Noninverting amplifier node voltage — A noninverting op-amp has RIN = 1.0 kΩ (from inverting node to ground) and RFB = 2.5 kΩ (from output to inverting node). If 1.42 mV is applied to the noninverting (+) input, what is the voltage at the RIN node (i.e., the inverting input node)?

Introduction:
In the noninverting amplifier, the inverting node is forced to follow the noninverting input via negative feedback. Recognizing this “virtual short” (V− ≈ V+) lets you immediately find node voltages and then, if needed, the closed-loop gain and output voltage.

Given Data / Assumptions:

• RIN = 1.0 kΩ from inverting node to ground
• RFB = 2.5 kΩ from output to inverting node
• Vin at noninverting (+) = 1.42 mV
• Ideal/high-gain op-amp with negative feedback and linear operation

Concept / Approach:

For a noninverting configuration, the closed-loop action makes V− ≈ V+. Therefore, the inverting node sits at approximately the same voltage as the applied noninverting input. This is independent of the actual RIN and RFB values so long as the loop is closed and not saturated.

Step-by-Step Solution:

Verification / Alternative check:

Solving the resistor divider with V− constrained to equal V+ yields the same node voltage; the feedback forces the equality to minimize differential input.

Why Other Options Are Wrong:

• 3.5 mV / 0.56 mV: Do not follow from the virtual short condition.
• ground: The node is at the input value, not 0 V.
• 4.97 mV: That is the approximate output voltage, not the inverting-node voltage.

Common Pitfalls:

• Confusing the inverting node voltage with the output voltage.
• Assuming RIN and RFB form a fixed divider to ground regardless of feedback (they do not; V− is constrained by the loop).

Final Answer:

1.42 mV