Fault effect — in a parallel resistive network, opening (going open-circuit) one branch resistor will cause the total equivalent resistance seen by the source to _____ .

Difficulty: Easy

Decrease
Remain exactly the same
Increase
Become zero
Become negative

Correct Answer: Increase

Explanation:

Introduction / Context:
Understanding how faults affect equivalent resistance helps with diagnostics. In parallel networks, each branch adds conductance (1/R). Removing a branch by opening it reduces total conductance.

Given Data / Assumptions:

Purely resistive DC network.
One branch goes open (infinite resistance).
All other branches remain unchanged.

Concept / Approach:
Total conductance G_total = Σ(1/Ri). If one branch opens, its 1/R term becomes 0, so G_total decreases, meaning R_total = 1/G_total increases. Source current drops accordingly for a fixed source voltage.

Step-by-Step Solution:

1) Express R_total via conductance sum.2) Remove one positive term from the sum due to open branch.3) Conclude G_total decreases → R_total increases.4) Expect lower total current and power draw.

Verification / Alternative check:
Try numbers: parallel of 6 kΩ and 12 kΩ is 4 kΩ; opening the 12 kΩ branch leaves 6 kΩ, which is larger than 4 kΩ.

Why Other Options Are Wrong:

Decrease/zero/negative: contradict conductance arithmetic and physical realizability.Remain same: only true if the opened branch carried negligible current (infinite R), which is the definition of already-open.

Common Pitfalls:
Thinking of voltage division (series behavior) instead of conductance addition in parallel.

Fault effect — in a parallel resistive network, opening (going open-circuit) one branch resistor will cause the total equivalent resistance seen by the source to _____ .

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