Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:A key hallmark of parallel connections is that the equivalent resistance is always less than the smallest individual branch resistance. This rule is frequently used to sanity-check calculations and quickly bound answers during design and exams.
Given Data / Assumptions:
Concept / Approach:The parallel identity is 1/R_eq = 1/R_1 + 1/R_2 + ... + 1/R_n. Since at least one positive conductance term equals 1/R_min and others are nonnegative, the total conductance exceeds 1/R_min, implying R_eq < R_min. Hence, the claim that R_eq is greater than the lowest-value branch is incorrect in all ideal cases.
Step-by-Step Solution:
1) Identify the smallest resistor R_min in the set. 2) Compute conductance sum S = sum(1/R_i) >= 1/R_min. 3) With any additional branch, S > 1/R_min. 4) Therefore R_eq = 1/S <= R_min, and with more than one branch, R_eq < R_min strictly.Verification / Alternative check:Example: R_min = 100 Ω in parallel with anything else. Even with a very large resistor in parallel, the equivalent becomes slightly less than 100 Ω. This corroborates the general rule for parallel networks.
Why Other Options Are Wrong:
Common Pitfalls:Comparing sums of resistances rather than conductances; applying series intuition to parallel networks; overlooking that R_eq is dominated by the lowest branch value downward.
Final Answer:Incorrect
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