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101. What is the 2’s complement of 0011 0101 1001 1100?
1100100111001011
1100101001100100
1100101001100011
1100101011111111
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Answer: 3. 1100101001100011
Explanation:
To find the 2’s complement:
Invert all bits: 1100101001100011
Add 1 to the least significant bit (LSB): 1100101001100011+1=1100101001100100.
102. A flip-flop is a:
I/O element
Memory element
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Answer: 3. Memory element
Explanation:
A flip-flop is a basic memory element that stores one bit of data.
103. With an RS latch, a high S and low R sets the output to:
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Explanation:
In an RS latch, when S (Set) is high and R (Reset) is low, the output is set to high.
104. With a NAND latch, a low R and a low S produce a:
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Explanation:
In a NAND latch, when both R and S are low, it creates a race condition, which is an invalid state.
105. Computers use thousands of flip-flops. To coordinate the overall action, a common signal called the:
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Explanation:
The clock signal synchronizes the operation of flip-flops and other components in a computer.
106. With positive clocking, the clock signal must be:
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Explanation:
In positive clocking, the flip-flop responds when the clock signal is high.
107. With a JK master-slave flip-flop, the master is clocked when the clock is:
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Explanation:
In a JK master-slave flip-flop, the master is clocked when the clock is high, and the slave is triggered when the clock is low.
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Explanation:
The input word is stored in the buffer register on the next positive clock edge when the LOAD input is active.
109. A shift register moves the bits:
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Explanation:
A shift register shifts bits either to the left or to the right.
110. One flip-flop divides the clock frequency by a factor of:
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Explanation:
A single flip-flop divides the clock frequency by 2.
111. Instead of counting with binary numbers, a ring counter uses words that have a single high:
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Explanation:
A ring counter uses words with a single high bit that circulates through the register.
112. The memory that is ultraviolet-light erasable and electrically programmable is:
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Explanation:
EPROM (Erasable Programmable Read-Only Memory) can be erased using ultraviolet light and reprogrammed electrically.
113. The memory cell of a dynamic RAM is simpler and smaller than the memory cell of a:
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Explanation:
Dynamic RAM (DRAM) uses a simpler and smaller memory cell compared to Static RAM (SRAM).
114. How many memory locations can 14 address bits access?
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115. The 2764 is a 65,536-bit EPROM organized as 8,192 words of 8 bits each. How many address lines does it have?
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116. The 2732 is a 4096x8 EPROM. How many address lines does it have?
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117. Address 200H contains the byte 3FH. What is the decimal equivalent of 3FH?
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119. Flip-flop outputs are always:
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Explanation:
The outputs of a flip-flop (Q and Q') are always complementary.
120. A combinational logic circuit used to send data from a single source to two or more separate destinations is called:
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Explanation:
A demultiplexer sends data from one input to one of several outputs based on the control signals.
121. In which of the following adder circuits is the carry look ripple delay eliminated?
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Explanation:
The carry-look-ahead adder eliminates the ripple delay by calculating carry signals in advance.
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Explanation:
In an RS flip-flop, applying a momentary '1' to the S (Set) input will set the output Q to 1.
123. How many full adders are required to construct an m-bit parallel adder?
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Explanation:
An m-bit parallel adder requires m full adders, one for each bit.
124. The dynamic race hazard problem occurs in:
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Explanation:
The dynamic race hazard problem occurs in sequential circuits due to timing issues in signal propagation.
125. A shift register can be used for:
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Explanation:
A shift register can perform parallel-to-serial conversion, serial-to-parallel conversion, and act as a digital delay line.
126. Which of the following flip-flops is free from the race-around problem?
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Explanation:
The master-slave JK flip-flop is designed to eliminate the race-around problem.
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Explanation:
An n-stage shift register introduces a delay of n clock periods (nT).
128. The master-slave JK flip-flop is effectively a combination of:
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Explanation:
The master-slave JK flip-flop combines the functionality of an SR flip-flop and a T flip-flop.
129. The functional difference between SR flip-flop and JK flip-flop is:
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Explanation:
The JK flip-flop has a feedback path that allows it to toggle its output when both inputs are high.
130. The term sum-of-products in Boolean algebra means:
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Explanation:
Sum-of-products (SOP) refers to the OR of multiple AND terms.
131. A positive AND gate is also a negative:
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Explanation:
A positive AND gate behaves like a negative NOR gate due to De Morgan's laws.
132. What table shows the electrical state of a digital circuit's output for every possible combination of electrical states in the inputs?
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Explanation:
A truth table shows the output of a digital circuit for all possible input combinations.
133. How many bits are required to encode all twenty-six letters, ten symbols, and ten numbers?
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Explanation:
To implement a two-input OR function using NAND gates, three NAND gates are required.
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Explanation:
Adding inverters to the inputs of an AND gate results in a NOR gate.
136. Which of the following Boolean algebra expressions is incorrect?
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137. Which gate is known as the universal gate?
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Explanation:
The NAND gate is a universal gate because it can be used to implement any other logic gate.
138. What logic function is produced by adding an inverter to the output of an AND gate?
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Explanation:
Adding an inverter to the output of an AND gate produces a NAND gate.
139. An OR gate can be imagined as:
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Explanation:
An OR gate behaves like switches connected in parallel, where the output is high if any input is high.
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Explanation:
Adding inverters to the inputs and output of an AND gate produces a NOR gate.
141. Which of the following algebra statements represents the commutative law?
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Explanation:
The commutative law states that the order of operands does not affect the result in addition or multiplication.
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Explanation:
The NOT gate outputs the complement of the input.
143. ASCII and EBCDIC differ in:
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Explanation:
ASCII and EBCDIC use different encoding sequences to represent characters.
144. In which code do successive code characters differ in only one bit position?
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Explanation:
Gray code ensures that successive values differ by only one bit.
145. Cyclic codes are used in:
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Explanation:
Cyclic codes are used in data transfer, signal processing, and computation.
146. The 2's complement of the binary number 010111.1100 is:
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Explanation:
To find the 2's complement:
147. The ASCII code:
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Explanation:
The ASCII code includes 128 characters, with 32 being control characters.
148. The Gray code for decimal 7 is:
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Explanation:
The Gray code for decimal 7 is 0100.
149. The octal equivalent of decimal 324.987 is:
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Explanation:
The octal equivalent of decimal 324.987 is 504.771.
150. When an odd number is converted into a binary number, the least significant digit (LSD) is:
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Explanation:
The least significant digit (LSD) of an odd binary number is always 1.
No change
Answer: 2. High
Race condition
Reset condition
Set condition
No change condition
Answer: 1. Race condition
Master
Answer: 3. Clock
Answer: 1. High
Set, reset
High, low
Race, no change
Set, race
Answer: 2. High, low
Clock edge
Register
Transistor
Answer: 1. Clock edge
Left or right
Up or down
Forward or backward
None of the above
Answer: 1. Left or right
Sixteen
Answer: 1. Two
Answer: 3. Bit
Answer: 1. EPROM
Volatile RAM
Semiconductor RAM
Static RAM
Bipolar RAM
Answer: 3. Static RAM
Answer: 1. 16,384
With 14 address bits, the number of memory locations is 214=16,384.
To address 8,192 words, the number of address lines required is log28192=13.
To address 4,096 words, the number of address lines required is log24096=12.
The hexadecimal value 3FH is equivalent to 3×16+15=63 in decimal.
7FFFH,64387
BFFFH,49,152
BFFFH,49,151
7FFFH,64,386
Answer: 3. BFFFH,49,151
48K memory corresponds to 48×1024=49,152 bytes.
The highest address is 49,152−1=49,151, which is BFFFH in hexadecimal.
Complimentary
Independent of each other
The same
Same as inputs
Answer: 1. Complimentary
Decoder
Multiplexer
Encoder
Demultiplexer
Answer: 4. Demultiplexer
Half adder
Parallel adder
Full adder
Carry-look-ahead adder
Answer: 4. Carry-look-ahead adder
Q will flip from 0 to 1
Q will flip from 0 to 1 and then back to 0
Q will remain unchanged
Q will flip from 1 to 0
Answer: 1. Q will flip from 0 to 1
Combinational circuits only
Sequential circuits only
Both combinational and sequential circuits
None of the above
Answer: 2. Sequential circuits only
Parallel to serial conversion
Digital delay line
Serial to parallel conversion
All of the above
Answer: 4. All of the above
T flip-flop
Master-slave JK flip-flop
SR flip-flop
All of the above
Answer: 2. Master-slave JK flip-flop
An SR flip-flop and a T flip-flop
An SR flip-flop and a D flip-flop
A T flip-flop and a D flip-flop
Two T flip-flops
Answer: 1. An SR flip-flop and a T flip-flop
JK flip-flop is faster than SR flip-flop
JK flip-flop has a feedback path
JK flip-flop accepts both input 1
JK flip-flop does not require an external clock
Answer: 2. JK flip-flop has a feedback path
The AND function of several OR functions
The OR function of several AND functions
The OR function of several OR functions
The AND function of several AND functions
Answer: 2. The OR function of several AND functions
NAND gate
AND gate
NOR gate
OR gate
Answer: 3. NOR gate
Function table
Routing table
Truth table
ASCII table
Answer: 3. Truth table
To encode 26 letters, 10 symbols, and 10 numbers (total 46 items), at least 6 bits are required (26=64).
Answer: 4. NOR
AB+A(B+C)+B(B+C)=B+AC
[AB(C+BD)+AB]C=BC
AB(C+D)=A+B+CD
(A+C)(ABC+ACD)=ABC+ACD
Answer: 3. AB(C+D)=A+B+CD
The expression AB(C+D) simplifies to ABC+ABD, not A+B+CD.
NOT gate
NAND gate
AND gate
XOR gate
Answer: 2. NAND gate
Answer: 1. NAND
Switches connected in parallel
MOS transistors connected in series
Switches connected in series
All of the above
Answer: 1. Switches connected in parallel
Answer: 3. NOR
(A+B)+C=A+(B+C)
A+B=B+A
A⋅(B+C)=(A⋅B)+(A⋅C)
Answer: 2. A+B=B+A
Answer: 1. NOT
Their efficiency in storing data
The random and sequential access method
The number of bytes used to store characters
Their encoding sequences
Answer: 4. Their encoding sequences
Gray code
8421 code
Excess-3 code
Algebraic code
Answer: 1. Gray code
Data transfer
Continuously varying signals
Arithmetic and logical computation
All of the above
Answer: 4. All of the above
101001.1100
010111.0011
101000.0100
101000.0011
Answer: 3. 101000.0100
Invert all bits: 101000.0011
Add 1 to the least significant bit (LSB): 101000.0100.
Is a subset of 8-bit EBCDIC
Is used only in Western countries
Is version II of the ASC standard
Has 128 characters, including 32 control characters
Answer: 4. Has 128 characters, including 32 control characters
Answer: 1. 504.771
0 or 1
All of the above