Hi I'm supposed to find Vo I the attached image.

I just want to confirm one thing. I labelled a node a V3 I know in an op amp the voltages of both leads going in to an op amp are the same.
The second branch has just a wire from ground to the op amp. I know the current of that is 0. Since the current is?0 can I say V3 is 0?

The original question asks to find v0 in terms of everything labelled (not v3)

 

ideal OpAmp:
1. no current flows into or out of input
2. voltage difference between inputs is zero volts

as a result of last point:
one of the inputs is grounded (0VDC) so your V3 must also be zero volts.

 

You are correct.
The V3 at the none inverting input to op-amp is 0 volts because it is connected to ground which we assume is 0 volts.
The V3 at the inverting input is also 0 volts because of the virtual ground concept.

So. For example if you want to find current in Rf, you would go If=(Vo-V3)/Rf=(Vo-0)/Rf

 

Hi I'm supposed to find Vo I the attached image.

I just want to confirm one thing. I labelled a node a V3 I know in an op amp the voltages of both leads going in to an op amp are the same.
The second branch has just a wire from ground to the op amp. I know the current of that is 0. Since the current is?0 can I say V3 is 0?

The original question asks to find v0 in terms of everything labelled (not v3)

Your conclusion is correct, but your reasoning is wrong.

The conclusion that V3 is 0 is NOT a consequence of the current into the opamp inputs being zero. It is because, as you stated, the opamp voltages at both inputs are the same. This is the "virtual short" principle. So the mere fact that the voltage at the non-inverting input is zero is sufficient to state that the voltage at the inverting input, also known as V3, is zero. Always with the caveat that the opamp is operating in its active region.

Also, in the future please try to minimize the size of your attachments. Some members have very slow internet connections (such as dial up or, often worse, satellite). Consider the following, which was done in Paint in less than 30 seconds by resizing it to 500 pixels wide and clipping the bottom part, which contained no relevant information. This file is only 38 KB, or about 5% of the size of the one you posted.

 

Hi,

I have to agree that large file sizes are a waste of server space, unless of course the diagram is very very complicated, but since this one like many others is not very complex, i think it should be a file much smaller. I could probably get this file down to 10k for example. 700k is unnecessary.

For the main question, the op amp can also be viewed as a current summer, where it sums the currents on the input and forces that current though the feedback path. For this view, you'd only have two currents summing, and the sum of the currents would flow through the feedback. So in this simpler view you'd have:
Vout=(I1+I2)*Rf

and adjust polarity for the inverting function (make that negative).

I1 and I2 are easy to calculate because of the virtual ground Wbahn was talking about. So they are equal to the input voltage divided by the input resistance on the respective input.

 

seal308, what you have asked is a very general question and different people would tend to answer it differently since they don't know your level of expertise on the subject matter. Op-amps are very important basic components and they have various modes which would to a different behavior from that op-amp. Although when you measure the voltage at the non-inverting input in a real lab environment, you might find some potential different; the virtual-short however assumes that both voltages at inverting and non-inverting re shorted. Ideally there should be a ground at the inverting input as well.

Regards,
Dave J
http://blog.7pcb.com