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Electronics (ECE)

Analog/digital electronics, communication

187 Q 4 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 131–140 of 187
Topics in Electronics (ECE)
Q.131 Medium Digital Electronics
A 4-bit binary subtractor using 2's complement method requires:
A 4 full adders and complementing circuit
B 4 full subtractors only
C 3 full adders and 1 full subtractor
D Separate subtraction logic circuits
Correct Answer:  A. 4 full adders and complementing circuit
EXPLANATION

2's complement subtraction (A-B) converts to addition A+2's complement of B. This requires 4 full adders (for addition) and XOR gates (for bit complementing), avoiding dedicated subtractors

Test
Q.132 Medium Digital Electronics
Which multiplexer configuration is used to implement any Boolean function with minimum gates?
A 2-to-1 multiplexer
B 4-to-1 multiplexer
C n-to-1 multiplexer with 2ⁿ inputs
D 8-to-1 multiplexer
Correct Answer:  C. n-to-1 multiplexer with 2ⁿ inputs
EXPLANATION

An n-variable Boolean function can be efficiently implemented using a 2ⁿ-to-1 multiplexer where variables form select lines and minterms form data inputs, achieving minimal gate implementation

Test
Q.133 Medium Digital Electronics
For a modulo-6 counter using JK flip-flops, how many flip-flops are required at minimum?
A 2 flip-flops
B 3 flip-flops
C 4 flip-flops
D 6 flip-flops
Correct Answer:  B. 3 flip-flops
EXPLANATION

For modulo-N counter, minimum flip-flops required = ⌈log₂(N)⌉. For modulo-6: ⌈log₂(6)⌉ = ⌈2.58⌉ = 3 flip-flops (counts 0-5)

Test
Q.134 Medium Digital Electronics
What is the setup time in a flip-flop?
A Time required for output to stabilize after clock pulse
B Minimum time input must be stable before clock edge
C Time taken to reset the flip-flop
D Delay between input and output
Correct Answer:  B. Minimum time input must be stable before clock edge
EXPLANATION

Setup time (tsu) is the minimum duration the input signal must remain stable and valid before the active clock edge for reliable capture and state change

Test
Q.135 Medium Digital Electronics
In Booth's multiplication algorithm, which sequence of multiplier bits requires the fewest operations?
A Alternating 1s and 0s
B Consecutive 1s
C All 1s
D Consecutive 0s
Correct Answer:  B. Consecutive 1s
EXPLANATION

Booth's algorithm replaces consecutive 1s with a single subtraction and addition operation. Consecutive 1s (like 0111) are treated as 1000-1, reducing computation steps compared to processing each 1 individually

Test
Q.136 Medium Digital Electronics
In a synchronous counter design, what is the main advantage over asynchronous counters?
A Lower power consumption
B Simpler circuit design
C Higher operating frequency without timing issues
D Fewer flip-flops required
Correct Answer:  C. Higher operating frequency without timing issues
EXPLANATION

Synchronous counters clock all flip-flops simultaneously, eliminating propagation delay problems. This allows higher operating frequencies without cumulative timing errors present in ripple counters

Test
Q.137 Medium Digital Electronics
In Mealy machine, the output depends on:
A Present state only
B Present input only
C Both present state and present input
D Previous state only
Correct Answer:  C. Both present state and present input
EXPLANATION

Mealy machine output depends on both present state and present input. Moore machine output depends only on present state. Mealy machines are more efficient.

Test
Q.138 Medium Digital Electronics
An asynchronous counter with n flip-flops counts from 0 to:
A n-1
B 2^n - 1
C 2^n
D n
Correct Answer:  B. 2^n - 1
EXPLANATION

An n-bit counter has 2^n possible states (from 0 to 2^n - 1). For 3-bit counter: states are 0 to 7 (000 to 111).

Test
Q.139 Medium Digital Electronics
What is the maximum frequency of a synchronous counter if each flip-flop has a propagation delay of 10 ns?
A 50 MHz
B 100 MHz
C 200 MHz
D 25 MHz
Correct Answer:  B. 100 MHz
EXPLANATION

In synchronous counters, all flip-flops change simultaneously, so maximum frequency is determined by one flip-flop delay. f_max = 1/t_pd = 1/10ns = 100 MHz.

Test
Q.140 Medium Digital Electronics
For a Karnaugh map of 4 variables, what is the minimum number of cells needed to cover all minterms?
A 4
B 8
C 16
D 32
Correct Answer:  C. 16
EXPLANATION

A 4-variable K-map has 2^4 = 16 cells, each representing one minterm. To cover all minterms, minimum 16 cells are needed.

Test
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