Solving a DC Circuit with OpAmp

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Discussion Overview

The discussion revolves around solving a DC circuit involving operational amplifiers (op-amps) and resistors. Participants are attempting to determine the potentials at various nodes and the currents flowing through the circuit, while addressing the complexities introduced by the configuration of resistors and the op-amp's characteristics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes that the op-amp does not draw current into its input and that nodes b and c are at the same potential, but struggles with the implications of the 6k resistor on the calculations.
  • Another participant challenges the initial assumption by suggesting that the current to ground through the 6kΩ resistor should not be ignored, proposing a potential divider approach for calculating voltages.
  • A different participant attempts to recalculate the potentials and currents based on the revised understanding, leading to a potential conclusion that the voltage at node d could be zero, questioning the feasibility of this outcome for op-amps.
  • Several participants engage in recalculating currents and voltages, with some expressing uncertainty about their calculations and the implications of excluding certain voltage sources from their equations.
  • There is a repeated focus on the relationship between the currents I_12 and I_6, with participants trying to reconcile their values to find a consistent solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or the implications of their results. Multiple competing views and uncertainties about the circuit behavior remain throughout the discussion.

Contextual Notes

Participants express uncertainty regarding the treatment of the 6k resistors and the impact of the op-amp's characteristics on the circuit analysis. There are unresolved mathematical steps and dependencies on assumptions that affect the conclusions drawn.

doublemint
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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!
 

Attachments

  • circuit.png
    circuit.png
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V_a=12V
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.
 
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?
 
V_d = V_c -3 +I_3*9k = 12-12I
12-12I https://www.physicsforums.com/images/icons/icon5.gif

I can't see it.
 
Last edited by a moderator:
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?
 
doublemint said:
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[/color] +I_3*9k = 6-3I-3+9(1-I) = 12-12I
Check this.
 
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.
 
doublemint said:
then I= I_12 +I_6 = 0.75
correct
 

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