RC integrator fault case: If the capacitor in an RC integrator goes open-circuit, the output will simply equal the input voltage. True or false?

Introduction / Context:
Troubleshooting analog signal-shaping circuits requires understanding failure modes. An open capacitor radically changes an RC integrator's behavior and output node impedance.

Given Data / Assumptions:

• Standard RC integrator: input → resistor → node → capacitor to ground; output is taken at the capacitor node.
• Capacitor fails open (infinite impedance).
• Measurement uses a high-impedance instrument (typical DMM/oscilloscope).

Concept / Approach:

With the capacitor open, no current flows through the branch, so the node becomes floating except for leakage and measurement loading. The network no longer forms a divider; there is no guaranteed equality between input and output—often the output is undefined or follows stray capacitances, not the intended signal.

Step-by-Step Solution:

Verification / Alternative check:

Replace the opened capacitor with a very large value (simulate open); observe Vout becoming noisy, drifting, or holding a random charge. In contrast, shorting the capacitor (not typical failure) would force Vout ≈ 0, not Vin.

Why Other Options Are Wrong:

• “True” assumes the node magically mirrors input; without a conductive path, the node cannot reproduce Vin and becomes high-impedance and indeterminate.

Common Pitfalls:

Confusing an open capacitor with removing the resistor; conflating probe-through effects (a scope probe can inadvertently provide a small capacitance that momentarily couples input to output).

Final Answer:

False