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The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated. In the case of a real opamp (with rails, that are not shown in the model above) then it can't source more voltage than it has available and the output will be limited by what the rails can source.

This circuit is simply a comparator. The open loop equation above applies to both opamps and comparators for ideal simple amplifiers, however with rails either amplifier will saturate.

Source: https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html

The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated. In the case of a real opamp (with rails, that are not shown in the model above) then it can't source more voltage than it has available and the output will be limited by what the rails can source.

This circuit is simply a comparator. The open loop equation above applies to both opamps and comparators for ideal simple amplifiers, however with rails either amplifier will saturate.

Source: https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html

In the early days of amplifiers, people would try and run them open loop, with little success because of variations in amplifiers. It was then that someone realized that negative feedback can take out the inconsistencies

The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated. In the case of a real opamp (with rails, that are not shown in the model above) then it can't source more voltage than it has available and the output will be limited by what the rails can source.

The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated. In the case of a real opamp (with rails, that are not shown in the model above) then it can't source more voltage than it has available and the output will be limited by what the rails can source.

This circuit is simply a comparator. The open loop equation above applies to both opamps and comparators for ideal simple amplifiers, however with rails either amplifier will saturate.

Source: https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html

The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated.

The equation that you need is:

\$ V_{out} = A_v (V_{+}-V_{-})\$

Av is the open loop gain.

The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them.

If this is a real op amp, an input bias current will be given and the voltage drop of the resistors can be calculated. In the case of a real opamp (with rails, that are not shown in the model above) then it can't source more voltage than it has available and the output will be limited by what the rails can source.