74LS138 decode index — Evaluate: “With A0 = 1, A1 = 1, A2 = 0 (device enabled), output 3 is the selected line.”

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context: Understanding the mapping between binary inputs and decoder outputs is essential for address decoding. The 74LS138 asserts one of eight active-LOW outputs based on A (LSB), B, C (MSB) when enabled. This item verifies correct output selection for a given input code.Given Data / Assumptions: A0 ≡ A = 1, A1 ≡ B = 1, A2 ≡ C = 0. Outputs are active-LOW; “selected” means driven LOW. Device is enabled (G1=1, G2A=0, G2B=0). Concept / Approach: The selected output index equals the binary number CBA, with C as MSB and A as LSB. For CBA = 0 1 1, the index is 3, so Y3 is asserted LOW. Therefore the statement is correct.Step-by-Step Solution: Arrange inputs: C B A = 0 1 1.Compute index: 04 + 12 + 1*1 = 3.Select line: Y3 goes LOW; other outputs remain HIGH. Verification / Alternative check: Datasheet truth tables confirm A=1, B=1, C=0 → Y3 LOW when enabled. Why Other Options Are Wrong: Incorrect: Conflicts with binary-to-one-hot mapping.True only if outputs are active HIGH: The 74LS138 uses active-LOW outputs; interpretation already accounts for this.Depends on strobe only: Enables must be correct, but the selected line still depends on C/B/A. Common Pitfalls: Swapping bit order or assuming A is MSB.Overlooking the active-LOW nature of outputs when reading timing diagrams. Final Answer: Correct

74LS138 decode index — Evaluate: “With A0 = 1, A1 = 1, A2 = 0 (device enabled), output 3 is the selected line.”

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