The person that wrote this problem needs to be beaten severely.
That it would make the diagram too simple is beside the point. In many cases "in the real world" the answer is right before your eyes just as much as this one would have been had the 5V actually been possible and it is not unreasonable to craft problems that embody this notion so that students can see that sometimes you don't have to slug through a problem to find a trivial answer. The problem here is that the value given is not consistent with the circuit given. But which has priority? The circuit, or the specific information given? It's not obvious which is to be taken as the given and which is inconsistent with the given information. It's a bad problem (and for bigger reasons than this).The problem is assuming that the voltage across the 10k resistor is 5 V. It's not. I know the diagram indicates that, but that would make the problem too simple.
You computation of V1 is correct, except that you're being sloppy with your units.For the problem 4 i have V1 which is node after op-amp V1= 20v but i can't figure out how to get V0
V1 = -10(10k/5k)+20(10k/5k) = 20V
Any help would be appreciated
It would really help if you would indicated what these new currents are that you've introduced, namely I1 through I8. In particular, what directions they are each assigned. We are not mind readers and you shouldn't make people that you are asking for help have to reverse engineer your equations.LDC3 if I do what you said I get the following equations
I3 = I1 - I2 = 10mA - 2mA = 8mA
I5 = I3 - I4 = 8mA - 5/20K = 7.75mA
I7 = I5 - I6 = 7.75mA - 10IL
IL = I8 + I7 = (5.1)*(5) +7.75mA -10IL
solve for IL =>
IL = (5.1*5 +7.75mA)/11 = 2.318Amp is that correct?
You really get that the V2 is a bit over 30,000V?You need to recalculate the last line. I get a bit over 3 amps.
Is IL positive or negative?
BecauseI'm getting confused. Why is it not
IL = I7 - I8 like all the previous equations?
IL = 7.75mA - 10IL - (5.1)*(5)
Sounds about right. I was guessing something in the 0.5mA/V range, but didn't look at it close enough to determine whether it would need to be positive or negative. Let's see... 8mA - 0.25mA - 0.5mA - 5mA = 2.45mA that still needs to be consumed. So -2.25mA/5V = -0.5mAV + .05mA/V = -0.45mA/V. Yep, I agree.Interestingly if one changes the "5.1V2" VCCS parameter to -450uA/V with V2 as the controlling voltage then V2 will actually be 5V.
I made a mistake with my calculations. I erroneously assumed 5.1 x V2 was in mamps and then labeled it as amps. 2 mistakes at once.You really get that the V2 is a bit over 30,000V?
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