In a simple RC integrator, if the capacitor accidentally becomes shorted (zero impedance), what happens to the output measured across the capacitor node?

Introduction:
An RC integrator typically consists of a series resistor feeding a capacitor to ground, with the output taken across the capacitor. This question explores the failure mode when the capacitor is shorted, a common troubleshooting scenario that produces a distinctive symptom at the output node.

Given Data / Assumptions:

• Passive RC integrator: input → series R → node → capacitor to ground, output at the node.
• Steady operation with a bounded input signal.
• "Shorted capacitor" means the capacitor's impedance has collapsed to ~0 Ω.

Concept / Approach:
If the shunt element (capacitor) is a short, the output node is directly clamped to ground potential. Regardless of the input waveform, the node voltage cannot rise above approximately 0 V because any attempted change is immediately diverted to ground through the short.

Step-by-Step Solution:
1) In a healthy RC integrator, the capacitor charges/discharges, creating an output proportional to the time-integral of the input.2) With a shorted capacitor, the output node is effectively a direct connection to ground.3) Therefore, the measured output voltage is ≈ 0 V for all practical purposes.4) The circuit no longer integrates; the input sees a series resistor into a ground short, which may stress the source.

Verification / Alternative check:
Measure with a multimeter or oscilloscope: the node reads near 0 V despite input changes. Upstream, current is limited only by the series resistor, confirming the fault signature.

Why Other Options Are Wrong:
Same as input: impossible; the node is clamped to ground.

None of the above: incorrect because 0 V is the expected reading.

Is at ground vs would measure zero volts: both descriptions point to 0 V, but the most precise measurement-oriented choice is "would measure zero volts".

Common Pitfalls:
Assuming the integrator will simply "stop integrating" but still pass some signal. A hard short forces the output to 0 V; any remaining signal appears only as current through the series resistor.

Final Answer:
would measure zero volts