Difficulty: Easy
Correct Answer: False
Explanation:
Introduction / Context:Understanding how faults affect equivalent resistance is key to troubleshooting. An open circuit in one branch of a parallel network removes that path entirely. This item asks whether the overall resistance goes down or up when a branch opens.
Given Data / Assumptions:
Concept / Approach:Parallel equivalent resistance is governed by conductance addition: G_eq = Σ (1/Rk). Opening a branch sets 1/R_open = 0, removing that term. Therefore total conductance decreases, and equivalently, total resistance Req = 1/G_eq increases. With higher Req at a fixed voltage, total current drawn from the source decreases. Hence, the statement that total resistance decreases is incorrect; it increases.
Step-by-Step Solution:
Initial: G_eq = 1/R1 + 1/R2 + … + 1/Rn.Fault: branch j opens → remove 1/Rj term → G_eq_fault < G_eq.Therefore Req_fault = 1 / G_eq_fault > 1 / G_eq = Req.Consequently, Itot_fault = V / Req_fault is smaller than before.Verification / Alternative check:Example: R1 = R2 = 10 Ω in parallel → Req = 5 Ω. Open R2 → only R1 remains → Req = 10 Ω, which is larger, not smaller.
Why Other Options Are Wrong:
Common Pitfalls:Confusing series and parallel responses to opens: an open in series breaks the circuit and sends resistance to infinity; an open in one parallel branch simply removes that branch while leaving other paths intact but increasing overall resistance.
Final Answer:False
Discussion & Comments