Difficulty: Easy
Correct Answer: increases
Explanation:
Introduction / Context:Understanding how total resistance changes with the number of parallel paths is essential for predicting current draw and power. Removing a branch changes the equivalent resistance and thus the load on the source.
Given Data / Assumptions:
Concept / Approach:For resistors in parallel: 1 / R_eq = Σ (1 / R_i). Removing a branch eliminates a positive term from the sum of conductances, so the total conductance decreases. Therefore, R_eq = 1 / (Σ conductance) increases.
Step-by-Step Reasoning:
Initial: 1 / R_eq(initial) = G1 + G2 + ... + Gn.Remove one: 1 / R_eq(new) = G1 + G2 + ... + G(n-1) < original sum.Thus R_eq(new) = 1 / (smaller sum) is larger → total resistance increases.Verification / Alternative check:Numerical example: two 100 Ω in parallel → R_eq = 50 Ω. Remove one branch → R_eq = 100 Ω, which is larger than 50 Ω.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:increases
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