Benchmark rule for parallels: evaluate the statement that the equivalent resistance of any nontrivial parallel network is always less than the smallest individual resistor value present.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
This benchmark rule is a quick mental check when simplifying circuits: the equivalent resistance of resistors in parallel should never exceed the smallest branch value. Recognizing this prevents calculation mistakes and helps validate calculator results before committing designs.


Given Data / Assumptions:

  • At least two resistors in parallel.
  • Resistors are linear and time invariant.
  • We ignore parasitic effects and measurement errors.


Concept / Approach:
For N parallel resistors, 1 / R_eq = sum(1 / R_k) with k from 1 to N. Because each term 1 / R_k is positive, the sum is greater than 1 / R_min, where R_min is the smallest resistor. Taking reciprocals, R_eq is less than R_min. This holds for any finite set of positive resistances and becomes a strict inequality when there are two or more branches.


Step-by-Step Solution:

1) Identify R_min among branch values. 2) Note that sum(1 / R_k) ≥ 1 / R_min + another positive term for each additional branch. 3) Take the reciprocal to obtain R_eq < R_min for any network with at least two branches.


Verification / Alternative check:
Numerical sanity check: 6 ohm in parallel with 3 ohm gives R_eq = 2 ohm, which is less than 3 ohm, the smallest branch.


Why Other Options Are Wrong:
Incorrect: contradicts the reciprocal property.
Only for two branches or small ratio spreads: the inequality is general for any count and ratios.


Common Pitfalls:
Applying series formula by mistake; forgetting that adding a parallel path lowers R_eq.


Final Answer:
Correct

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