# Find IL need help

Discussion in 'Homework Help' started by kzdizelkz, May 19, 2013.

May 19, 2013
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Feb 5, 2010
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3. ### kzdizelkz Thread Starter New Member

May 19, 2013
6
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For the first problem. With nodal analysis I have two nodes with V1 on top and V3 at the bottom first Voltage controlled current source is 25.5A when voltage across 20k is 5V current across it will be 0.25mA so from what i understand IL should be V2/10k=0.5mA but this doesn't work rest of it out. Any help would be appreciated.

4. ### LDC3 Active Member

Apr 27, 2013
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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.

Start by working out the current in each node from the right (after all we are given the current flow in these lines).

5. ### kzdizelkz Thread Starter New Member

May 19, 2013
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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

6. ### kzdizelkz Thread Starter New Member

May 19, 2013
6
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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?

7. ### LDC3 Active Member

Apr 27, 2013
920
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You need to recalculate the last line. I get a bit over 3 amps.
Is IL positive or negative?

8. ### kzdizelkz Thread Starter New Member

May 19, 2013
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its positive and i still get 2.318A (7.75 is in mA)

9. ### LDC3 Active Member

Apr 27, 2013
920
160
I'm getting confused. Why is it not
IL = I7 - I8 like all the previous equations?
IL = 7.75mA - 10IL - (5.1)*(5)

10. ### WBahn Moderator

Mar 31, 2012
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The person that wrote this problem needs to be beaten severely.

So they specify that the voltage across from top node to bottom node is 5V. That directly tells you what the answer is, namely IL=5V/10kΩ=0.5mA. But, this answer is not consistent with the circuit. But it's actually impossible to solve the circuit because the equation governing one of the dependent sources is nonsensical. The left-most source is a voltage-controlled current source, but the controlling equation is given as 5.1V2. Well, 5.1 multiplied by a voltage is a voltage and not a current. So now you have to assume the units on the 5.1. Most people that don't think that units matter will jump in and say that you use the base units, which in this case would be converting from volts to amps. So if V2 is 5A, then the current must be 25.5A. Many people will be happy assuming that it must be 25.5mA since that is on the same order as the other currents in the circuit. But on what basis can you support such a claim? It's lousy engineering and the kind of sloppiness that eventually gets people killed. It should not be tolerated.

In this problem, specifying V2 overspecifies the problem. For the given circuit, and assuming that the transconductance of the leftmost dependent current source is 5.1mA/V, you get that IL is slightly more than 0.13mA, which makes V2 only 1.3V.

If we assume that it is 5.1A/V, then the load current comes out to be about 0.16μA.

11. ### WBahn Moderator

Mar 31, 2012
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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).

12. ### WBahn Moderator

Mar 31, 2012
18,087
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You computation of V1 is correct, except that you're being sloppy with your units.

V1 = - 10kΩ [10V/5kΩ - 20V/5kΩ] = 20V

Note how not only do the units work out, but the form of the equation also matches the analysis of the circuit that produces it. Thus it is much easier to verify that it is correct.

Q1) Now, given V1, what is the voltage across the two resistors to the right of it?

Q2) What is the current flowing in those two resistors?

Q3) What is the voltage drop across the 1kΩ resistor?

Q4) What is the voltage at the node between the two resistors?

Q5) What is Vo?

13. ### WBahn Moderator

Mar 31, 2012
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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.

Notice that you are falling into the trap that I mentioned in my first response. Consider your equation:

IL = I8 + I7 = (5.1)*(5) +7.75mA -10IL

You don't know what the units on the 5.1 are and so you just ignore the issue, throw them away, and assume that whatever number you get you can just add it to what ever number happens to be next to it. That's sloppy and that is how engineers kill people (and do so in job lots).

If you can't put consistent units to something, then something is wrong! Now, how you handle it is a judgement call. In a case like this, my recommendation would be to put a big note off to the side saying that critical units were not supplied and that you are assuming them to be such and such. Then proceed to solve the problem with your assumed units.

The second cardinal rule of engineering that won't kill people unnecessarily is to always ask if the answer makes sense. Notice that you have written your equations still accepting that V2=5V. Well, that means that the voltage across the load resistor is also 5V, which we know results in IL=0.5mA. Yet you come up with 2.318A, which would result in V2 being 23,180V. Does this make sense?

14. ### WBahn Moderator

Mar 31, 2012
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You really get that the V2 is a bit over 30,000V?

15. ### WBahn Moderator

Mar 31, 2012
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Because

1) He has defined I8 to be a current flowing up instead of a current flowing down, as he did the previous even-numbered currents. But you can't know that unless you reverse engineer his equations, which is poor form on his part.

2) Because you still don't have units on the 5.1, so is that (5.1)*(5) going to give you 26.5A or 26.5mA or what? Sloppy, sloppy, sloppy.

3) Because you both are still insisting on using V2=5V even after you stated that it was incorrect. Why are you still using something that you know is wrong?

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Interestingly if one changes the "5.1V2" VCCS parameter to -450uA/V with V2 as the controlling voltage then V2 will actually be 5V.

17. ### WBahn Moderator

Mar 31, 2012
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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.

Makes you wonder how the person putting this together ever came up with 5V. My guess is that they didn't even stop to thing whether they were overdefining the system and just assumed that they could specify whatever voltage they wanted to anywhere they wanted to.

18. ### LDC3 Active Member

Apr 27, 2013
920
160
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.

19. ### kzdizelkz Thread Starter New Member

May 19, 2013
6
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WBahn,
But if the this problem AS-IS with out any changes I was correct that IL would be equal to IL = 5V/10K = 0.5mA I have to use what I was given.

20. ### WBahn Moderator

Mar 31, 2012
18,087
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The problem, AS IS, has no solution. Period. End of paragraph. End of chapter. End of story. Any solution you come up with that assumes V2=5V will violate both KVL and KCL and be physically impossible.