Ok this is part of step function. I'm just going to do the basic analysis of these things.

i(0), Rth etc and ask my lecturer when I get back about how to complete the functions for the variations.

For the moment I want to ask a series of what may seem to you guys to be the most stupid questions ever posed by a human being, but to me are going to make things so much clearer.
This thing down here is driving me absolutely nuts. I can't find a loop. I can see that it's probably very simple to solve but I need to know some things..
All things in blue are given in the circuit. The red and green stuff is mine.

Q1. N2 - is that the ground node? is it equal to zero?
Q2. If q1 is true. Is that because that power source between N2 and N1 is making it 0?
Q3. I know this may seem silly but is that Amp source 2ix supplying any voltage to N1? Do amp sources in general supply any voltage?
Q4 Is N1 equal to 12v no matter what?
Q5. Is the current between N2 and point B = ix?
Or does some of ix go up from N2 to N1? Or can no current travel up that leg because the negative end of the power source is sitting there?

 

A1. NO. NO. (The ground node is wherever you attach the ground symbol.)
A2. NO
A3. Maybe. YES.
A4. NO
A5. Maybe. Maybe. NO.

Note: An answer of "Maybe" indicates that I have not solved the circuit so I don't know what the exact answer is.

 

I think I might have had a bit of a eureka moment or perhaps not. Is this right? Couldn't be? Could it?

 

If ISC is the current through the switch at BC, then I see a current of 8A flowing through the parallel combination of a 1Ω and 3Ω resistors. So by the current divider rule, 6A must flow through the 1Ω resistor and 2A through the 3Ω resistor.

 

Sorry that is Isc or Io in this case. Thats the inductor between A and B. There's actually two switches in this circuit. There's one on the same branch, just above the 12v. Closed at this point. There's another open switch to the left of B. I just didn't include it in this diagram cause it helps in think through it better

 

Ok this is part of step function. I'm just going to do the basic analysis of these things.

i(0), Rth etc and ask my lecturer when I get back about how to complete the functions for the variations.

For the moment I want to ask a series of what may seem to you guys to be the most stupid questions ever posed by a human being, but to me are going to make things so much clearer.
This thing down here is driving me absolutely nuts. I can't find a loop. I can see that it's probably very simple to solve but I need to know some things..
All things in blue are given in the circuit. The red and green stuff is mine.

Q1. N2 - is that the ground node? is it equal to zero?
Q2. If q1 is true. Is that because that power source between N2 and N1 is making it 0?
Q3. I know this may seem silly but is that Amp source 2ix supplying any voltage to N1? Do amp sources in general supply any voltage?
Q4 Is N1 equal to 12v no matter what?
Q5. Is the current between N2 and point B = ix?
Or does some of ix go up from N2 to N1? Or can no current travel up that leg because the negative end of the power source is sitting there?

RBR1317 has already mostly answered things, but I'm going to make a few additional observations and it's just as easy to answer all of the questions from scratch.

A1) The "ground" node is whatever node you decide to call ground (a better term is "common" since "ground" implies something that is not necessarily true). What you are really doing is picking one node and declaring to the world what the voltage on that node is. You can do both of those randomly, but after that the value of the voltages AT all of the other nodes will be fixed due to your choices. This is because the notion of the voltage at a certain point is always, in actuality, the voltage difference between that point and some point that we happen to label as zero volts.

A2) The 12 V source does nothing other than supplies whatever current is required in order to force Node N1 to be 12 V higher than Node N2. So if you know the voltage at one of those two nodes (relative to the common reference node), then you know the voltage at the other. But if you don't know one, then you don't know either, you only know the relationship between them.

A3) Just as a voltage source will provide whatever current is required in order to impose the required voltage difference between its terminals, a current source will provide whatever voltage across its terminal that is needed to impose the required current through it.

A4) Only if you CHOOSE to assign a voltage of 0 V to Node N2.

A5) Only if the current in the 12 V source happens to be zero, which it probably isn't.

Think about your reasoning -- you are claiming that no current can flow into the negative terminal of a battery. So what happens to the current that flows out of the positive terminal of a battery?

 

Beautiful... That clears things up so much. Thank you sir. What about my reasoning in post #3. Can I do that? Would -2ix be flowing across that 4 ohm resistor and into N1. I can't think why not. Is that at least a correct way to solve this circuit?

 

Beautiful... That clears things up so much. Thank you sir. What about my reasoning in post #3. Can I do that? Would -2ix be flowing across that 4 ohm resistor and into N1. I can't think why not. Is that at least a correct way to solve this circuit?

Well, the first step is to see whether the method you used yielded a correct answer. If it didn't, then you know it doesn't work (unless you made a mistake somewhere). If it did, then there is a good likelihood that it is valid.

So assume your answer is correct and see if it results in a circuit that is self-consistent.

In doing so, you will hopefully notice a few things. First and foremost, the majority of the circuit is superfluous to the question of what the short-circuit current between B and C is.

You have an 8 A current source feeding a current divider formed by a 1 Ω and a 3 Ω resistor. How much current is flowing in the 1 Ω resistor?

Notice that the 12 V source, the 2·ix CCCS, and the 4 Ω resistor are completely irrelevant. As long as a path exists from Node D back to Node N2 (other than the one through the current source), you only need to consider the current source and the current divider.

And you STILL need to start tracking your units properly.

 

hmmm... Which would be correct -8 (3/4) = -6... Now that should make me really happy but it actually just makes my brow curl up in confusion. Like why for example considering your analysis does the 2ix not pump in some current to that equation?

About tracking my units. You've said that to me three times. I could see the error on the first. Missed it on the second. With this the only problem I can see is maybe I should have put the negative in front of the 1 instead of the brackets. What exactly do you mean by 'not tracking my units?'

 

It's impossible to say whether +6 A or -6 A is correct for Isc because your diagram never defines what Isc is. Your title implies that it is the current through the closed switch, but you never indicate what direction it is flowing in. Whether it is positive or negative depends entirely on that definition. Both voltage and current have both magnitude and direction and the sign of the quantity only tells us the direction if it is defined what direction is considered positive.

Almost all physical quantities are made up of two parts -- a magnitude and a unit. A distance has units of length, such as inches, meters, or light-years. Current has units of charge per second. A current of 1 A is a charge flow of 1 C/s (coulomb/second). Voltage has a unit of energy per unit charge. A voltage of 1 V is 1 J/C (joule/coulomb). A resistance has units of voltage per unit current. A one ohm resistor produces one volt across it per ampere of current flowing through it.

The unit associated with a physical quantity is fundamentally part of that quantity.

If I tell you that the person on the left is 76 tall and the person on the right 72 tall, which one is taller? What if I tell you that the person on the left is 76 cm tall and the person on the right is 72 inches?

Does 72 = 6 or does 6 = 2?

Of course not.

But 72 inches = 6 feet and 6 feet = 2 yards.

Does 5 = 5?

Of course.

But 5 km does not equal 5 miles.

The units are fundamentally part of nearly any physical quantity.

The numerical part and the dimensional part behave as if they are multiplied together and they can (and should) be manipulated exactly that way.

Tracking and checking units is perhaps the single most powerful error detection and correction tool available to the scientist or engineer. Most mistakes you make (certainly not all) will mess up the units allowing you to catch them almost immediately. At any point if the units don't work out, then the answer is guaranteed to be wrong.

Unfortunately, most textbooks and most instructors are hideously sloppy with units and, in most cases, for a very simple reason: Most textbook authors and most instructors have little or no experience in the real world. If they get an answer wrong because they couldn't be bothered to track units then there is little consequence beyond, perhaps, some embarrassment. In the real world, when people get sloppy with units airliners with a hundred people on them run out of fuel in mid air, billion dollar space probes get slammed into planets by mistake, and people with pregnant wives and small children don't come home because the text fixture they designed fails shooting a piece of steel through their chest.

Sadly, most engineers are sloppy with units because they learned from textbooks and instructors that were sloppy with units.

But if you will develop the habit of always, always, ALWAYS tracking your units, you will discover that you will spend a lot less time working problems and will get much higher grades because the work you do will far more often be correct.

 

Hmmm I shall consider this in detail my friend both the consequences to myself and the people around me. It's probably good practice and when these equations start 'coming into focus,' I'll definately start making a point of it. For the moment i'll add as much detail as possible, but I like to keep the equations and diagrams as clean as I can, for me, because the over complication is actually much more likely at this time, to cause me to make mistakes. Stretch your mind back to when you first seen all this. Trying to understand it all. Now you're infinity likely, to have a much higher IQ than me. I'm not the sharpest tool in the box. I got myself the qualifications for this degree just by brute effort. These things look like mountains to me. Monsters with fangs n claws.

A favor I would ask. Can you rewrite one of the lines of my equations, with proper unit tracking so I can see what you mean? When the things become more familiar I'll make it my practice.

And btw thanks for the great advice with everything. You'll probably get frustrated with me sometimes but understand, it's a mechanical problem with me. I can't make myself as naturally intelligent as you. I could maybe get to where you're at but it takes me a much longer time. Like a nice linear graph. I can get there but it's much further down the X axis...if you get me haha. Thanks my friend.

 

I think you are short changing yourself -- you are choosing to forego the ability to not make many of the mistakes you'll make (and that we all make) and the ability to catch most of the ones you do. You don't want to take the time to do it right, but you don't mind taking the time to do it over or to turn in errors that you didn't catch and lose the corresponding points. Does that really sound like a rational approach to you? Would you go to a doctor that took that approach to their medical training? Most people wouldn't knowingly do so on the basis that a doctor that won't exercise due diligence to catch and correct errors is grossly negligent and might kill someone. But consider that most negligent doctors are limited to killing people one at a time, while negligent engineers can and do kill people in job lots.

I remember full well back when I was first struggling with this stuff. Specifically, I remember a watershed day when, on the first day of class (I still remember the room number and where I was sitting in the room and where the professor was standing), my professor opened up the course with a tongue-in-check comment that went pretty much exactly like this: There are two things that separate people like you and me from Nobel Prize winning physicists; the person walking across the stage in Stockholm always does two things: they always, always, always track their units and they always, always, always ask if the answer makes sense. That day the light went on and I started applying that advice diligently (though not perfectly) to everything I did. That semester was my first ever 4.0 in college. I made a point years later of thanking that professor (who had gone on to become the president of the university) for what was truly life-changing advice. In any course I teach that involves dimensional computations I require, with very severe penalties, that work properly track units and, invariably, most of the students hate it and complain loudly -- yet within a month or so they have adapted and the quality of their work skyrockets, along with their grades. It's not just a technique that yields results, it is a transformational change in attitude the reaps benefits far beyond the effort involved. I've had perhaps a dozen students come back years later and thank me for forcing them to become units-aware. I remember one student in particular that was a C to D student on Academic Probation that suddenly got a 3.5 and was on the Dean's List every semester from then until he graduated.

As for showing you an example, I believe I have uploaded solutions to at least one of your earlier problems. I'll try to find the post and link to it.

 

Look at my solution here: https://forum.allaboutcircuits.com/...ources-thevenin-eq.134134/page-3#post-1119612

Each line of the solution is dimensionally sound.

The first line has a resistance on the left side and the ratio of a voltage to a current on the other, which has units of resistance (volts/amp).

The second line has units of voltage on the left side and the product of a current times a resistance on the other, which has units of volts ( (amps)·(volts/amp) ).

The third line has current on the left and two terms that are each current on the right. The same for the next line.

Each and every line of the Rth calculation that follows has units of resistance on the right hand side.

 

hmm interesting my initial feeling is one of repulsion but that's because I'm not used to seeing equations like that. I think i'll maybe try a half way approach for now. Clearly define what I'm doing in definitional things but I do like my sums nice n clean. I like the diagram labelling. There was one thing I couldn't understand at line 5. Why is it
IT= ix - 1/2(ix)
and not
IT = ix + 1/2(ix)?

 

hmm interesting my initial feeling is one of repulsion but that's because I'm not used to seeing equations like that.

That's exactly what it is -- lack of familiarity. It doesn't take long before it is the other way around. I have a very hard time working with equations that are not dimensionally sound because my mind simply balks at adding 8 to Ix or something similar. I have to go in and correct the units otherwise I am very uncomfortable proceeding since the units don't work out properly and I long ago trained myself to check the units at each step and stop right then and there anytime that was the case. This is usually subconscious so I don't sense any effort when the units are correct, but it creates a huge amount of effort to slug through something when the units are all botched up.

I think i'll maybe try a half way approach for now. Clearly define what I'm doing in definitional things but I do like my sums nice n clean. I like the diagram labelling. There was one thing I couldn't understand at line 5. Why is it
IT= ix - 1/2(ix)
and not
IT = ix + 1/2(ix)?

Consider KCL at node 'a':

1) You have It entering the node from the left.
2) You have ix/2 entering the node from the right
3) You have ix leaving the node upward (upper branch)

KCL says that the sum of the currents entering a node equals the sum of the current leaving the node.

Sum entering: It + ix/2
Sum leaving: ix

It + ix/2 = ix
It = ix - ix/2

 

Thanks again. I'll take a close look at both those threads once I get home.