Construct the logic symbol, Boolean expression and truth table of AND gate.

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Q. Construct the logic symbol, Boolean expression and truth table of AND gate.

Chhattisgarh CgbseCGBSE Intermediate Board 2020Subjective· 3mImportance★★★★★

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Concept understanding — Logic Gate AND

The AND Gate: The "All-or-Nothing" Gate

Imagine you are trying to start a car. Two things must happen: you must turn the key, and you must press the brake pedal. If either one is missing — you turn the key without pressing the brake, or you press the brake without the key — the engine stays off. Only when both actions happen together does the car start.

That is exactly what an AND gate does. It is the "all-or-nothing" gate: the output is 1 only when every single input is 1. If even one input is 0, the output is 0.

The Precise Statement

An AND gate takes two (or more) binary inputs — each either 0 (false, off) or 1 (true, on) — and produces one binary output. The rule is:

Y = A \cdot B

The dot ( $\cdot$ ) is the Boolean multiplication symbol. It does not mean ordinary arithmetic multiplication; it means logical AND. The output $Y$ is 1 if and only if $A = 1$ and $B = 1$ .

Truth Table (for two inputs)

Input A	Input B	Output Y = A · B
0	0	0
0	1	0
1	0	0
1	1	1

Only the last row — both inputs high — gives a high output.

Why "Multiplication"?

The name "AND" comes from everyday language: "I will go to the park if it is sunny and I have free time." Both conditions must be true. The symbol $\cdot$ (dot) is used because, in Boolean algebra, AND behaves like multiplication in ordinary arithmetic when you only use 0 and 1:

$0 \times 0 = 0$
$0 \times 1 = 0$
$1 \times 0 = 0$
$1 \times 1 = 1$

So $A \cdot B$ literally gives the same result as multiplying the two bits. That is why the AND gate is also called a product gate.

Watch out

Do not confuse this with the OR gate, which outputs 1 if any input is 1. AND is stricter — it demands all inputs be 1.

More Than Two Inputs

An AND gate can have three, four, or more inputs. The rule stays the same: output is 1 only when every input is 1. For a 3-input AND gate:

Y = A \cdot B \cdot C

Truth table (abbreviated):

A	B	C	Y
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	0
1	0	0	0
1	0	1	0
1	1	0	0
1	1	1	1

Only the very last combination — all three inputs 1 — lights up the output.

Real-World Analogy …

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The AND gate is a basic digital logic gate whose output is 1 only when all its inputs are 1; symbol is D-shaped, Boolean expression $Y=A\cdot B$ .

Logic symbol: The AND gate for two inputs A and B is drawn as a D-shaped (flat-back, rounded-front) block with the two inputs A and B entering on the flat side and a single output Y leaving from the rounded side:

A ----\
       )---- Y = A·B
B ----/

Boolean expression:

$Y = A \cdot B \quad (\text{read as "A AND B"})$

Truth table (for a 2-input AND gate):

A	B	Y = A·B
0	0	0
0	1	0
1	0	0
1	1	1
…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set A1 markMCQ
$Q. If in a logic gate output (Y) is obtained by the product of its both inputs (A.B), then the gate is (A) AND (B)$
(OR)
$(C) NOR (D) NOT$
›Reveal solutionSolution

Output Y = A·B is the logical multiplication of inputs, which defines the AND gate → option (A).

The Boolean relation Y = A · B means the output is HIGH (1) only when both inputs A and B are HIGH; this is exactly the truth table of an AND gate. The OR gate correspo …

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CBSE 2020Set ANNUAL1 mark
$Q. Write the truth table of AND gate.$
›Reveal solutionSolution

AND gate: output is 1 only if every input is 1 — it implements the logical multiplication $Y = A\cdot B$ .

A logic gate is a digital circuit that follows a certain logical relationship between its input(s) and output, based on binary (0/1) logic. The AND gate is a two- (or more-) input gate whose Boolean expression is

$Y = A \cdot B$

Its output $Y$ is HIGH (logic 1) only when all its inputs are HIGH; if even one input is LOW (0), the output is LOW.

Truth table for a 2-input AND gate:

| Input A | Input B | Output Y = A·B |

|:---:|:---:|:---:| …

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CBSE 2019Set 55/5/11 mark
$Q. Identify the equivalent gate for the circuit of a combination of gates shown in the figure. Write its symbol.$
(OR)
$Draw the logic symbol of the gate shown by the combination of gates and write its name.$
›Reveal solutionSolution

Main question: three NAND gates arranged as NAND(A,A)= $\bar A$ , NAND(B,B)= $\bar B$ , then NAND( $\bar A,\bar B$ ) — by De Morgan's theorem this equals $A+B$ , an OR gate.

OR alternative: two NOT gates feeding a NOR gate give, by De Morgan's theorem, $\overline{\bar A+\bar B}=A\cdot B$ — an AND gate.

Figure — CBSE 2019 55/5/1 Q5

Main Question — three-NAND network

The real figure wires $A$ into one NAND gate with both its inputs tied together (so it outputs $\overline{A\cdot A}=\bar A$ ), and similarly $B$ into a second NAND gate wired as an inverter ( $\overline{B\cdot B}=\bar B$ ). These two inverted signals then feed a third NAND gate:
$X=\overline{\bar A\cdot\bar B}.$
Apply De Morgan's theorem $\overline{X\cdot Y}=\bar X+\bar Y$ :
$X=\overline{\bar A}+\overline{\bar B}=A+B.$
The output $A+B$ is exactly the OR operation, so this three-NAND combination behaves as one OR gate. …
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Q.Construct the logic symbol, Boolean expression and truth table of AND gate.

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