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simulate this circuit – Schematic created using CircuitLab
Vout should be -10 volts according to the voltage gain equation of inverting amplifier which is based on the rule of op amp that A is at zero volts. Here the output is grounded and the node A becomes 0.91 V (i.e) it is not at virtual ground. What would happen now? would the output voltage change? if it does change I am having a hard time understanding how it would?
rule of op amp that A is at zero volts
This is your basic misconception. That's not what opamps do. It is bad to think of opamps this way. Unfortunately, we see this and other "rules" too often that are really just something complicated dumbed down by throwing out a bunch of necessary conditions. That doesn't make the conditions go away, only your understanding of them. Don't fall for "rules of thumb". There is no substitute for actually understanding what is going on.
Opamps take the voltage difference between their inputs, multiply that by a large number, like 100,000 or more, and make that their output within the voltage and current capability of the output. When the rest of the circuit manages the opamp such that it's output tries to be within it's active output range, then by necessity its two inputs will be very close in voltage. For example, when the opamp output is at 10 V but could go higher if it wanted to, then the inputs are only (10 V)/100,000 = 100 µV different at a gain of 100,000. 100 µV is close enough to be "same" for many practical purposes.
In the circuit you show, the positive input is being held at 0 V. Therefore, whenever the opamp output is within its "linear" active range (not clipped to either extreme), it's negative input is very close to 0 V. That the opamp stays in the linear region is only due to the feedback around the opamp.
If you force the two inputs further apart, the opamp will try to slam its output full high or full low.
When you hold its output at a fixed value when it's trying to drive to a different value, there will be a large difference between the two inputs. This difference is 910 mV in your example. That means the opamp will try to drive its output low. However, since you're holding it fixed at 0 V it can't. It will sink as much current as it can. What happens then depends on how well the opamp is protected from excessive output current. It could get hot and blow up, or it could just sit there sinking the maximum current it can, but not being able to change its output voltage.
What happens now is because the A input is more positive than input B, the output tries to go low, but can't because you've grounded the output.
This is from the TL081 data sheet.
When the two inputs are at the same voltage, the FETs' tail current source is split equally, balancing the current mirror consisting of the diode and transistor. When the input voltages are different, the current mirror is unbalanced, and will source or sink current to the Miller integrator that drives the output totem pole stage. In the case of the A input being higher, current leaves the current mirror, into the Miller integrator which drives it low, turning the lower output transistor on, which draws current between the output terminal and the V- supply.
As the output is connected to ground, its voltage does not change, while the lower transistor draws 10s of mA from it. This current is limited to non-destructive levels by the design of the TL081 op-amp (not shown on this simplified diagram). Most other small signal opamps also have safe output short circuit currents. Some higher current opamps might not have well defined short circuit output current, and would overheat and die under such abuse.
If you were using a power opamp, and the output was grounded by a small fuse, then the output would try to go low, pull enough current to blow the fuse, and then go to the low voltage that would cause the feedback network to be able to equalise its input voltages.
As a detail worth noting on that opamp diagram, you notice that in tiny print at the current mirror resistors it says 'internally trimmed'? These resistors are laser trimmed during manufacturing test to make the current mirror balance when the inputs are equal voltage, to get the input offset voltage as close to zero as possible.
How does the output voltage of an Op-Amp swing to keep the inverting terminal at virtual ground in this circuit?
It doesn't.
Because you've grounded the op-amp output, you've broken the feedback loop and the circuit no longer maintains a virtual ground at the inverting input.
The gain can no longer be calculated from the inverting op-amp amplifier formula because this is no longer an inverting op-amp amplifier. In fact the gain is 0, because the output voltage never changes (\$\Delta{}V_{out}=0\$) in response to changes in the input voltage.
If you break the rules of zero differential DC input, it can no longer be a "virtual ground with 0 diff input and is no longer linear output. The same applies to AC but declines as gain reduces with rising frequency. Negative feedback provides this linear null input voltage as long as the output is in the linear range.
Remember that an Op Amp is a differential to single-ended converter.
The term "virtual ground" is somewhat of a misnomer or easily misunderstood. But if you understand ground is just a local reference voltage and that reference can be floating above any other isolated ground, then it is accurate. For Op Amps (OA) we prefer input bias near the mid-range, other times near ground if allowed and other times near Vcc, if allowed in datasheet. ( or often just Vin+/- = Vcc/2 )
If you think it only means you have 0V on either input, then it is incorrect.
Why do I say isolated?
Because the output impedance of an Op Amp (OA) is often effectively a million times lower than the input impedance of either input ( without feedback ) we say the input is isolated from the output until you add feedback.
What does it really mean ?
We know the open loop gain > 1e6 , thus the only time you have linear operation is when we consider the differential voltage is effectively zero.
It is better to consider it like a floating voltage source with a differential impedance of zero but a very high common mode impedance.
The other factor is the common mode input range is not always Vee to Vcc. Some other performance aspects are traded off when when an Op Amp is considered Rail-to Rail input, while others make it work to the full positive rail (Vcc) or the full negative rail (Vee) which could be ground or 0V.
We do not consider each input as a virtual ground because they aren't. We consider only the differential voltage and not the common mode input as a virtual differential ground. Thus for differential gain, we must consider balanced impedances on each input to avoid common mode errors (e.g. (CMRR) and input bias current offsets.
The OA is a high gain differential-to-single-ended output. Now in reality there is some common mode (CM) connection but at DC it is attenuated by the order of magnitude open loop gain of the OA which is often > 1 million or 1e6 or 120 dB. In datasheets this rejection ratio is called the "Common Mode Rejection Ratio" or CMRR. This means if you change the input DC reference Voltage to both Vin+ and Vin- AND the differential input voltage is still Zero meaning the output is not saturated, then the output is not affected by the DC common mode (CM) input.
The CM is rejected or "isolated" as long as the assumptions for linear input and output range are satisfied. This ratio declines with rising frequency due to the gain-bandwidth product limitations of OA with internal compensation.
If your Op Amp ever has a differential DC voltage more than the rated Vio input offset (uv~mV range), the output WILL be saturated at either supply rail. Then it no longer has ANY linear properties. Always check for DC bias to each input to see how they match and the effects of input current creating an offset.
