Op-amp output: resistors in parallel

[Mechatron]

In the inverting operational amplifier circuit I've drawn below I am trying to calculate the output voltage and the voltage at the blue spot. The resistors are of the magnitude of several K ohm.
I have done the following:

Uout = Rf * (I1+I2+I3) = Rf * (U1/R1 + U2/R2 + U3/R3)

The voltage Ux before the negative input pin of the operational amplifier:
Ux - Uout = Rf * (I1+I2+I3)
Ux = Rf * (I1+I2+I3) + Uout

The circuit:
http://s8.postimg.org/s6yj53wcl/Circuit.png

 

[Jony130]

But if an ideal op-amp is used the voltage at point X will always be equal to 0V.
So why you want to find the voltage at point X ?

 

[Mechatron]

But if an ideal op-amp is used the voltage at point X will always be equal to 0V.
So why you want to find the voltage at point X ?

So there is no voltage drop over the resistors, although there is a current generated through the resistor? So the point X is just like a ground? I just find it difficult to understand how there can be a flow of electrons, when there is no voltage.

 

[Mechatron]

But if an ideal op-amp is used the voltage at point X will always be equal to 0V.
So why you want to find the voltage at point X ?

Should I consider the voltage on the positive pin of the op-amp as a voltage reference?
So if there was a battery attached between the ground and the positive input pin, the voltage at point X would be equal to the voltage of the battery (voltage reference)?

 

[Jony130]

So there is no voltage drop over the resistors, although there is a current generated through the resistor? So the point X is just like a ground? I just find it difficult to understand how there can be a flow of electrons, when there is no voltage.

For suer there will be a current flow through resistors.

Can you solve for Va in this simply circuit ??

Next as yourself a question, the voltage at point Va is equal to 0V but current still flow in this circuit why?

So there is no voltage drop over the resistors, although there is a current generated through the resistor? So the point X is just like a ground? I just find it difficult to understand how there can be a flow of electrons, when there is no voltage.

Exactly, it is true as long as we have a negative feedback in the circuit.
https://www.physicsforums.com/showthread.php?t=596704#post3866238

 

[meBigGuy]

The basic operation of an ideal op-amp is that the output will do the best it can to cause the + and - pins to be equal. When you write the circuit equations, you start with that assumption, that + and - are equal.

 

[Mechatron]

The basic operation of an ideal op-amp is that the output will do the best it can to cause the + and - pins to be equal. When you write the circuit equations, you start with that assumption, that + and - are equal.

It's strange how you and the_emi_guy seem to be the only two guys who answer me, in general.

 

[sophiecentaur]

I just found this thread. Did no one introduce the term 'virtual Earth summing junction' yet? I was taught the concept and the term very early in my OP Amp work. If you google that term, you may come across lots more to read.
The point about OpAmps is that they can be assumed, very often, to have infinite voltage gain (at low frequency). As with all negative feedback circuits, the overall gain of the amplifier tends to be defined by the feedback components and not the individual device gain (assuming it is 'high enough'). If you look at the volts on the virtual earth, for a number of different amplifier chips, the actual voltage variations will all be tiny but they will vary from device to device - although the overall circuits will have the same performance. Hence, you do not care about that signal (except to check that it is 'zero' when you are fault finding).

 

[meBigGuy]

pretty old fashioned, misleading, inaccurate, distracting, confusing term. (notice that I am trolling. I wonder if I will catch anything?)

 

[sophiecentaur]

Your amplifiers will all hoot!
Poetic justice.
And what's wrong with 'virtual earth'? Is it any worse than claiming that an amplifier will "do the best it can"? I would never ever attribute 'wanting' to any circuit (except, possibly a neural network).

 

[sophiecentaur]

The basic operation of an ideal op-amp is that the output will do the best it can to cause the + and - pins to be equal. When you write the circuit equations, you start with that assumption, that + and - are equal.

Whilst we are being picky, it is the negative feedback and not the Op Amp that achieves what you claim. An Op Amp just amplifies the difference between its inputs.
You should sort out your own problems before throwing brickbats.

 

[meBigGuy]

Lol ...