Virtual ground in op-amp analysis: a “virtual ground” node is considered a ______ ground because, although it sits at (approximately) 0 V potential, it is not physically connected to earth or chassis ground and cannot sink infinite current.

Introduction / Context:
Op-amp circuits are often analyzed using the “virtual ground” concept, especially in inverting configurations. Students commonly mistake this node for an actual ground connection. Clarifying what a virtual ground means prevents design errors, such as trying to draw heavy current from the summing junction or confusing reference and return currents.

Given Data / Assumptions:

• Operational amplifier operating in linear region with negative feedback.
• Inverting configuration where the summing junction is held near 0 V.
• No explicit physical tie from the summing node to earth or chassis.

Concept / Approach:
A virtual ground is a node maintained at approximately 0 V (ground potential) by the action of feedback, not by a direct connection to ground. Thus, it behaves as a voltage ground (correct potential) but not a current ground (cannot source/sink arbitrary current; the op-amp input ideally draws negligible current). This distinction explains why the inverting input remains near 0 V while the output supplies the necessary current through the feedback network.

Step-by-Step Solution:

Verification / Alternative check:
Measuring continuity from the summing node to chassis shows open circuit; yet a meter reads ~0 V during operation. If the loop saturates or opens, the node no longer sits at 0 V—another sign it is merely virtual, enforced by feedback when linear operation holds.

Why Other Options Are Wrong:

• Ordinary: implies a normal physical ground node, which it is not.
• Chassis: tied to metal enclosure; not applicable here.
• Earth: safety/utility ground; not connected in this context.
• Protective safety: a specific grounding scheme for safety; unrelated to op-amp summing nodes.

Common Pitfalls:
Attempting to draw load current from the virtual ground; forgetting that if feedback is broken or the op-amp saturates, the node will not stay at 0 V; confusing reference potential with return paths.

Final Answer:
Voltage