Difficulty: Easy
Correct Answer: Shorted capacitor
Explanation:
Introduction / Context:
The time constant of a simple RC stage is τ = R * C. Faults that effectively reduce R or C will reduce τ and undermine the intended timing/shape of the response. Identifying which failures drive τ downward helps with troubleshooting pulse-shaping networks.
Given Data / Assumptions:
Concept / Approach:
A shorted capacitor effectively makes C → 0 in the time-constant product, collapsing τ toward zero and destroying the intended time-based behavior (the node becomes a near-short for AC transients in many topologies). By contrast, an open resistor removes the current path entirely (no defined exponential); an open capacitor removes C from the circuit (again no defined RC exponential). Increased leakage (finite parallel resistance) modifies the effective network but does not predictably “decrease τ” in the intended series path sense; instead it distorts the response and DC bias.
Step-by-Step Solution:
Verification / Alternative check:
Lab observation: with a shorted capacitor, edge responses vanish into immediate jumps limited only by wiring; with an open capacitor, the node stops integrating/differentiating entirely rather than becoming “faster.”
Why Other Options Are Wrong:
Common Pitfalls:
Assuming any fault that “speeds up” a response means τ simply decreased; some faults destroy the RC mechanism instead of altering τ cleanly.
Final Answer:
Shorted capacitor.
Discussion & Comments