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Two resistors R1 = 36 Ω ± 5% and R2 = 75 Ω ± 5% are connected in series. What is the combined resistance with its worst-case tolerance?

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:

  • R1 = 36 Ω with ±5% tolerance → ±1.8 Ω.
  • R2 = 75 Ω with ±5% tolerance → ±3.75 Ω.
  • Series connection: nominal R_total = R1 + R2.


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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