Digital Circuits — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Digital Circuits MCQs with step-by-step solutions (21 questions). Part of Semiconductor Electronics. Practise online on Prepizo — no login needed.
▶ Practise Digital Circuits online (free)
Questions with solutions
Q1 — Digital Circuits · easy · theory
Which of the following gate is called universal gate?
A. OR gate
B. AND gate
C. NAND gate ✓ Correct
D. NOT gate
Solution: NAND (and NOR) are universal gates — every logic gate and Boolean expression can be realised using only NAND (or only NOR) gates.
Q2 — Digital Circuits · medium · numerical
In a logic circuit, inputs A and B are each passed through a NOT gate, and the two inverted outputs feed a NOR gate whose output is Y. The truth table is
A. Y = 0 only for A = 0, B = 0 (otherwise 1)
B. Y = 1 except when A = 1, B = 1 (then 0)
C. Y = 1 only for A = 0, B = 0 (otherwise 0)
D. Y = 1 only for A = 1, B = 1 (otherwise 0) ✓ Correct
Solution: $Y = \overline{\bar{A} + \bar{B}} = A \cdot B$ (De Morgan) — the circuit behaves as an AND gate, so Y = 1 only when A = B = 1.
Q3 — Digital Circuits · medium · theory
In a circuit, a +6 V supply feeds an LED (output Y) through a resistor. Two switches A and B are connected in parallel between the supply line and ground — closing either switch (input = 1) diverts the current away from the LED. The circuit corresponds to the logic gate
A. NOR ✓ Correct
B. AND
C. OR
D. NAND
Solution: The LED glows (Y = 1) only when both switches are open (A = B = 0); closing either one grounds the current. This is the NOR truth table.
Q4 — Digital Circuits · medium · theory
In a circuit, a +6 V supply feeds an LED (output Y) through a resistor. Two switches A and B are connected in series between the LED line and ground — the current is diverted away from the LED only when both switches are closed (input = 1). The correct Boolean operation represented by the circuit diagram drawn is
A. OR
B. NAND ✓ Correct
C. NOR
D. AND
Solution: The LED glows (Y = 1) unless both A and B are 1 (both switches closed). This is the NAND truth table.
Q5 — Digital Circuits · medium · numerical
In a combination of gates, inputs A and B feed two AND gates: the first AND gate receives A and $\bar{B}$ (B inverted), the second receives $\bar{A}$ and B. The outputs of the two AND gates feed an OR gate whose output is Y. The output Y can be written in terms of inputs A and B as
A. $\overline{A \cdot B} + A \cdot B$
B. $A \cdot \bar{B} + \bar{A} \cdot B$ ✓ Correct
C. $\overline{A \cdot B}$
D. $\overline{A + B}$
Solution: First AND: $A \cdot \bar{B}$; second AND: $\bar{A} \cdot B$; OR combines them:
$Y = A \cdot \bar{B} + \bar{A} \cdot B$ (the XOR expression)
Q6 — Digital Circuits · medium · numerical
In an electrical network, inputs A and B feed a NOR gate; its output is fed to both inputs of a second NOR gate, whose output passes through a NOT gate to give Y. The given electrical network is equivalent to
A. AND gate
B. OR gate
C. NOR gate ✓ Correct
D. NOT gate
Solution: $Y_1 = \overline{A + B}$; the second NOR with tied inputs inverts it: $Y_2 = A + B$; the final NOT gives $Y = \overline{A + B}$ — a NOR gate overall.
Q7 — Digital Circuits · medium · numerical
In a circuit, inputs A and B feed an OR gate; the OR output and input C feed an AND gate whose output is Y. To get output 1 for the circuit, the correct choice for the input is
A. A = 1, B = 0, C = 0
B. A = 1, B = 1, C = 0
C. A = 1, B = 0, C = 1 ✓ Correct
D. A = 0, B = 1, C = 0
Solution: $Y = (A + B) \cdot C$ — Y = 1 requires C = 1 and at least one of A, B equal to 1. Only A = 1, B = 0, C = 1 satisfies this.
Q8 — Digital Circuits · medium · numerical
Inputs A and B are each inverted by NOT gates ($y_1 = \bar{A}$, $y_2 = \bar{B}$), and the two inverted signals feed a NOR gate whose output is y. Which logic gate is represented by this combination of logic gates?
A. OR
B. NAND
C. AND ✓ Correct
D. NOR
Solution: $y = \overline{\bar{A} + \bar{B}} = A \cdot B$ — the combination represents an AND gate.
Q9 — Digital Circuits · medium · numerical
Inputs A and B feed a NAND gate; its output goes to a second NAND gate whose two inputs are joined together. The output (X) of this logic circuit will be
A. $\bar{A} \cdot \bar{B}$
B. $\overline{A \cdot B}$
C. $A \cdot B$ ✓ Correct
D. $\overline{A + B}$
Solution: The second NAND with tied inputs acts as a NOT gate:
$X = \overline{\overline{A \cdot B}} = A \cdot B$ — an AND gate.
Q10 — Digital Circuits · easy · theory
The device that can act as a complete electronic circuit is
A. Junction diode
B. Integrated circuit ✓ Correct
C. Junction transistor
D. Zener diode
Solution: An integrated circuit is a miniature electronic circuit built on a single semiconductor crystal, containing transistors, resistors and capacitors — a complete circuit in itself.
Q11 — Digital Circuits · medium · numerical
In a circuit, inputs A and B feed an OR gate (gate I); the OR output and input C feed an AND gate (gate II) whose output is Y. To get an output Y = 1, the inputs (A, B, C) must be
A. 0, 1, 0
B. 0, 0, 1
C. 1, 0, 1 ✓ Correct
D. 1, 0, 0
Solution: $Y = (A + B) \cdot C$ — with A = 1, B = 0, C = 1: $Y = 1 \cdot 1 = 1$
Q12 — Digital Circuits · medium · numerical
In a circuit, inputs A and B feed a NOR gate; its output is fed to both inputs of a NAND gate, whose output then passes through a NOT gate. The circuit is equivalent to
A. AND gate
B. NAND gate
C. NOR gate ✓ Correct
D. OR gate
Solution: $Y_1 = \overline{A + B}$; NAND with tied inputs inverts: $Y_2 = A + B$; final NOT: $Y = \overline{A + B}$ — a NOR gate.
Q13 — Digital Circuits · medium · numerical
Inputs A and B feed a NOR gate (gate I); its output $\overline{A + B}$ is fed to both inputs of a second NOR gate (gate II) whose output is Y. The output Y for all possible inputs A and B is expressed by the truth table
A. Y = 1 only when A = 1, B = 1 (otherwise 0)
B. Y = 1 except when A = 1, B = 1 (then 0)
C. Y = 1 only when A = 0, B = 0 (otherwise 0)
D. Y = 0 only when A = 0, B = 0 (otherwise 1) ✓ Correct
Solution: The second NOR with tied inputs inverts the first:
$Y = \overline{\overline{A + B}} = A + B$ — an OR gate: Y = 0 only when both inputs are 0.
Q14 — Digital Circuits · easy · theory
The output of OR gate is 1
A. only if both inputs are zero
B. if either or both inputs are 1 ✓ Correct
C. only if both inputs are 1
D. if either input is zero
Solution: The OR operation $Y = A + B$ gives 1 if either or both inputs are 1.
Q15 — Digital Circuits · medium · numerical
Inputs A and B feed a NAND gate; its output goes to a second NAND gate with both of its inputs joined together, giving output Y. This diagram performs the logic function of
A. OR gate
B. AND gate ✓ Correct
C. XOR gate
D. NAND gate
Solution: $X = \overline{A \cdot B}$; the second NAND acts as NOT: $Y = \overline{X} = A \cdot B$ — an AND gate.
Q16 — Digital Circuits · easy · theory
The truth table given below — Y = 0 for (A, B) = (0,0), (1,0), (0,1) and Y = 1 for (1,1) — represents
A. AND gate ✓ Correct
B. NOR gate
C. OR gate
D. NAND gate
Solution: Output 1 only when both inputs are 1 is the AND operation: $Y = A \cdot B$
Q17 — Digital Circuits · medium · numerical
In a circuit, inputs A and B feed two AND gates: one receives $\bar{A}$ and B, the other receives A and $\bar{B}$; the outputs of both AND gates feed an OR gate whose output is Y. This circuit represents
A. OR gate
B. XOR gate ✓ Correct
C. AND gate
D. NAND gate
Solution: $Y = \bar{A}B + A\bar{B}$ — the Boolean expression of the XOR gate.
Q18 — Digital Circuits · medium · numerical
Which one of the following gates will have an output of 1? (A) NAND gate with inputs 1 and 1; (B) NOR gate with inputs 1 and 1; (C) NAND gate with inputs 0 and 1; (D) XOR gate with inputs 0 and 0.
A. A
B. B
C. C ✓ Correct
D. D
Solution: (A) $\overline{1 \cdot 1} = 0$; (B) $\overline{1 + 1} = 0$; (C) $\overline{0 \cdot 1} = 1$; (D) $0 \oplus 0 = 0$
Only gate C outputs 1.
Q19 — Digital Circuits · easy · theory
The following truth table — Y = 1 only for (A, B) = (0,0), and Y = 0 for (1,1), (1,0), (0,1) — belongs to which of the following four gates?
A. NOR ✓ Correct
B. XOR
C. NAND
D. OR
Solution: $Y = \overline{A + B}$ — output 1 only when both inputs are 0 is the NOR gate.
Q20 — Digital Circuits · easy · theory
Which of the following gates corresponds to the truth table: Y = 0 for (A, B) = (1,1) and Y = 1 for (1,0), (0,1), (0,0)?
A. NAND ✓ Correct
B. OR
C. AND
D. XOR
Solution: Output is always 1 except when all inputs are 1 — the NAND gate.
Q21 — Digital Circuits · easy · theory
The following truth table — Y = 0 for (A, B) = (0,0) and Y = 1 for (0,1), (1,0), (1,1) — corresponds to the logical gate
A. NAND
B. OR ✓ Correct
C. AND
D. XOR
Solution: Output high if any or all inputs are high, low only when all inputs are low — the OR gate.