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Solving a DC Circuit with OpAmp

by doublemint
Tags: circuit, opamp, solving
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doublemint
#1
Mar13-12, 11:28 PM
P: 142
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 Thumbnails
circuit.png  
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NascentOxygen
#2
Mar14-12, 08:47 AM
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P: 5,146
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.
doublemint
#3
Mar14-12, 10:07 PM
P: 142
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?

NascentOxygen
#4
Mar14-12, 10:28 PM
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P: 5,146
Solving a DC Circuit with OpAmp

V_d = V_c -3 +I_3*9k = 12-12I
12-12I

I can't see it.
doublemint
#5
Mar14-12, 11:17 PM
P: 142
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?
NascentOxygen
#6
Mar15-12, 05:16 AM
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P: 5,146
Quote Quote by doublemint View Post
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
Check this.
doublemint
#7
Mar15-12, 11:39 AM
P: 142
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.
NascentOxygen
#8
Mar17-12, 07:25 PM
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P: 5,146
Quote Quote by doublemint View Post
then I= I_12 +I_6 = 0.75
correct


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