# Simple Circuit Problem (Clarification of doubts)

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1. Nov 26, 2014

### galaxy_twirl

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I did a source conversion on the right-most 1A current source to get a voltage source to work with for this question. However, when I tried to work out the solution, my values seem wrong, but I don't understand why, especially for the part where I had ix = 4A. May I know where is my mistake? Thank you! :D

(Pardon for the poor image. If it cannot be seen clearly, feel free to ask me to another shot. :) )

My solution (partial, stuck):

2. Nov 27, 2014

### GreenPrint

When you did your conversion, point B shouldn't be connected to ground, it wasn't in the original problem before the conversion.

3. Nov 27, 2014

### galaxy_twirl

Oh, Point B isn't connected to the ground. I accidentally drew a stray line there and the zig-zag things were meant to 'cancel' the stray line. Sorry. :X

4. Nov 27, 2014

### GreenPrint

Oh ok let me look through this again.

5. Nov 27, 2014

### galaxy_twirl

Thank you! :D

6. Nov 27, 2014

### GreenPrint

The value for I_x doesn't seem to be correct. I'm not exactly sure how you got the value that you did. The node to the left of the 5 ohm resistor there's only three paths for current to flow through. It looks like your summing the currents at node A when you solve for Ix? If this is the case there shouldn't be four currents.

7. Nov 27, 2014

### galaxy_twirl

Yupp, my value of ix is wrong but I don't know why. ><

Yup, I only handled 3 branches for A, which I got this equation:

(5ix - v1/5) - 1 - 2 = 0

8. Nov 27, 2014

### Staff: Mentor

You have made the same mistake as you did with an earlier circuit analysis. When you lasso a block of any circuit and say the sum of currents into that block = 0, you must include all currents in all wires cut. Here, your lasso is cutting across 5 wires, so you should account for those 5 currents. You have a current (5 Ix - 20)/30 leaving the top, but you also need to show it reentering at the bottom on the earth wire. (If you were to erase one of your earth wires, that current would join with the current in the other earth wire so you'd still need to account for it.)

When you do it correctly you'll get the equation you already have using Σ currents to an ordinary node. So your SN idea achieves nothing additional.

9. Nov 27, 2014

### Staff: Mentor

Galaxy_twirl, you really should forget about making the reference node into a supernode and writing a node equation for it. It's almost never a profitable approach and just leads to more work for little benefit.

A reference node is usually chosen because it has the most common connections. Since you do not have to write the node equation for the reference node, this eliminates having to deal with the most complicated equation pertaining to the circuit.

In your circuit with the 1 A source converted to its Thevenin equivalent, you have $i_x$ running between two nodes of "fixed" potentials (with respect to the reference node). One is $5i_x$ and the other 20 V. You know the values of the resistors in between. So just write the equation for that current.

10. Nov 27, 2014

### galaxy_twirl

Oh dear. I'm terrible with supernodes. T.T I think I will clarify with my teacher again tomorrow. Dx

I see.. Hmm. Does it then mean that, assume currents entering the supernode is -ve, while currents leaving the node is +ve, I will have: (5 Ix - 20)/30 - (5 Ix - 20)/30? And is this why you said that my SN achieves nothing additional?

May I know, what do you mean by this?

I see. I actually got the answer after my friend helped me out with the question but I don't understand why my answer is wrong and hence my post here.

Thank you so much for your help! :D:D

11. Nov 27, 2014

### galaxy_twirl

I see. Thanks! I think I shall try and avoid supernodes for now, unless I really need to use them. :)

I see. I didn't know that reference nodes are chosen because of common connections. What I learnt is that, if I have an (independent) voltage source, I just make the -ve end of it GND, especially if the -ve of the (independent) voltage source is connected to a wire that has nothing connected across it. May I know if what I had said is correct? :)

I also learnt that I have to use supernodes if I am doing node voltage analysis and I need to write KCL for nodes having a dependent or independent voltage source. Mirroring this, side-tracking a little, I must link branch current to mesh current for mesh current analysis if a mesh has an independent or dependent current source. May I know if what I had said is correct? :)

I see. I think that was what my friend did, so in the end, I had:

$5i_x - 20$/$(20+10) = 0$
$i_x = -0.8A$

12. Nov 27, 2014

### Staff: Mentor

You can't avoid them if your circuit contains voltage sources connecting arbitrary nodes together. But you CAN avoid trying to write node equations for the reference node. It just not necessary and it makes things harder in the end.
It's fine if your circuit contains just one voltage source. Sometimes circuits will have several voltage sources and not all of their -ve terminals will be connected to the same node.

With experience you will develop an intuition for choosing an appropriate node as the reference node in order make solving the circuit for what you want easier. It really does take practice, and an exhaustive list of things to look for would not be particularly useful. You'll get used to picking a node that leads to fewer and simpler equations.
Yup. That's pretty much the picture. If you don't use supernodes or supermeshes, then for some circuits trying to write nodal or mesh equations becomes very nasty, where you have to introduce a bunch of new variables and equations to account for unknown voltages across current sources or currents through voltage sources.
Sure. Isn't that better than writing a supernode equation for the reference node? :)

13. Nov 27, 2014

### galaxy_twirl

I see. Just wondering, assuming reference nodes mean ground nodes, I thought it would be easier to write node equations with them, because they will be 0V and I can do, for example, $(v_2 - 0)/R = 0$ which looks simpler.

I see. Yes I agree, as the saying goes "practice makes perfect". Just wondering, do you know of any resources that provide good practice questions of different difficulty levels for circuit analysis, such as node voltage analysis, mesh current analysis, Thevenin and Norton equivalents? :)

I see. This reminds me of something I read when I first started learning this a few months ago. Haha. :)

Yup, definitely! ^_^

Thanks for all your help. Just wondering, are you an electrical engineer? :) I have a feeling that NascentOxygen and you are electrical engineers or engineers because both of you often reply to my posts. :)

14. Nov 27, 2014

### Staff: Mentor

I don't keep a list, but web searches on terms like "circuit examples" or "mesh analysis tutorial" or "Thevenin example" should turn up lots of stuff.

I have a background in Pure & Applied Sciences, Computer Science, and Electronic Systems (Electrical engineering with an emphasis on control systems).

15. Nov 27, 2014

### galaxy_twirl

I see. Alright, I shall search for them when I am free. :D Thank you~ :)

I see. Wow! That's like 3 degrees! Haha. Cool~ I'm doing electrical engineering now, just one bachelor's degree, and I am already struggling in my first year in the first semester. Oh dear. Haha. I must buck up. :(

Control systems sound like.. robots! Am I right? :) I have to choose my 'specialisation' in Year 3 though right now, I am thinking of not spcialising but taking modules across the whole EE field, like power systems, micro-electronics, etc. I think Computer Science is cool, but I just can't program. Gahhh. I guess everything takes time and practice, and we can't rush things that are meant to be taken slowly.

16. Nov 27, 2014

### Staff: Mentor

Any type of control system, really. Industrial processes have lots of electronic, hydraulic, pneumatic, etc., control systems. Robots certainly use them for smooth motions and accurate positioning.

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