Difficulty: Easy
Correct Answer: Approximately 0 V in potential, but it cannot source or sink load current like a real ground
Explanation:
Introduction / Context:The phrase “virtual ground” is used constantly when analyzing inverting op-amp circuits. Students often misinterpret it as a literal ground connection that can sink or source current. In reality, it is “virtual” only in voltage, not in current-handling capability. This question confirms the correct, practical meaning so you can size feedback networks and return paths correctly.
Given Data / Assumptions:
Concept / Approach:With negative feedback, the op-amp drives its output until v− ≈ v+ (the “golden rule”). If v+ is grounded, the inverting node sits at approximately 0 V, hence “virtual ground.” However, that node is not a physical return capable of carrying arbitrary current. Ideally, no current flows into the op-amp input; instead, the current arriving through the input resistor leaves through the feedback path and ultimately returns to the real ground via the source and supply rails.
Step-by-Step Solution:
Recognize: v+ = 0 V → feedback makes v− ≈ 0 V (virtual ground in voltage).Apply KCL at the summing node: i_in ≈ i_fb because i_opamp_in ≈ 0 A.Conclude: the node is near 0 V but is not a current sink/source; currents circulate through the feedback network and supply returns.Therefore, the best description is “0 V in potential but not a power ground.”Verification / Alternative check:Simulate an inverting amplifier: measuring node voltage shows ~0 V while op-amp input current remains near zero. If you attempt to draw load current from the summing node, the amplifier saturates or behaves incorrectly, proving it is not a power return.
Why Other Options Are Wrong:
True chassis ground: incorrect—no direct low-impedance connection to the return plane exists at the summing node.Floating/undefined: false—the node voltage is well-defined by feedback.Exactly 0 V and perfect return: unrealistic; exact equality requires infinite gain, and current-sinking ability is absent.Forces supply currents: misunderstanding; supply currents are determined by output drive, not by “equal and opposite” at the node.Common Pitfalls:Attempting to connect loads to the virtual ground node; ignoring finite open-loop gain and input bias currents; forgetting that “virtual” denotes approximate voltage equality, not current capability.
Final Answer:Approximately 0 V in potential, but it cannot source or sink load current like a real ground
Discussion & Comments