
#1
Mar1312, 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_c3V)/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! 



#2
Mar1412, 08:47 AM

HW Helper
P: 4,716

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 opamp + input is so tiny. 



#3
Mar1412, 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 = 6I*3k V_d = V_c 3 +I_3*9k = 1212I V_aI_12 = V_d > I_12=0.5I V_b = V_a I_6*6k > I_6 = 10.5I Therefore I = I_12 + I_6 = 1 mA This would mean that the potential at node d is zero. Can opamps do that? 



#4
Mar1412, 10:28 PM

HW Helper
P: 4,716

Solving a DC Circuit with OpAmpI can't see it. 



#5
Mar1412, 11:17 PM

P: 142

well i calculated I_3 as (V_c3V)/3k = (63I3)/3 = 1I
so subbing that into V_d = V_c 3 +I_3*9k = 63I3+9(1I) = 1212I unless i calculated I_3 wrong? 



#6
Mar1512, 05:16 AM

HW Helper
P: 4,716





#7
Mar1512, 11:39 AM

P: 142

I see, I could exclude the 3V source from the calculation of V_d which would make it 1512I
So now, V_aI_12*12k = V_d = 1512I 126I  I_12*12k = 1512I I_12 = 0.5I0.25 Since V_b = V_a I_6*6k > I_6 = 10.5I 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
Mar1712, 07:25 PM

HW Helper
P: 4,716




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