Difficulty: Easy
Correct Answer: It increases.
Explanation:
Introduction / Context:
Understanding how faults affect network behavior is essential for troubleshooting. In parallel resistor networks, an “open” means one branch no longer conducts. Knowing how this changes the equivalent resistance helps predict current changes and diagnose failures.
Given Data / Assumptions:
Concept / Approach:
The equivalent resistance of parallel branches is determined by the sum of branch conductances: 1/Req = 1/R1 + 1/R2 + ... . Removing a branch by opening it eliminates its conductance contribution (1/Rbranch → 0). With fewer conducting paths, the total conductance decreases, so the equivalent resistance increases.
Step-by-Step Solution:
Start with: 1/Req = Σ(1/Ri) over all parallel branches.When a branch opens: its term 1/Ropen = 0 (since Ropen → ∞), reducing Σ(1/Ri).With smaller Σ(1/Ri), the reciprocal Req = 1 / Σ(1/Ri) becomes larger.Therefore, the total (equivalent) resistance increases when any conducting branch opens.
Verification / Alternative check:
Example: two 100 Ω resistors in parallel yield Req = (100*100)/(100+100) = 50 Ω. If one opens, only one 100 Ω path remains, so Req jumps from 50 Ω to 100 Ω, clearly an increase.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
It increases.
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