Identify the equivalent gate for the circuit of a combination of gates shown in the figure. Write its symbol.

Question

Figure — Figure — CBSE 2019 55/5/1 Q5 — FigureFigure — CBSE 2019 55/5/1 Q5

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.

CBSECBSE Class XII Board 2019Subjective· 1mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Logic Gate AND

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

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

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"?

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

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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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.

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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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.Identify the equivalent gate for the circuit of a combination of gates shown in the figure. Write its symbol.

Main Question — three-NAND network

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