How Do You Calculate the Voltage at Node C in an Op Amp Circuit?

[lam58]

Hello, I'm stuck on this question, it asks to find the voltage at node c (on the circuit below) then use this value to work out other values of the op amp circuit.

I tried using kirchhoffs laws but that didn't work, then I used thevenins. Basically I took out the 6k and 8k resistors and calculated the voltage difference between the two 3k resistors and got 10.5V. Then I took away the two 3k resistors and put back the 8k and 6k then using the 10.5v found the voltage drop across the 6k resistor i.e. (6k x 10.5v)/(6k + 9k) and got 4.5v. I then took 4.5 from 10.5 and got 6 volts. However I don't have answers to this question so I'm unsure if I've got the right answer and/or my method is correct, which I doubt.

Any help on how to approach this question would be much appreciated.

 

[Jony130]

The correct answer is 6V. Why you don't use nodal analysis? Also are you aware that thanks to negative feedback action the voltage at point B must be equal the voltage at point A? If so VA = 0V so the VB also must be equal to 0V.

 

[lam58]

The correct answer is 6V. Why you don't use nodal analysis? Also are you aware that thanks to negative feedback action the voltage at point B must be equal the voltage at point A? If so VA = 0V so the VB also must be equal to 0V.

Mainly because I'm an idiot.

 

[Jony130]

Mainly because I'm an idiot.

Don't be so harsh on yourself, new year is coming.

I think that you should try again but this time use KCL.

For this diagram

We can write for node C the KCL

I1 = I2 + I3 + I4 and additional

I1 = (21V - Vc)/3K

I2 = Vc/6K

I3 = (Vc - Vd)/8K

I4 = ??

Next do the same thing for B node and next instead VB put VB = 0V and solve this simultaneous equations.

 

[lam58]

Don't be so harsh on yourself, new year is coming.

I think that you should try again but this time use KCL.

For this diagram

We can write for node C the KCL

I1 = I2 + I3 + I4 and additional

I1 = (21V - Vc)/3K

I2 = Vc/6K

I3 = (Vc - Vd)/8K

I4 = ??

Next do the same thing for B node and next instead VB put VB = 0V and solve this simultaneous equations.

I'm missing something here I'm putting I4 as (Vc-Vb)/3k

and for node B

(Vc-Vb)/3k = (Vb-Vd)/5k.

But this doesn't seem to be right.

 

[Jony130]

I'm missing something here I'm putting I4 as (Vc-Vb)/3k

Very good

and for node B

(Vc-Vb)/3k = (Vb-Vd)/5k.

But this doesn't seem to be right.

Why? I don't see any mistake. The ideal opamp has infinite input impedance. That means its inputs don't draw any current at all. And this is why I4 = I5
So you have all information needed to solve this problem.