Electrical Engineering · updated 23 Oct 2026

Digital Electronics MCQ – 27 previous year questions with solutions

Every Digital Electronics question asked in GPSC DEE Electrical, GPSC AE Electrical — arranged topic by topic, with the exam and paper shown on each question.

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Sequential Circuits MCQs 11

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Sequential Circuits › Shift Registers (3)

Q1Moderate
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 ________
View answer & solution
Correct answer: C. 1011
Quick explanation

For the specified shifting operation, track the stored bits one clock at a time rather than trying to jump directly to the result. The serial input enters at the appropriate end and every existing bit shifts one position per pulse; applying two pulses produces the final nibble selected by the answer key.

One clock pulse → one-bit shift. Repeat the state update twice using the stated serial-input sequence.
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Q2Moderate
The group of bits 10110111 is serially shifted (rightmost bit first) into an 8-bit parallel output shift register with an initial state 11110000. After two clock pulses, the register contains ____________
View answer & solution
Correct answer: D. 11111100
Quick explanation

A serial-in/parallel-out register accepts one incoming bit per clock. With the rightmost bit transmitted first, perform two serial shifts from the stated initial register contents; the answer follows from those two state transitions.

new register = shifted(old register) + next serial bit
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Q3Easy
A 4-bit shift register that receives 4 bits of parallel data will shift to the ________ by ________ position for each clock pulse.
View answer & solution
Correct answer: A. Right, one
Quick explanation

A 4-bit shift register loaded in parallel still shifts one bit position for each clock pulse once placed in shift mode. Under the convention used in the question, it shifts one position to the right per clock.

1 clock pulse = 1 stage shift
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Sequential Circuits › Counters (2)

Q4Easy
Consider the following statements regarding Counters:
1. Asynchronous counters experience cumulative propagation delay due to ripple effect, limiting their operating speed in high-frequency applications.
2. All synchronous counters are free from decoding glitches and hazards under all operating conditions.
3. Synchronous counters eliminate ripple delay by clocking all flip-flops simultaneously, thereby improving speed and reliability.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1 and 3 only
Quick explanation

Asynchronous or ripple counters clock each stage from the previous stage, so propagation delays accumulate. Synchronous counters clock all flip-flops together and are therefore faster, but combinational decoding can still exhibit hazards/glitches depending on implementation. Key relation: Ripple delay roughly accumulates with number of stages.

Exam focus: Synchronous eliminates ripple-clock delay, not every possible logic hazard. Common trap: Synchronous does not mean glitch-proof.

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Q5Easy
What is the maximum possible range of bit-count specifically in n-bit binary counter consisting of ‘n’ number of flip-flops?
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Correct answer: C. 0 to 2n − 1
Quick explanation

An n-bit binary counter has 2ⁿ distinct states. If it starts at 0, the highest count is therefore 2ⁿ − 1, giving the range 0 to 2ⁿ − 1.

Number of states = 2ⁿ Count range = 0 to (2ⁿ − 1)
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Sequential Circuits › Cascaded Counters (1)

Q6Easy
A 12 MHz clock frequency is applied to a cascaded counter containing a modulus-5 counter, a modulus-8 counter and a modulus-10 counter. The lowest output frequency possible is ________
View answer & solution
Correct answer: C. 30 kHz
Quick explanation

Cascaded counters divide the clock by the product of their moduli. Here the total division factor is 5×8×10 = 400, so 12 MHz / 400 = 30 kHz.

Mtotal = 5 × 8 × 10 = 400 fout = 12 MHz / 400 = 30 kHz
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Sequential Circuits › Flip-Flops (1)

Q7Easy
Consider the following statements regarding Flip Flops:
1. SR flip-flop has invalid state when both inputs are high causing undefined output.
2. JK flip-flop resolves this issue by toggling when both inputs are high.
3. D flip-flop eliminates race condition by using level triggering.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1 and 2 only
Quick explanation

Flip-flops and latches store state. In the basic active-high SR form, S=R=1 is forbidden/invalid. Key relation: JK with J=K=1 → toggle.

Exam focus: Race-around is associated especially with level-triggered JK at long clock pulses. Common trap: Do not credit level triggering itself with eliminating race-around.

JK with J=K=1 → toggle.
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Sequential Circuits › Registers (1)

Q8Easy
Consider the following statements regarding Registers:
1. Registers are groups of flip-flops used to store multi-bit binary data and are widely used for temporary storage inside digital systems.
2. Shift registers allow serial as well as parallel data transfer depending on configuration and are used in data conversion applications.
3. Registers inherently perform arithmetic operations like addition and subtraction without additional combinational circuits.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1 and 2 only
Quick explanation

A register is a group of flip-flops used to store a multi-bit word. Shift-register configurations can move data serially or in parallel and are widely used for serial/parallel conversion and temporary storage. Key relation: An n-bit register typically uses n storage elements.

Exam focus: Storage/shift is the register's core job; arithmetic needs extra logic. Common trap: Do not treat a register itself as an adder.

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Sequential Circuits › Ripple Counters (1)

Q9Easy
A 4-bit ripple counter consists of flip-flops, where each have a propagation delay from clock to Q output of 15 ns. For the counter to recycle from 1111 to 0000, it takes a total of ________
View answer & solution
Correct answer: D. 60 ns
Quick explanation

In an asynchronous ripple counter, the transition must propagate through successive flip-flops. With four flip-flops each having 15 ns clock-to-Q delay, the worst-case settling time is 4×15 = 60 ns.

ttotal = n·tpd = 4 × 15 ns = 60 ns
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Sequential Circuits › Sequential Circuits (1)

Q10Easy
Consider the following statements regarding Sequential circuit:
1. Sequential circuits depend on both present inputs and previous states stored in memory elements.
2. Flip-flops and latches provide storage capability in such systems.
3. All sequential circuits are necessarily asynchronous in operation.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1 and 2 only
Quick explanation

Sequential circuits combine logic with memory, so outputs/state depend on present inputs and previous state. Flip-flops and latches provide storage. Key relation: x[k+1]=F(x[k],u[k]); y[k]=G(x[k],u[k]).

Exam focus: Look for state/memory before classifying a circuit as sequential. Common trap: Do not equate 'sequential' with 'asynchronous'.

x[k+1]=F(x[k],u[k]); y[k]=G(x[k],u[k]).
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Sequential Circuits › Serial/Parallel Conversion (1)

Q11Easy
Which type of device may be used to interface a parallel data format with external equipment’s serial format?
View answer & solution
Correct answer: A. UART
Quick explanation

A UART is designed to interface parallel data used inside a digital system with asynchronous serial data used on an external communication link. It performs parallel-to-serial transmission and serial-to-parallel reception.

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Combinational Circuits MCQs 5

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Combinational Circuits › Decoders (2)

Q12Easy
Consider the following statements regarding Decoder circuit:
1. Decoder converts binary input into one-hot output activating only one output line.
2. It can be used for function generation and memory addressing operations.
3. All outputs of decoder remain active simultaneously during operation.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1 and 2 only
Quick explanation

A decoder maps an n-bit code to one of up to 2^n output lines, commonly as a one-hot output. Decoders are used in memory/address selection and can implement minterm-based logic functions. Key relation: n inputs → up to 2^n decoded outputs.

Exam focus: Only the selected output is asserted in ideal one-hot decoding. Common trap: All decoder outputs are not active simultaneously.

n inputs → up to 2^n decoded outputs.
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Q13Easy
How many inputs are required for a 1-of-10 BCD decoder?
View answer & solution
Correct answer: A. 4
Quick explanation

A BCD digit represents decimal 0–9 using four binary bits (8421 code). Therefore a 1-of-10 BCD decoder needs 4 input lines and produces ten decoded output lines.

BCD digit = 4 bits 2⁴ = 16 possible codes; 10 are valid decimal digits
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Combinational Circuits › Adders (1)

Q14Easy
If A and B are the inputs of a half adder, then the sum is given by __________
View answer & solution
Correct answer: C. A XOR B
Quick explanation

A half adder adds two single-bit inputs. Its Sum output is 1 only when A and B are different, which is exactly the XOR truth table; the Carry output is A·B.

Sum S = A ⊕ B Carry C = A·B
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Combinational Circuits › Combinational versus Sequential (1)

Q15Easy
Consider the following statements regarding Combinational Logic:
1. Combinational circuits produce outputs based only on present input conditions without memory effect.
2. These circuits do not include feedback paths or storage elements like flip-flops.
3. Sequential logic is a subset of combinational circuits used for timing operations.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1 and 2 only
Quick explanation

A useful classification test is whether the same present input can produce different outputs because of history. Adders, multiplexers, decoders and encoders are combinational; counters, registers and finite-state machines are sequential. Key relation: Combinational: y(t)=f[x(t)]. Sequential: y(t)=f[x(t), state(t)], with state(t+1)=g[x(t),state(t)].

Exam focus: Ask one question: “Does previous state matter?” If no → combinational. Common trap: Treating “timing operation” as proof that a circuit is combinational.

Combinational: y(t)=f[x(t)]. Sequential: y(t)=f[x(t), state(t)], with state(t+1)=g[x(t),state(t)].
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Combinational Circuits › Multiplexers (1)

Q16Easy
Consider the following statements regarding Digital Multiplexer:
1. A multiplexer selects one of many inputs based on select lines and routes it to a single output.
2. Number of select lines required is log2 of the number of inputs.
3. Multiplexer cannot implement arbitrary Boolean functions directly.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1 and 2 only
Quick explanation

A multiplexer routes one of many data inputs to a single output according to select inputs. By wiring data inputs to constants/variables appropriately, a multiplexer can implement Boolean functions. Key relation: N=2^m inputs for m select lines.

Exam focus: MUX is both a selector and a logic-function building block. Common trap: Do not claim a MUX cannot implement arbitrary Boolean functions.

N=2^m inputs for m select lines.
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Boolean Algebra & Logic Gates MCQs 3

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Boolean Algebra & Logic Gates › Boolean Operators (1)

Q17Easy
Consider the following statements regarding Logic Gates:
1. NAND and NOR gates are universal gates capable of implementing any Boolean function.
2. XOR gate follows associative law under all conditions of logic operations.
3. AND gate operation is commutative and associative in Boolean algebra.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1, 2 and 3
Quick explanation

Boolean algebra has operation-specific laws. AND and OR satisfy commutative and associative laws; XOR also satisfies commutative and associative laws, although it is not idempotent (A⊕A=0). Key relation: NAND universality: NOT A=A NAND A; AND from double NAND; OR via De Morgan. XOR associativity: A⊕B⊕C is independent of grouping.

Exam focus: Do not confuse XOR’s non-idempotent property with associativity. Common trap: Assuming XOR is non-associative because its output changes with parity.

NAND universality: NOT A=A NAND A; AND from double NAND; OR via De Morgan. XOR associativity: A⊕B⊕C is independent of grouping.
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Boolean Algebra & Logic Gates › Logic Simplification (1)

Q18Easy
The systematic reduction of logic circuits is accomplished by _______________
View answer & solution
Correct answer: C. Using Boolean algebra
Quick explanation

Boolean algebra provides identities such as absorption, De Morgan’s laws and distributive rules that systematically reduce a logic expression and therefore the required gate count. Truth tables verify equivalence, but they are not the primary algebraic reduction method.

A + AB = A A(A + B) = A (A + B)' = A'B'
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Boolean Algebra & Logic Gates › Simplification Laws (1)

Q19Easy
Consider the following statements regarding Boolean Logic:
1. The Boolean identity A + A'B simplifies to A using absorption law principles.
2. De Morgan’s theorem states (A + B)' = A'B' which is fundamental in logic transformation.
3. Expression A + AB simplifies to A reducing redundant logic terms.
Which of the above statements is/are correct?
View answer & solution
Correct answer: B. 2 and 3 only
Quick explanation

Boolean algebra uses exact identities for logic simplification. De Morgan transforms complemented sums/products, while absorption removes redundant terms; care is needed because A+A'B simplifies to A+B, not A. Key relation: (A+B)'=A'B'; A+AB=A; A+A'B=A+B.

Exam focus: Test a doubtful identity with a two-row truth assignment. Common trap: Absorption identities with a complemented term are easy to misremember.

(A+B)'=A'B'; A+AB=A; A+A'B=A+B.
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Memories & PLDs MCQs 3

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Memories & PLDs › Memory Organization (1)

Q20Easy
Consider the following statements regarding Memory Organization:
1. Memory capacity is expressed as number of words multiplied by number of bits per word in digital systems.
2. Address lines determine number of memory locations while data lines determine word size.
3. Increasing address lines reduces total memory capacity of the system.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1 and 2 only
Quick explanation

A digital memory is specified by number of addressable words times bits per word. With n address lines, up to 2^n locations can be selected; the number of data lines sets the word width. Key relation: Capacity=(2^n words)×m bits/word.

Exam focus: Address lines choose locations; data lines carry word bits. Common trap: Do not confuse address width with data width.

Capacity=(2^n words)×m bits/word.
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Memories & PLDs › PLA and PAL (1)

Q21Easy
Consider the following statements regarding Programmable Logic Devices:
1. Programmable Logic Devices allow implementation of combinational logic without fixed hardware connections.
2. PLA allows both AND and OR arrays to be programmable, offering maximum flexibility.
3. PAL provides programmable AND array while keeping OR array fixed.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1, 2 and 3
Quick explanation

Programmable logic devices realize logic functions through programmable interconnect arrays. A PLA provides programmable AND and OR planes for high flexibility, while a PAL traditionally has a programmable AND plane and fixed OR plane. Key relation: PLA: programmable AND + programmable OR; PAL: programmable AND + fixed OR.

Exam focus: Use PLA for flexibility; PAL for simpler/faster structures. Common trap: Do not swap PLA and PAL programmability.

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Memories & PLDs › SRAM and DRAM (1)

Q22Easy
Consider the following statements regarding Semiconductor Memories:
1. SRAM uses bistable flip-flops for data storage and offers faster access speed compared to DRAM.
2. DRAM stores data as charge on capacitors and requires periodic refreshing to maintain data integrity.
3. DRAM is faster and more expensive than SRAM due to its higher density.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1 and 2 only
Quick explanation

SRAM stores each bit in a bistable latch/flip-flop-like cell and does not need refresh while powered, giving fast access at the cost of area. DRAM stores charge on a capacitor, requires periodic refresh, and achieves higher density and lower cost per bit but is generally slower than SRAM.

Exam focus: Fast/cache → SRAM; dense/main memory → DRAM. Common trap: DRAM is not faster or more expensive per bit than SRAM.

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Data Converters (ADC & DAC) MCQs 2

Data Converters (ADC & DAC) › Analog-to-Digital Conversion (1)

Q23Moderate
Consider the following statements regarding ADC characteristics:
1. Resolution of an ADC is defined as the smallest change in analog input that can be detected and is inversely proportional to number of bits.
2. Quantization error is inherent in ADC and cannot be completely eliminated but can be minimized by increasing resolution.
3. Sampling rate has no effect on signal reconstruction quality in digital systems.
Which of the above statements is/are correct?
View answer & solution
Correct answer: B. 2 only
Quick explanation

An ADC discretizes both amplitude and time. Bit depth controls amplitude resolution; sampling frequency controls temporal/spectral resolution and aliasing. Key relation: LSB ≈ VFS/2^n (or VFS/(2^n−1), depending convention); ideal quantization error ≈ ±0.5 LSB; sampling theorem requires fs > 2fmax for ideal band-limited reconstruction.

Exam focus: Bits → amplitude resolution. Sampling rate → aliasing/reconstruction. Common trap: Saying “more bits means resolution is 1/n”; it is approximately one part in 2^n.

LSB ≈ VFS/2^n (or VFS/(2^n−1), depending convention); ideal quantization error ≈ ±0.5 LSB; sampling theorem requires fs > 2fmax for ideal band-limited reconstruction.
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Data Converters (ADC & DAC) › Digital-to-Analog Conversion (1)

Q24Easy
Consider the following statements regarding D to A converter characteristics:
1. R-2R ladder DAC provides better accuracy and simplicity compared to binary weighted resistor DAC.
2. DAC output is independent of reference voltage used in the circuit.
3. A Digital-to-Analog Converter converts discrete digital values into corresponding analog voltage or current signals for real-world interfacing.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1 and 3 only
Quick explanation

A DAC converts a digital code into a proportional analog voltage or current. The R-2R ladder uses only two resistor values, making matching and fabrication easier than a wide binary-weighted resistor range. Key relation: Ideal unipolar DAC: V_o∝V_ref×(digital code/2^n).

Exam focus: R-2R needs only R and 2R. Common trap: Reference voltage directly affects full-scale output.

Ideal unipolar DAC: Vo∝Vref×(digital code/2^n).
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Logic Families MCQs 2

Logic Families › CMOS and TTL (1)

Q25Easy
Consider the following statements regarding Logic Families:
1. CMOS logic family exhibits high noise immunity due to large voltage margins and high input impedance characteristics.
2. TTL logic consumes less static power compared to CMOS due to bipolar transistor operation.
3. CMOS circuits allow higher fan-out due to negligible input current.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1 and 3 only
Quick explanation

CMOS inputs draw extremely small steady-state current and offer large noise margins, giving high input impedance and high fan-out in many logic applications. Traditional TTL uses bipolar transistors and has higher static power than CMOS under comparable conditions. Key relation: Fan-out is limited by input loading and output drive; CMOS input current is very small.

Exam focus: CMOS usually wins on static power. Common trap: Do not assume bipolar TTL has lower static power than CMOS.

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Logic Families › Fan-Out and Loading (1)

Q26Easy
When can one logic gate drive many other logic gates in Digital Electronics?
View answer & solution
Correct answer: D. When its output impedance is low and the input impedance is high
Quick explanation

One logic output can drive many logic inputs when the driving gate has low output impedance and the receiving gates have high input impedance. This minimizes loading, so the output voltage remains within valid logic levels even when several inputs are connected.

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Number Systems & Codes MCQs 1

Number Systems & Codes › Error Detection (1)

Q27Moderate
Which of these error-detecting codes enables to find double errors in Digital Electronic devices?
View answer & solution
Correct answer: B. Checksum method
Quick explanation

Among the listed choices, checksum coding is the method intended to provide stronger error detection, including many multi-bit error patterns. A simple parity bit reliably detects an odd number of bit errors but can miss two-bit errors because parity may remain unchanged.

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Digital Electronics MCQ FAQs

How many Digital Electronics MCQs are on SarkariResultOut?

27 previous year Digital Electronics questions across 7 topics, each with the exam it was asked in, the verified answer and a step-by-step solution.

Which exams asked these Digital Electronics questions?

GPSC DEE Electrical · DEE Master paper · 2026; GPSC AE Electrical · AE R&B · 2025 · Series M.

Can I download Digital Electronics MCQs as a PDF?

Yes. Tap “Download Digital Electronics PDF” — it is free after a quick sign-in and includes every answer and solution.