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!
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?
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?
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.