# Homework Help: Solving a DC Circuit with OpAmp

1. Mar 13, 2012

### doublemint

Hey everyone,

I am trying to solve for the circuit that I have attached.

So I know that for a opamp, it does not draw any currents into its input side and the nodes b and c are the same potential. To figure out the potential at node b, i need to know I_6, but that 6k resistor is messing everything up.

If I assume that the 6k resistor is parallel to other 6k resistors then I can solve for I_6=1 mA
However, if that is the case then I_12=1 mA as well. But that means that the resulting voltage at node d is 0...which does not make sense..
Furthermore, from node c, I can use that to solve for the potential at node d.

I cannot say all three 6k resistors are in series since from that diagram it does not seem to be unless the ground has to do something with it.

Am I missing something here?

Here is my math:

V_a=12V
I_6=12V/12k = 1mA
therefore: V_b=V_c=6V
I_3=(V_c-3V)/3k = 1mA
V_d=V_c+9I_3=6+9=15V

this does not make sense since from V_a, I_12=1mA and that would result in V_d=0V...

DM
P.S sorry the diagram is so small!

#### Attached Files:

• ###### circuit.png
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2. Mar 14, 2012

### Staff: Mentor

This can't be right. Why ignore the current I to ground through the 6kΩ?
V_a=12 – I*6k

The potential divider with a pair of 6kΩ resistors gives
V_b=V_a / 2
This potential divider is unloaded because the current into the op-amp + input is so tiny.

3. Mar 14, 2012

### doublemint

Ok, that kinda makes sense to me.

So if I care my calculations with V_a=12 – I*6k,
V_c = V_b = 6-I*3k
V_d = V_c -3 +I_3*9k = 12-12I
V_a-I_12 = V_d --> I_12=0.5I
V_b = V_a -I_6*6k --> I_6 = 1-0.5I
Therefore I = I_12 + I_6 = 1 mA

This would mean that the potential at node d is zero. Can opamps do that?

4. Mar 14, 2012

### Staff: Mentor

12-12I https://www.physicsforums.com/images/icons/icon5.gif [Broken]

I can't see it.

Last edited by a moderator: May 5, 2017
5. Mar 14, 2012

### doublemint

well i calculated I_3 as (V_c-3V)/3k = (6-3I-3)/3 = 1-I

so subbing that into V_d = V_c -3 +I_3*9k = 6-3I-3+9(1-I) = 12-12I

unless i calculated I_3 wrong?

6. Mar 15, 2012

### Staff: Mentor

Check this.

7. Mar 15, 2012

### doublemint

I see, I could exclude the 3V source from the calculation of V_d which would make it 15-12I
So now,
V_a-I_12*12k = V_d = 15-12I
12-6I - I_12*12k = 15-12I
I_12 = 0.5I-0.25

Since V_b = V_a -I_6*6k --> I_6 = 1-0.5I
then I= I_12 +I_6 = 0.75

As for I_2,
It should simply just be I_2 = V_d/2k
and I could just find the value of every other current and voltage by subbing in I.

8. Mar 17, 2012

correct