For a 2-input XNOR gate, what will be the output if input A is 1 and input B…
2025
For a 2-input XNOR gate, what will be the output if input A is 1 and input B is 0?
Answer: B. 0 — Concept: An XNOR (exclusive-NOR) gate implements logical equivalence. For two inputs its output is Y = A ⊙ B = (A ⊕ B)′ = A·B + A′·B′, so the output level is…
- A.
Output is the same as an OR gate
- B.
0
- C.
1
- D.
Output cannot be predicted
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Correct answer: B
Concept: An XNOR (exclusive-NOR) gate implements logical equivalence. For two inputs its output is Y = A ⊙ B = (A ⊕ B)′ = A·B + A′·B′, so the output level is high when the two inputs carry the same level and low when the two inputs carry different levels.
Application: The stem fixes A = 1 and B = 0, so the two input levels are unlike.
Evaluate the XOR term first: A ⊕ B = 1 ⊕ 0 = 1, because exclusive-OR is high exactly when the inputs differ.
XNOR is the complement of XOR, so Y = (A ⊕ B)′ = (1)′ = 0.
The sum-of-products form gives the same value: Y = A·B + A′·B′ = (1·0) + (0·1) = 0 + 0 = 0.
Cross-check: The complete two-input truth table agrees with that derivation.
A | B | A ⊕ B | Y = A ⊙ B |
|---|---|---|---|
0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
Contrast: An OR gate driven by the same A = 1, B = 0 would give 1, because OR only asks whether at least one input is high; XNOR instead compares the two levels with each other. A logic gate is also a combinational element — every one of the four input combinations maps to one fixed output level — so no input pattern leaves the result indeterminate.
Result: For A = 1 and B = 0, the XNOR output is 0.