Logic Circuits Quiz: Interpret Gates And Build Circuit Logic

1) For a truth table with n Boolean input variables, how many rows (input combinations) are needed?

2^n rows

2 rows

N rows

N^2 rows

Every input variable can be 0 or 1, so n variables give 2 × 2 × … × 2 = 2^n possible combinations, one per row in the truth table. Each Boolean variable represents an independent binary choice, and multiplying the number of possibilities reflects how many distinct input states the function must account for. Thus, a truth table with n inputs always contains 2^n rows.

Explanation

Every input variable can be 0 or 1, so n variables give 2 × 2 × … × 2 = 2^n possible combinations, one per row in the truth table. Each Boolean variable represents an independent binary choice, and multiplying the number of possibilities reflects how many distinct input states the function must account for. Thus, a truth table with n inputs always contains 2^n rows.

2) How many possible input combinations does a logic gate with 4 inputs have?

4

8

16

32

With 4 inputs, each can be 0 or 1, giving 2^4 = 16 different input patterns. The total grows exponentially as inputs increase, because every added variable doubles the number of possible input states. A 4-variable function must therefore evaluate 16 unique combinations.

Explanation

With 4 inputs, each can be 0 or 1, giving 2^4 = 16 different input patterns. The total grows exponentially as inputs increase, because every added variable doubles the number of possible input states. A 4-variable function must therefore evaluate 16 unique combinations.

3) A Boolean function of n variables formally maps:

Constants to variables

Variables to constants

Input tuples {0,1}^n to an output in {0,1}

Operators to variables

A Boolean function takes an n-tuple of bits (each 0 or 1) as input and produces a single bit (0 or 1) as output, so its domain is {0,1}^n and codomain is {0,1}. This mathematical description captures digital circuits: many input wires enter a gate network, and exactly one output wire carries the final 0 or 1 value.

Explanation

A Boolean function takes an n-tuple of bits (each 0 or 1) as input and produces a single bit (0 or 1) as output, so its domain is {0,1}^n and codomain is {0,1}. This mathematical description captures digital circuits: many input wires enter a gate network, and exactly one output wire carries the final 0 or 1 value.

4) A 3-variable Boolean function f(A,B,C) has how many distinct input variables?

2

3

6

8

The name ‘3-variable function’ means it depends on three separate Boolean inputs: A, B, and C. Each variable contributes independently to the input space, so the truth table must account for 2^3 = 8 possible input combinations.

Explanation

The name ‘3-variable function’ means it depends on three separate Boolean inputs: A, B, and C. Each variable contributes independently to the input space, so the truth table must account for 2^3 = 8 possible input combinations.

5) A 2-input gate has this truth table: (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→1. Which gate is it?

AND gate

XOR gate

NAND gate

OR gate

The output is 1 whenever at least one input is 1; this is exactly the behavior of the OR gate. The OR operation represents logical inclusion: the output becomes true if any condition among the inputs holds, making it the simplest multi-input “allow” condition.

Explanation

The output is 1 whenever at least one input is 1; this is exactly the behavior of the OR gate. The OR operation represents logical inclusion: the output becomes true if any condition among the inputs holds, making it the simplest multi-input “allow” condition.

6) A 2-input gate has this truth table: (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→0. Which gate is it?

AND gate

XOR gate

OR gate

XNOR gate

This output is 1 precisely when the inputs differ and 0 when they are the same, which defines the XOR gate. XOR evaluates the inequality of bits; it acts like the logical version of “either, but not both,” making it key for parity checks and digital addition.

Explanation

This output is 1 precisely when the inputs differ and 0 when they are the same, which defines the XOR gate. XOR evaluates the inequality of bits; it acts like the logical version of “either, but not both,” making it key for parity checks and digital addition.

7) Which logic gate corresponds to the Boolean operation of multiplication (·) between bits?

AND gate

OR gate

NOT gate

XOR gate

In Boolean algebra, the product A·B is 1 only when both A and B are 1, which is exactly the truth table of an AND gate. Multiplication and conjunction share identical truth behavior, so the algebraic notation mirrors the functionality of physical AND gates in circuits.

Explanation

In Boolean algebra, the product A·B is 1 only when both A and B are 1, which is exactly the truth table of an AND gate. Multiplication and conjunction share identical truth behavior, so the algebraic notation mirrors the functionality of physical AND gates in circuits.

8) The Boolean expression A + B represents the behavior of which basic logic gate?

NOT gate

AND gate

OR gate

NAND gate

The + symbol in Boolean algebra denotes OR, so A + B is true if A or B (or both) are true—matching the OR gate. Unlike arithmetic addition, Boolean OR outputs 1 without carrying or summing; it simply checks for the presence of at least one true input.

Explanation

The + symbol in Boolean algebra denotes OR, so A + B is true if A or B (or both) are true—matching the OR gate. Unlike arithmetic addition, Boolean OR outputs 1 without carrying or summing; it simply checks for the presence of at least one true input.

9) A 3-input AND gate has inputs A, B, and C. What is its Boolean expression?

A + B + C

(A · B · C)'

A · B · C

A ⊕ B ⊕ C

An AND gate outputs 1 only if all inputs are 1; for three inputs this is written as the product A·B·C. This expresses a strict requirement that every condition must be satisfied. If any variable is 0, the entire product collapses to 0.

Explanation

An AND gate outputs 1 only if all inputs are 1; for three inputs this is written as the product A·B·C. This expresses a strict requirement that every condition must be satisfied. If any variable is 0, the entire product collapses to 0.

10) A circuit takes inputs X and Y into an OR gate, and the result is then ANDed with Z. Which expression describes the output?

X + (Y · Z)

X · Y + Z

(X + Y) · Z

X + Y + Z

First X and Y go into an OR gate, giving X + Y. That output is then ANDed with Z, giving the overall expression (X + Y)·Z. The circuit therefore outputs 1 only when Z is 1 and at least one of X or Y is also 1. The AND stage restricts the otherwise permissive OR output.

Explanation

First X and Y go into an OR gate, giving X + Y. That output is then ANDed with Z, giving the overall expression (X + Y)·Z. The circuit therefore outputs 1 only when Z is 1 and at least one of X or Y is also 1. The AND stage restricts the otherwise permissive OR output.

11) Which pair of logic gates are called 'universal gates' because either one alone can be used to build any combinational logic circuit?

AND and OR

XOR and XNOR

NAND and NOR

NOT and AND

NAND and NOR are universal: given enough of either gate and proper wiring, you can construct all basic gates (NOT, AND, OR) and thus any Boolean function. This universality underlies hardware design: complex logical circuitry can be built entirely from one gate type, simplifying manufacturing.

Explanation

NAND and NOR are universal: given enough of either gate and proper wiring, you can construct all basic gates (NOT, AND, OR) and thus any Boolean function. This universality underlies hardware design: complex logical circuitry can be built entirely from one gate type, simplifying manufacturing.

12) How can you implement a NOT gate using only a 2-input NAND gate?

Connect one input to 0 and use the other as the input.

Connect one input to 1 and use the other as the input.

Connect both inputs of the NAND gate to the same signal.

It is not possible with NAND alone.

If both inputs are A, the NAND output is ¬(A·A) = ¬A, so tying the two inputs together makes the gate behave as a NOT gate. Since A·A = A, feeding the same signal twice lets the NAND gate invert A, demonstrating how NAND alone can generate inversion.

Explanation

If both inputs are A, the NAND output is ¬(A·A) = ¬A, so tying the two inputs together makes the gate behave as a NOT gate. Since A·A = A, feeding the same signal twice lets the NAND gate invert A, demonstrating how NAND alone can generate inversion.

13) How can you implement a NOT gate using only a 2-input NOR gate?

Connect both inputs of the NOR gate to the same signal.

Connect one input to 1 and use the other as the input.

Connect one input to 0 and use the other as the input.

It is not possible with NOR alone.

With both inputs equal to A, the NOR gate output is ¬(A + A) = ¬A, so feeding the same input into both pins yields a NOT operation. Because A + A = A, NOR also becomes a NOT gate under this configuration, showing its universality.

Explanation

With both inputs equal to A, the NOR gate output is ¬(A + A) = ¬A, so feeding the same input into both pins yields a NOT operation. Because A + A = A, NOR also becomes a NOT gate under this configuration, showing its universality.

14) According to De Morgan’s Laws, a NAND gate output (A · B)' is equivalent to:

An AND gate with inverted inputs, A' · B'

An OR gate with inverted inputs, A' + B'

A NOR gate with inverted inputs

An OR gate with normal inputs, A + B

De Morgan’s law gives ¬(A·B) = ¬A ∨ ¬B. In Boolean notation this is A' + B', meaning an OR gate fed by the complements of A and B. Thus, NAND logic can be decomposed into NOT operations followed by an OR, clarifying how compound gates simulate simpler ones.

Explanation

De Morgan’s law gives ¬(A·B) = ¬A ∨ ¬B. In Boolean notation this is A' + B', meaning an OR gate fed by the complements of A and B. Thus, NAND logic can be decomposed into NOT operations followed by an OR, clarifying how compound gates simulate simpler ones.

15) According to De Morgan’s Laws, a NOR gate output (A + B)' is equivalent to:

An OR gate with inverted inputs, A' + B'

An AND gate with inverted inputs, A' · B'

A NAND gate with inverted inputs

An AND gate with normal inputs, A · B

De Morgan’s law gives ¬(A∨B) = ¬A ∧ ¬B. In Boolean notation this is A' · B', so it behaves like an AND gate whose inputs have been inverted. NOR therefore represents a NOT applied to an OR, which mathematically equals the AND of negated inputs.

Explanation

De Morgan’s law gives ¬(A∨B) = ¬A ∧ ¬B. In Boolean notation this is A' · B', so it behaves like an AND gate whose inputs have been inverted. NOR therefore represents a NOT applied to an OR, which mathematically equals the AND of negated inputs.

16) An OR gate can be built using only NOR gates by:

Using a single NOR gate

Using a NOR gate followed by another NOR gate acting as an inverter

Connecting two NOR gates in parallel

It is not possible to build OR from NOR

First, NOR the inputs to get ¬(A∨B). Then feed that result into both inputs of a second NOR gate, which inverts it back to A∨B. The first NOR creates the negated OR, while the second NOR (acting as a NOT) reverses the negation, reconstructing the OR gate using only NOR components.

Explanation

First, NOR the inputs to get ¬(A∨B). Then feed that result into both inputs of a second NOR gate, which inverts it back to A∨B. The first NOR creates the negated OR, while the second NOR (acting as a NOT) reverses the negation, reconstructing the OR gate using only NOR components.

17) Why is an XOR gate NOT considered a universal gate?

It can only have two inputs.

It cannot be used to construct all other logic gates by itself.

It cannot produce a 1 output.

It cannot accept binary inputs.

A universal gate must be sufficient on its own to realize any Boolean function. XOR alone cannot generate all basic gates (for example, it cannot easily create a constant output), so it is not universal. XOR lacks the ability to directly force its output to 0 or 1 without additional gates, preventing it from forming the full set of logical operations required.

Explanation

A universal gate must be sufficient on its own to realize any Boolean function. XOR alone cannot generate all basic gates (for example, it cannot easily create a constant output), so it is not universal. XOR lacks the ability to directly force its output to 0 or 1 without additional gates, preventing it from forming the full set of logical operations required.

18) Which combination of gates correctly implements the expression A' + B?

Inputs A and B into an OR gate, then the output into a NOT gate

Input A into a NOT gate, then its output and B into an OR gate

Input A into a NOT gate, then its output and B into an AND gate

Inputs A and B into an AND gate, then the output into a NOT gate

The expression A' + B means 'NOT A OR B', so you first invert A with a NOT gate and then OR that result with B. This captures the logical structure directly: a single negation applied to A combined with a permissive OR condition depending on B.

Explanation

The expression A' + B means 'NOT A OR B', so you first invert A with a NOT gate and then OR that result with B. This captures the logical structure directly: a single negation applied to A combined with a permissive OR condition depending on B.

19) A circuit shows inputs A and B going into an AND gate, and that output going into a NOT gate. Which Boolean expression does this implement?

A' · B'

(A + B)'

(A · B)'

A' + B'

The AND gate computes A·B, and the NOT gate inverts the result, giving ¬(A·B), which is written as (A · B)'—this is a NAND implementation. NAND encapsulates the idea of requiring both inputs not to be simultaneously true, a fundamental building block in digital logic.

Explanation

The AND gate computes A·B, and the NOT gate inverts the result, giving ¬(A·B), which is written as (A · B)'—this is a NAND implementation. NAND encapsulates the idea of requiring both inputs not to be simultaneously true, a fundamental building block in digital logic.

20) A 2-input XOR gate has its output fed into a NOT gate. What Boolean expression represents the final output?

A ⊕ B

A + B

A · B

(A ⊕ B)'

The XOR gate computes A ⊕ B; inverting that output gives ¬(A ⊕ B), written as (A ⊕ B)'. This is the behavior of an XNOR function. XNOR outputs 1 exactly when the inputs match, a crucial function in equality tests and comparator circuits.

Explanation

The XOR gate computes A ⊕ B; inverting that output gives ¬(A ⊕ B), written as (A ⊕ B)'. This is the behavior of an XNOR function. XNOR outputs 1 exactly when the inputs match, a crucial function in equality tests and comparator circuits.

Logic Circuits Quiz: Interpret Gates and Build Circuit Logic