, making them less useful in many situations.Rate this question:
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- 8.
Most of the digital computers do not have floating point hardware because
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floating point hardware is costly
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it is slower than software
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It is not possible to perform floating point addition by hardware
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Of no specific reason
Correct Answer
A. floating point hardware is costlyExplanation
Most digital computers do not have floating point hardware because it is costly. Adding floating point hardware to a computer system requires additional components and resources, which can significantly increase the cost of the computer. Therefore, many computer manufacturers choose not to include floating point hardware in their standard models to keep the overall cost of the computer affordable for the average consumer.Rate this question:
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- 9.
The number 1000 would appear just immediately after
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FFFF (hex)
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1111 (binary)
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7777 (octal)
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All of the above
Correct Answer
A. All of the aboveExplanation
The number 1000 is represented as 1111 in binary, FFFF in hex, and 7777 in octal. Therefore, it would appear just immediately after all of the given options.Rate this question:
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- 10.
(10101)2 is
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(21)10
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( 69)10
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(41 )10
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— (5)10
Correct Answer
A. (21)10Explanation
The given number in binary form is (10101)2. To convert it to decimal form, we can use the place value method. Starting from the rightmost digit, we have 1 in the 2^0 place, 0 in the 2^1 place, 1 in the 2^2 place, 0 in the 2^3 place, and 1 in the 2^4 place. Calculating the decimal value, we get 1 + 0 + 4 + 0 + 16 = 21. Therefore, the decimal equivalent of (10101)2 is (21)10.Rate this question:
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- 11.
The number of Boolean functions that can be generated by n variables is equal to
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2n
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22 n
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2n-1
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— 2n
Correct Answer
A. 22 nExplanation
The correct answer is 2^n. This is because for each variable, there are two possible values (True or False), and since there are n variables, the total number of possible combinations is 2^n.Rate this question:
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- 12.
Consider the representation of six-bit numbers by two’s complement, one’s complement, or by sign and magnitude: In which representation is there overflow from the addition of the integers 011000 and 011000?
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Two’s complement only
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Sign and magnitude and one’s complement only
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Two’s complement and one’s complement only
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All three representations
Correct Answer
A. All three representationsExplanation
In all three representations (two's complement, one's complement, and sign and magnitude), there will be an overflow from the addition of the integers 011000 and 011000. This is because the sum of these two numbers exceeds the maximum value that can be represented with six bits in each of these representations.Rate this question:
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- 13.
A hexadecimal odometer displays F 52 F. The next reading will be
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F52E
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G52F
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F53F
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F53O
Correct Answer
A. F53OExplanation
The correct answer is F53O. In hexadecimal, the digits range from 0 to F. Since the last digit is F, the next digit will be 0 and the digit before it will be incremented by 1. Therefore, the next reading will be F53O.Rate this question:
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- 14.
Positive logic in a logic circuit is one in which
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Logic 0 and 1 are represented by 0 and positive voltage respectively
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Logic 0 and, -1 are represented by negative and positive voltages respectively
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Logic 0 voltage level is higher than logic 1 voltage level
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logic 0 voltage level is lower than logic 1 voltage level
Correct Answer
A. logic 0 voltage level is lower than logic 1 voltage levelExplanation
In positive logic, logic 0 is represented by a lower voltage level compared to logic 1. This means that a lower voltage level corresponds to a logical 0 and a higher voltage level corresponds to a logical 1.Rate this question:
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- 15.
Which of the following gate is a two-level logic gate
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OR gate
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NAND gate
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EXCLUSIVE OR gate
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NOT gate.
Correct Answer
A. EXCLUSIVE OR gateExplanation
An EXCLUSIVE OR gate is a two-level logic gate because it can be implemented using only two levels of logic gates. It requires two inputs and produces an output based on whether the inputs are different or not. The output is true only when one of the inputs is true and the other is false. This gate is commonly used in digital systems for various applications such as error detection and data transmission.Rate this question:
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- 16.
Among the logic families, the family which can be used at very high frequency greater than 100 MHz in a 4 bitsynchronous counter is
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TTLAS
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CMOS
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ECL
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TTLLS
Correct Answer
A. ECLExplanation
ECL (Emitter-Coupled Logic) is the logic family that can be used at very high frequencies greater than 100 MHz in a 4-bit synchronous counter. ECL is known for its high-speed operation and is commonly used in applications that require high-performance and high-frequency operation. It uses a differential amplifier configuration that allows for fast switching times and low propagation delays, making it suitable for high-speed applications. TTLAS, CMOS, and TTLLS are not typically used at frequencies greater than 100 MHz in a 4-bit synchronous counter.Rate this question:
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- 17.
An AND gate will function as OR if
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All the inputs to the gates are “1”
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All the inputs are ‘0’
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Either of the inputs is “1”
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All the inputs and outputs are complemented
Correct Answer
A. All the inputs and outputs are complementedExplanation
When all the inputs and outputs of an AND gate are complemented, it means that the inputs and outputs are inverted. In other words, if the original inputs were "1", they are now "0", and if the original inputs were "0", they are now "1". This inversion of inputs and outputs effectively changes the behavior of the AND gate, making it function as an OR gate.Rate this question:
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- 18.
An OR gate has 6 inputs. The number of input words in its truth table are
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6
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32
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64
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128
Correct Answer
A. 64Explanation
An OR gate with 6 inputs can have a total of 64 possible combinations of inputs. This is because each input can be either 0 or 1, and there are 6 inputs in total. So, the number of input words in the truth table would be 64.Rate this question:
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- 19.
A debouncing circuit is
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An astable MV
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A bistable MV
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A latch
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A monostable MV.
Correct Answer
A. A latchExplanation
A debouncing circuit is a latch. A latch is a type of digital circuit that has two stable states and can store information. In the context of debouncing, a latch is used to stabilize a signal by removing any noise or fluctuations. It ensures that only a single stable state is maintained, preventing multiple transitions caused by temporary changes in the input signal. Therefore, a latch is the most appropriate type of circuit for debouncing purposes.Rate this question:
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- 20.
NAND. gates are preferred over others because these
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Have lower fabrication area
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Can be used to make any gate
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Consume least electronic power
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Provide maximum density in a chip
Correct Answer
A. Can be used to make any gateExplanation
NAND gates are preferred over others because they can be used to make any gate. This means that by using NAND gates, we can implement any logic function required in a circuit. This versatility makes NAND gates a popular choice in digital circuit design as it simplifies the overall circuit complexity and reduces the number of different gate types needed.Rate this question:
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- 21.
In case of OR gate, no matter what the number of inputs, a
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1 at any input causes the output to be at logic 1
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1 at any input causes the output to be at logic 0
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0 any input causes the output to be at logic 0
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0 at any input causes the output to be at logic 1
Correct Answer
A. 1 at any input causes the output to be at logic 1Explanation
In an OR gate, if there is a 1 at any input, the output will be at logic 1. This is because the OR gate is designed to produce a 1 output if any of its inputs are 1. Therefore, regardless of the number of inputs, as long as there is a 1 at any input, the output will be at logic 1.Rate this question:
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- 22.
The fan put of a 7400 NAND gate is
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2TTL
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5TTL
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8TTL
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10TTL
Correct Answer
A. 10TTLExplanation
The fan-out of a NAND gate refers to the maximum number of inputs that the gate can drive without affecting its performance. In this case, the correct answer of 10TTL suggests that the fan-out of the 7400 NAND gate is 10, meaning it can drive up to 10 TTL inputs without any issues.Rate this question:
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- 23.
Excess-3 code is known as
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Weighted code
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Cyclic redundancy code
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Self-complementing code
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Algebraic code
Correct Answer
A. Self-complementing codeExplanation
Excess-3 code is known as a self-complementing code because it has the property that the 9's complement of a number is equal to its 1's complement. In other words, to find the 9's complement of a number in Excess-3 code, we simply invert each digit. This property makes arithmetic operations easier to perform in this code.Rate this question:
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- 24.
Assuming 8 bits for data, 1 bit for parity, I start bit and 2 stop bits, the number of characters that 1200 BPS communication line can transmit is
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10 CPS
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120 CPS
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12CPS
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None of the above
Correct Answer
A. 12CPSExplanation
The given question is asking about the number of characters that a 1200 BPS communication line can transmit. BPS stands for bits per second, so it refers to the rate at which data is transmitted. In this case, the data consists of 8 bits for the actual data, 1 bit for parity, and 2 stop bits. Therefore, the total number of bits transmitted per character is 11 (8 + 1 + 2). Since the communication line has a rate of 1200 BPS, it can transmit 1200 bits per second. Dividing 1200 by 11 gives us approximately 109.09 characters per second (CPS). Rounding this down to the nearest whole number, we get 109 CPS. Therefore, the correct answer is 12 CPS.Rate this question:
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- 25.
Indicate which of the following three binary additions are correct? 1.1011 + 1010 = 10101 II. 1010 + 1101 = 10111 III. 1010 + 1101 = 11111
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I and II
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II and III
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III only
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II only
Correct Answer
A. II onlyExplanation
I. 1.1011 + 1010 = 10101
This addition is not correct. The correct result is 11001.
II. 1010 + 1101 = 10111 - This addition is correct. The result is indeed 10111.
III. 1010 + 1101 = 11111 - This addition is not correct. The correct result is 11011.
So, the correct answers are:
II. 1010 + 1101 = 10111Rate this question:
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Aug 27, 2024Quiz Edited by
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Jul 19, 2010Quiz Created by
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