Sequential Circuits MCQ Quiz - Objective Question with Answer for Sequential Circuits - Download Free PDF
Last updated on Jun 10, 2025
Latest Sequential Circuits MCQ Objective Questions
Sequential Circuits Question 1:
In a J-K flip flop, when J = 1 and K = 1 then it will be considered as:
Answer (Detailed Solution Below)
Sequential Circuits Question 1 Detailed Solution
Concept:
JK flip flop:
The truth table of JK flipflop:
J |
K |
Q |
\(\bar Q\) |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
T flip-flop is formed by combining both J and K inputs of the JK-flipflop
In the above truth table when J = K = 1, its output is toggled.
Characteristic Table of JK flip flop
J |
K |
Qn |
Qn+1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
Qn+1 = JQ̅n + K̅Qn
Sequential Circuits Question 2:
Which of the followings are sequential circuit?
A. Flip flop
B. Multiplexer
C. Counter
D. Decoder
E. Adder
Choose the correct answer from the options given below :
Answer (Detailed Solution Below)
Sequential Circuits Question 2 Detailed Solution
The correct answer is 2) A, C Only.
Key Points
- Sequential circuits are types of circuits where the output depends not just on the current inputs, but also on the history of inputs. This means they have memory elements that store previous states.
- Flip flop is a basic memory element used in sequential circuits, which can store one bit of data.
- Counter is a sequential circuit that goes through a prescribed sequence of states upon the application of input pulses.
Additional Information
- Multiplexer (MUX) is a combinational circuit that selects one of many input signals and forwards the selected input into a single line. It does not store any state information, hence it is not a sequential circuit.
- Decoder is also a combinational circuit that converts binary information from n input lines to a maximum of 2^n unique output lines. It does not have memory, so it is not a sequential circuit.
- Adder is a combinational circuit used for adding binary numbers. It does not store any state information, hence it is not a sequential circuit.
Sequential Circuits Question 3:
What is the difference between edge-triggering and level-triggering?
Answer (Detailed Solution Below)
Sequential Circuits Question 3 Detailed Solution
The correct answer is: 1) Edge-triggering responds to the transition of the clock, while level-triggering responds to the level of the clock.
Triggering methods of flip-flop:
1.) Edge Triggering:
- In a sequential circuit, if the output changes when the signal transits from a high level to a low level or from a low level to a high level, we call it edge triggering.
- Here, the edge that changes the voltage from the low level to the high level is called the positive edge.
- And, the edge that changes the voltage from the high level to the low level is called the negative edge.
2.) Level Triggering:
- In the sequential circuit, if the output changes during the high voltage period or low voltage period, it is called level triggering.
- The output of the circuit is high for a positive level and low for a negative level.
Sequential Circuits Question 4:
How many distinct states are there are in an n-bit ring counter?
Answer (Detailed Solution Below)
Sequential Circuits Question 4 Detailed Solution
Ring Counter:
The Ring shift counter is a recirculating register in which the serial output is connected back to the serial input as shown:
A Straight ring counter with ‘n’ flip-flops will have n states.
Example:
A Ring counter of 5 bits has 5 states.
Sequential Circuits Question 5:
No. of Flip flop required for designing synchronous counter having 5 states with counting sequence : 0 = > 1 = > 6 = > 10 = > 4 will be :
Answer (Detailed Solution Below)
Sequential Circuits Question 5 Detailed Solution
Concept:
The number of flip-flops required for a synchronous counter is determined by the highest binary value in the counting sequence. The formula is:
n ≥ log2(max state value)
where n is the number of flip-flops needed.
Calculation:
Given:
Counting sequence: 0 → 1 → 6 → 10 → 4
Solution:
1. Identify the highest state value in the sequence: 10
2. Convert 10 to binary: 1010 (4 bits)
3. Calculate minimum number of flip-flops:
n ≥ log2(10) → n ≥ 3.32
Since we can't have a fraction of a flip-flop, we round up to the next integer: n = 4
Verification:
4 flip-flops can represent up to 24 = 16 states (0-15), which covers all required states in the sequence (0,1,6,10,4).
Final Answer:
The number of flip-flops required is 3) 4.
Top Sequential Circuits MCQ Objective Questions
Which condition is shown in J-K flip flop as no changes next state from the present state?
Answer (Detailed Solution Below)
Sequential Circuits Question 6 Detailed Solution
Download Solution PDFTruth Table of J-K flip flop
J |
K |
Qn+1 |
0 |
0 |
No change |
0 |
1 |
Reset |
1 |
0 |
Set |
1 |
1 |
Toggle |
At J = 0, K = 0, the J-K flip flop has no changes in the next state from the present state.
Truth Table of S-R flip flop
S |
R |
Qn+1 |
0 |
0 |
No change |
0 |
1 |
Reset |
1 |
0 |
Set |
1 |
1 |
Invalid |
Truth Table of D flip flop
D |
Qn+1 |
0 |
0 |
1 |
1 |
Truth Table of T flip flop
T |
Qn+1 |
0 |
No change |
1 |
Toggle |
D flip flop can be made from a J-K flip flop by making
Answer (Detailed Solution Below)
Sequential Circuits Question 7 Detailed Solution
Download Solution PDFD flip flop:
D flip flop has only one input terminal. The output of the D flip flop will be the same as the input. Hence, it is used in delay circuits.
The circuit is as shown below.
Logic symbol:
Truth table:
D |
Qn (Present state) |
Qn+1 (Next state) |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
Characteristic equation: Qn+1 = D
The D flip flop may be obtained from an S-R flip flop by just putting one inverter between the S and R as shown in the figure below.
S = R̅
The D flip flop may be obtained from a J-K flip flop by just putting one inverter between the J and K as shown in the figure below.
K = J̅
T flip flop:
T flip flop has only one input terminal. The output of the T flip flop will be toggled when the input is high on every new clock pulse. The output will be the same as the previous state when the input is low.
The circuit is as shown below.
Logic symbol:
Truth table:
T |
Qn (Present state) |
Qn+1 (Next state) |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
Characteristic equation: Qn+1 = TQ̅n + T̅Qn
The T flip flop may be obtained from a J-K flip flop by making both the inputs are the same i.e. J = K.
Which is used for storing the one-bit digital data?
Answer (Detailed Solution Below)
Sequential Circuits Question 8 Detailed Solution
Download Solution PDFFlip-Flop
- A flip-flop is the basic storage element in sequential logic.
- A flip-flop is a device that stores a single bit (binary digit) of data.
- The stored data can be changed by applying varying inputs.
- Flip Flops are edge-triggered while the latch is level-triggered.
- Flip Flops are of 4 types: SR, JK, T, and D flip-flops.
Register
- A Register is a collection of flip-flops.
- For storing n-bit data, a register comprising of n number of flip-flops is used.
GATE
- A logic gate is a device that acts as a building block for digital circuits.
- They perform basic logical functions that are fundamental to digital circuits.
- Example: AND, OR, NOT, NAND, NOR, XOR, XNOR
Master-slave configuration is used in FF to
Answer (Detailed Solution Below)
Sequential Circuits Question 9 Detailed Solution
Download Solution PDFRace around condition:
For JK flip-flop if J, K, and Clock are equal to 1 the state of flip-flop keeps on toggling which leads to uncertainty in determining the output of the flip-flop. This problem is called Race around the condition.
This can be eliminated by using the following methods.
- Increasing the delay of flip-flop
- Use of edge-triggered flip-flop
- Use of master-slave JK flip flop
The Master-slave configuration is used in a flipflop to eliminate the race around condition but not to store two bits of information.
Race around condition is associated with ______.
Answer (Detailed Solution Below)
Sequential Circuits Question 10 Detailed Solution
Download Solution PDFExplanation:
1. Practically, we don’t get toggling in sequential circuits. Since clock pulse is more than the propagation delay, so within one clock pulse the output will keep on toggling again and again and it may become indeterminate. This is known as the race-around condition.
2. Race Around condition occurs because of the feedback connection.
Race around condition (RAC)
- Race around condition occurs only in level-triggered flip flop
- Level triggered is transparent.
- Even though input is constant, output continuously toggles. Changes for some time continuously.
- RAC is when J = 1 and K = 1 [flip flop in toggling mode] and tp > tff
- Output toggles many times instead of once in one clock pulse. It is undesirable and called as “Race Around Condition”
Race around condition can further be explained with SR latch example:
A clocked latch is shown below
- When S = R = CLK = 1
- The outputs Q and Q’ are maintained at logic 0
- There is no race around condition
- However, when clocked SR latch is converted to JK flip flop by providing feedback as shown below
- Then when J = K = CLK = 1, then the output Q and Q’ keeps toggling between 0 and 1, thus, there is a race around condition there
So we can conclude that racing condition occurs in sequential circuits with the level-triggered clock.
Hence option (2) is the correct answer.
Important Points
The race around condition can be eliminated by using the following methods:
- Increasing the delay of flip-flop
- Use of edge-triggered flip-flop
- Use of master-slave JK flip flop
The Master-slave configuration is used in a flipflop to eliminate the race-around condition but not to store two bits of information.
The one input RS flip flop is the ______ flip flop
Answer (Detailed Solution Below)
Sequential Circuits Question 11 Detailed Solution
Download Solution PDFD flip flop :
D flip flop has only one input terminal. The output of the D flip flop will be the same as the input. Hence, it is used in delay circuits.
The circuit is as shown below.
Logic symbol:
Truth table:
D |
Qn (Present state) |
Qn+1 (Next state) |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
The D flip flop may be obtained from an S-R flip flop by just putting one inverter between the S and R as shown in the figure below.
S = R̅
∴ The one input RS flip flop is the D flip flop.
A register capable of incrementing and/or decrementing its contents.
Answer (Detailed Solution Below)
Sequential Circuits Question 12 Detailed Solution
Download Solution PDFThe correct answer is option 1): Counter
Concept:
Flip-flop is a 1-bit memory cell which can be used for storing digital data.- To increase the storage capacity in terms of the number of bits. A group of flip-flops is used. Such a group of flip-flops is known as a Register.
- A register capable of incrementing and/or decrementing its contents is called a counter
- A counter is a register capable of counting the number of clock pulses arriving at its clock input.
- A register is a group of flip-flops used to store multiple bits of data. An adder is a digital logic circuit in electronics that is extensively used for the addition of numbers.
- A latch is a circuit that has two stable states and can be used to store state information.
- The circuit can be made to change state by signals applied to one or more control inputs and will have one or two outputs.
Shifting a register content to left by one bit is equivalent to____
Answer (Detailed Solution Below)
Sequential Circuits Question 13 Detailed Solution
Download Solution PDFConcept:
Shifting a register content to left by one bit is equivalent to multiplication by 2.
Shifting a register content to right by one bit is equivalent to division by 2.
Explanation:
Decimal |
Binary |
Operation |
73 |
0100 1001 |
Original number |
73*2 = 146 |
1001 0010 |
left shift by |
When two asynchronous active low inputs PRESET and CLEAR are applied to a J-K flip flop the output will be
Answer (Detailed Solution Below)
Sequential Circuits Question 14 Detailed Solution
Download Solution PDFThe PRESET and CLEAR inputs of the JK Flip-Flop are asynchronous, which means that they will have an immediate effect on the Q and Q’ outputs regardless of the state of the clock and / or the J and K inputs.
1.When the preset input is activated, the flip-flop will be set (Q=1, not-Q=0) regardless of any of the synchronous inputs or the clock.
2.When the clear input is activated, the flip-flop will be reset (Q=0, not-Q=1), regardless of any of the synchronous inputs or the clock.
3.When preset and clear inputs are activated we get an invalid state on the output, where Q and not-Q go to the same state.
Important Points-
JK Flip-Flop Truth Table- From truth table it can be seen that the CLEAR (CLR) and PRESET inputs are active at a low logic level and put on the Q output of the Flip-Flop, a high logic level regardless of the state of the clock and / or the state of the J and K inputs.
|
Input |
Output |
|||||
|
Preset |
Clear |
CLK |
J |
K |
Q |
Q̅ |
Invalid |
0 |
0 |
✕ |
✕ |
✕ |
1* |
1* |
Preset |
0 |
1 |
✕ |
✕ |
✕ |
1 |
0 |
Clear |
1 |
0 |
✕ |
✕ |
✕ |
0 |
1 |
No change |
1 |
1 |
✕ |
✕ |
✕ |
Q0 |
Q̅0 |
No change |
1 |
1 |
↓ |
0 |
0 |
Q0 |
Q̅0 |
Reset |
1 |
1 |
↓ |
0 |
1 |
0 |
1 |
Set |
1 |
1 |
↓ |
1 |
0 |
1 |
0 |
Toggle |
1 |
1 |
↓ |
1 |
1 |
Q̅0 |
Q0 |
A 4-bit modulo 16 ripple counter uses JK flip flop. If the propagation delay of each flip flop is 50 ns, The maximum clock frequency that can be used is
Answer (Detailed Solution Below)
Sequential Circuits Question 15 Detailed Solution
Download Solution PDFConcept:
In Ripple counters, the carry ripples through, or propagates through every flip-flop, i.e. the propagation delays of all the flip-flops are added to get the overall delay in the counter.
For the counter to work properly, the next clock pulse must arrive when all the carry's generated are propagated through all the flip-flops and the output is stable. This can be mathematically stated as:
For the ripple counter to count properly:
TCLK ≥ n (tpd)FF
TCLK = Clock Interval, and is the inverse of fCLK, i.e.
\(f_{CLK}≤ \frac{1}{n\times (t_{pd})_{FF}}\)
Calculation:
For 4-bit modulo 16 ripple counter has 4 JK flip flops. The total propagation delay will be:
\(T = 4 \times 50\;n\;sec \)
T = 200 nsec
∴ The maximum clock frequency will be:
\(f = \frac{1}{{200\;n}} = 5\;MHz\)