Inverting OP AMP with 'hanging' resistor in feedback network

[Learnphysics]

Homework Statement

Find Vo/Vin.I understand perfectly how to do this question. It simply involves an application of ohm's law to find i1, then KVL around the first loop. KCL at the top, and KVL around the second loop. Then to solve simultaneously.

My issue however Is why the mechanics of KVL work in this case. We were taught that KVL required a closed loop, and the voltages needed to sum to zero. The diagram seems not be a closed loopIn the textbook example we say the KVL of those loops are:
Vin + i2(R2) -i3*R3 = 0.
(Vin = 0, due to a virtual short circuit)

I understand i2*R2 is the voltage drop across resistor 2. But for -i3*r3, wouldn't we also need to factor in the current i4 that's circulating around the neighboring loop? (as we'd do in mesh analysis?).

Also why are we adding the voltage of Vin when it isn't in the loop at all? It seems like we're simply adding the voltage of a certain point, as opposed to the voltage drop across a certain element.

I understand Op-amps, virtual short circuits, and such, this problem is really just focused around the circuit analysis of the feedback loop.

Thank you.

 

[gneill]

I understand i2*R2 is the voltage drop across resistor 2. But for -i3*r3, wouldn't we also need to factor in the current i4 that's circulating around the neighboring loop? (as we'd do in mesh analysis?).

Current i3 is the net current running through R3. If you like, it's the sum of the mesh currents I1 and I2 that pass through R3. Since i3 is the current through R3, the voltage drop is i3*R3 no matter what.

Also why are we adding the voltage of Vin when it isn't in the loop at all? It seems like we're simply adding the voltage of a certain point, as opposed to the voltage drop across a certain element.

Yeah, adding Vin there seems rather dubious. Unless it's a typo and they meant Vi, the voltage across the op-amp inputs which, for an ideal op-amp, will be essentially zero.

 

[CWatters]

My issue however Is why the mechanics of KVL work in this case. We were taught that KVL required a closed loop, and the voltages needed to sum to zero. The diagram seems not be a closed loop

We have had a similar question before.

Where do you think the open circuit is? None of the components in the diagram have an unconnected terminal.

I understand i2*R2 is the voltage drop across resistor 2. But for -i3*r3, wouldn't we also need to factor in the current i4 that's circulating around the neighboring loop? (as we'd do in mesh analysis?).

If you wanted to you could write KCL for the top end of R3, eg I2 + I3 - I4 = 0

Also why are we adding the voltage of Vin when it isn't in the loop at all? It seems like we're simply adding the voltage of a certain point, as opposed to the voltage drop across a certain element.

I agree with gneil. I think that's a typo and they mean Vi not Vin.

That makes sense if the loop is..

Earth
V-
V+
R2
R3
Earth