Gain of Electric Circuit: Vo/Vs and Resistance R

[TwoEG]

Homework Statement

So this problem is on Introduction to Electric Circuits 9th Edition (Richard C.Dorf)
This asks me to express the gain of this circuit Vo/Vs by terms of the resistance of R.

Homework Equations

KCL, KVL, some ideal OP AMP characteristics...

The Attempt at a Solution

Look at the undermost (horizontal) wire. Leftmost (perpendicular) wire has escaping current Vs/20(A). Middle (perpendicular) wire has no current (connected to OP AMP). Rightmost (perpendicular) wire has entering current Vo/80(A). Apply supernode KCL. Vs/20=Vo/80. Vo=4Vs. So gain is 4. Huh, this is weird. There is no variable R in this gain. Actually, provided answer sheet says gain is -4R/(R+80). But I cannot understand this answer. This gain has minus sign, so Vo and Vs must have different sign. This means undermost wire can only have entering current or escaping current. What's the problem with my logic?

 

[gneill]

The ground connection on the bottom rail is another current path, as is the output of the first op-amp. So you can't assume that the current that flows through Vs must be the same as the current that flows through the 80k resistor.

 

[TwoEG]

The ground connection on the bottom rail is another current path, as is the output of the first op-amp. So you can't assume that the current that flows through Vs must be the same as the current that flows through the 80k resistor.

Ahh... I forgot about that. But one more question. Is this mean that if I remove ground in this circuit, than the gain of circuit will become 4? I think gain of circuit must be same whether ground exists or not, since calculation using KCL and KVL on other parts of the circuit will be same.

 

[LvW]

Are you required to start the gain calculation with the basic KCL and KVL rules?
Why not using the known formulas for an inverting opamp adder?
That`s what the 1st stage does: It simply calculates the weighted sum of two input signals (Vs and Vo): V1=-[Vs(100k/20k)+Vo(R/20k)

 

[gneill]

Ahh... I forgot about that. But one more question. Is this mean that if I remove ground in this circuit, than the gain of circuit will become 4? I think gain of circuit must be same whether ground exists or not, since calculation using KCL and KVL on other parts of the circuit will be same.

The op-amps will still provide "hidden" current paths to ground, even if the bottom rail is made into an essential node by removing its ground connection. Think of them as controlled voltage sources with one "leg" of the source connected to ground.

My suggestion is to write KCL at the first op-amp's negative input and recognize how V1 is related to Vo.

 

[TwoEG]

Thanks for your replies! Helped me a lot :)