Difficulty: Easy
Correct Answer: 111 ± 5.55 Ω
Explanation:
Introduction / Context:Tolerance analysis ensures that a design functions over component variations. For series resistors, the nominal value is the sum of resistances, while the worst-case absolute tolerance is the sum of the individual absolute tolerances. This problem reinforces translating percentage tolerances into absolute ohmic tolerances and combining them correctly.
Given Data / Assumptions:
Concept / Approach:For worst-case analysis, add absolute tolerances algebraically for components in series: ΔR_total = ΔR1 + ΔR2. Percentage tolerances do not average; they accumulate as absolute terms in the same direction for worst-case bounds.
Step-by-Step Solution:
Compute nominal: R_total = 36 + 75 = 111 Ω.Compute absolute tolerances: ΔR1 = 0.05 * 36 = 1.8 Ω; ΔR2 = 0.05 * 75 = 3.75 Ω.Worst-case total tolerance: ΔR_total = 1.8 + 3.75 = 5.55 Ω.Therefore R_total = 111 ± 5.55 Ω.Verification / Alternative check:
Lower bound: 111 − 5.55 = 105.45 Ω; Upper bound: 116.55 Ω. Both are consistent with extremal combinations.Why Other Options Are Wrong:
±0 Ω ignores tolerance; ±2.778 Ω and ±7.23 Ω mis-handle percentage-to-absolute conversion; ±10% is not the worst-case rule for unequal series parts.Common Pitfalls:
Adding percentage tolerances directly instead of absolute values; forgetting to compute each part's absolute error first.Final Answer:
111 ± 5.55 Ω
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