# Opamps -- Unknown values

1. Jun 20, 2016

### j172

• Poster has been reminded to post schoolwork in the Homework Help forums (moved from EE)
Hi everybody!

This is my first post as I am needing some help in a question I am struggling with. Not looking for answers, just some guidance in the right direction.

1. The problem statement, all variables and given/known data

I need to calculate the unknown, highlighted, voltages on the op amp circuits (attached).

Assume all op-amps are ideal

2. Relevant equations

See below

3. The attempt at a solution

I have done the following so far:

a) Inverting
A=-Rf/Ri =(-100)/10=-10
Vi =I1*R1 =0.00001*100000=1 V
Vo=(-Rf/Ri )*Vi
Vo =1*-10
Vo =-10 V

b) Summing
Vo=-(Rf/R_1 V_1+Rf/R_2 V_2 )
-2=(Rf/R_1 1+Rf/(R/2) V_2 )

c) Non-inverting
V1 =V2
V1 =100 mV
V2 =100 mV

d) Current to Voltage
Vo=-I1*Rf
Vo =-0.0002*10000=-2
Vo =-2 V

I am struggling the (b). I cant see how to calculate given only one value, and also with (c)

Many thanks in advance, happy to discuss the methods.

Joel

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Last edited: Jun 20, 2016
2. Jun 20, 2016

### .Scott

For A, Note that the voltage at both inputs to the op amp are equal, and at ground.
For B, I have 1R-2R+x(R/2)=0
For C & D, we agree.

3. Jun 22, 2016

### rude man

You have R1 wrong.
correct (I took out the "Rf" , should be just "R" everywhere)
Right.
Right
The R's cancel. There is nothing more to do with (c) than you have already done.

4. Jun 22, 2016

### j172

Thank you very much for your help

5. Oct 9, 2016

### rob1985

b) Summing
Vo=-(R/R*V_1+R/R*V_2 )
-2=(R/R*1+R/(R/2)*V_2 )

So is there another step to this because this equation doesn't give V_2. do you transpose it ?

Last edited: Oct 9, 2016
6. Oct 9, 2016

It doesn't?

7. Oct 9, 2016

8. Oct 9, 2016

### rude man

Well let's see - you have one equation with 1 unknown.
If this is difficult I really am at a loss as to what to hint next, so I must perforce cease from further such.

9. Nov 1, 2016

### rob1985

b is 1/2 volt

10. Nov 2, 2016

### Staff: Mentor

As a check, substitute your answer into the original equation to ensure that your answer does indeed make both sides of the equation equal.