The version almost everyone reaches for first: the two input pins of an op-amp are shorted together, which is why they're at the same voltage. Or, a close relative: current flows from one input into the other to equalise them. Both are wrong. The inputs are not connected, and essentially no current enters either of them. The voltage equality is enforced by the feedback network — it's a consequence of the circuit's behaviour, not a physical wire.
Why the mistake is the natural reading
The term "virtual short" sounds like a short circuit. If two nodes are at the same voltage you'd normally conclude that something is connecting them, the way two ends of a wire are always at the same potential. It's a reasonable inference in almost every other context in circuit analysis.
The "virtual ground" version of the same confusion is just as natural. When the inverting terminal sits at 0 V in a standard inverting amplifier, the temptation is to treat it as ground — which implies it sinks current, the way a real ground connection does.
Neither picture is right, because the word "virtual" is doing a lot of work that the word alone doesn't communicate.
What is actually happening
An ideal op-amp has two properties that together create the virtual short:
-
Infinite open-loop gain (A_OL → ∞). The output is A_OL × (V+ − V−). If the output is a finite voltage — say, 5 V — and the gain is effectively infinite, then the differential input (V+ − V−) must be essentially zero. The two inputs are therefore at the same voltage, but only while the feedback loop is intact and the output is not saturated.
-
Infinite input impedance. No current enters either input terminal. The op-amp draws nothing from the nodes it measures.
Negative feedback is the mechanism that enforces rule one. If V+ drifts above V−, the output swings positive, which the feedback network uses to pull V− back up until the difference collapses. The loop settles when V+ ≈ V−. Remove the feedback, or saturate the output, and the virtual short disappears immediately.
The current question follows directly: because the input impedance is infinite, all the current that flows in the feedback resistors flows through those resistors from the output back to the source — none of it enters the op-amp pin. The virtual ground node (V−) is a junction point for external currents, but it neither sources nor sinks any current through the op-amp itself.
A worked example: the inverting amplifier
Consider a standard inverting amplifier with a 10 kΩ input resistor R1 and a 20 kΩ feedback resistor Rf. The non-inverting input (V+) is tied to ground (0 V). A 1 V input is applied at Vin.
Step 1 — apply the virtual short. Because V+ = 0 V and the feedback loop enforces V+ = V−, the inverting terminal is also at 0 V. This is the virtual ground. It is 0 V, but nothing flows into the op-amp pin.
Step 2 — find the current through R1. The voltage across R1 is Vin − V− = 1 V − 0 V = 1 V.
I₁ = 1 V / 10 kΩ = 0.1 mA
Step 3 — apply KCL at the V− node. No current enters the op-amp input. So all of I₁ must flow through Rf.
I_f = I₁ = 0.1 mA
Step 4 — find Vout. The right end of Rf is the output. The left end is at 0 V (virtual ground).
Vout = 0 V − I_f × Rf = 0 − (0.1 mA × 20 kΩ) = −2 V
Gain = Vout / Vin = −2 V / 1 V = −2 = −Rf / R1. Exactly the textbook formula — and the only algebra used was Ohm's law plus the two ideal op-amp rules.
Notice what did not appear: the op-amp's internal gain, any current entering the input pins, or any short circuit between V+ and V−. The virtual short is a constraint you impose on the node equations, not a wire you draw on the schematic.
How to internalize it
Treat the virtual short as a boundary condition, not a circuit element. Before writing any KCL equation, write down: V+ = V− (virtual short) and I_in+ = I_in− = 0 (infinite input impedance). These two lines are the entire analysis toolkit for ideal op-amp circuits.
Ask "what would the output have to be to make V+ = V−?" That question reframes the virtual short as an output that the feedback loop hunts for, rather than a fixed property of the input pins. If you ever get confused, trace the feedback path: the output drives Rf, Rf drives V−, and V− is pulled toward V+ until the differential collapses.
The virtual short vanishes at the rail. If the input signal is large enough that the required output would exceed the supply voltage, the op-amp saturates, the feedback breaks, and V+ ≠ V−. Every ideal-op-amp rule assumes linear operation — check that assumption before applying the analysis.
The same discipline matters when you're working out why a transistor's bias point shifts with beta: in both cases the trap is misidentifying where current actually flows and letting an intuitive but incorrect picture drive the algebra. When you move from resistive op-amp circuits into AC analysis, you first have to be precise about what an AC voltage even is — why a "120 V" outlet actually peaks at 170 V is the amplitude half of that picture, and the same care then applies to phase: why voltage leads or lags current in capacitors and inductors is the natural next concept as you start analysing op-amp frequency response.
Check yourself
An inverting op-amp circuit has R1 = 5 kΩ and Rf = 15 kΩ. The non-inverting input is grounded. You apply Vin = 2 V. What current flows into the inverting input pin of the op-amp?
A. 0.4 mA — the full current from R1 enters the pin
B. 0.3 mA — the current divides between Rf and the pin
C. 0 mA — no current enters either input
D. It depends on the op-amp's open-loop gain
Correct answer: C. An ideal op-amp has infinite input impedance, so no current enters either terminal regardless of the surrounding resistor values. All 0.4 mA from R1 (2 V / 5 kΩ) flows through Rf to the output.
Close the gap
The virtual short is one of those concepts where reading the rule is quick but knowing when it silently breaks — output saturation, AC gain roll-off, non-inverting topologies that look symmetric but aren't — takes working through circuits until the feedback loop becomes something you can trace, not just recite. A tutor that watches you set up the node equations catches the moment you accidentally draw current into the input pin and asks you to justify it, before the error propagates through the rest of the algebra.