2 ideal op amp circuit design, find voltage Vp
(Edit R1/(R1+R2)= k3 should be 2/3 not 1/3 and makes only minor changes to Vp) Thanks to @G36 but this is still correct simplified approach.


Oh I see it's homework. (hint was ideal OA's) 

Not so simple with 2 feedback loops.

R7 limits current but not gain,  k1=R6/R5=6.91, k2=R3/R4=2.78, k3 =R1/(R1+R2)=1/3 . The U2  introduces offset ( Vp) which also affects U1 out. 


Vo=k1* (Vp-Vi)  or \$Vp= \dfrac{Vo+k1*Vi}{k1} \$
Vp=k2k3* Vo=\$ \dfrac{Vo+k1*Vi}{k1} \$ thus \$Vo =Vi*\dfrac{k1}{(k1*k2*k3)-1}=\$
.....


Good luck , I think answer is 768mV... But it's definitely &lt; 1V.

PROOF

Your Equivalent Circuit with gain blocks for each stage except for U1, R6/R5 ratio = k1 then U2 gain=k2,  R1/(R1+R2) = k3,  







simulate this circuit &ndash; Schematic created using CircuitLab

The solution to the problem must match Vin+ to Vp for ideal OA's have 0V differential (virtual gnd).

Intuitive answer

If k1R is removed, then Vp = Vi (ok?) = Vin+ ( 0v diff) then Vo=(-)Vi due to 0V Gnd reference.  Removing k1 causes an infinite negative feedback loop like and unity gain non-inverting amp.

if kiR is shorted then Vp=Vin+=Vo=0V(gnd)

Therefore Vp must lie between 0 and Vin+ for any value of k1, k2 k3 due to ideal gain of U1 in the loop regardless of k2 , k3  ( k*0=0, k*∞=∞) which are distractions that serve no real purpose but may simulate real systems.

comments


R9 = 91 is irrelevant since ideal OA can drive any current and does not attenuate any signals.
this answer relies on learning the equations for k of an inverting OA and learning Control System transfer function reduction for simple solution. Otherwise KVL gets big... which can be done, but I'm not going to try Mathjax on it... ;)


Ask any Control System Prof to validate. I am very confident it is correct.