Engineering Simple Circuit Problem (Clarification of doubts)

[galaxy_twirl]

Homework Statement

Homework Equations

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 i_x = 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):

 

[GreenPrint]

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

 

[galaxy_twirl]

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

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

 

[GreenPrint]

Oh ok let me look through this again.

 

[galaxy_twirl]

Oh ok let me look through this again.

Thank you! :D

 

[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.

 

[galaxy_twirl]

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.

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

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

(5i_x - v₁/5) - 1 - 2 = 0

 

[NascentOxygen]

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 I_x - 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.

 

[gneill]

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.

 

[galaxy_twirl]

You have made the same mistake as you did with an earlier circuit analysis.

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

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 I_x - 20)/30 leaving the top, but you also need to show it reentering at the bottom on the Earth wire.

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 I_x - 20)/30 - (5 I_x - 20)/30? And is this why you said that my SN achieves nothing additional?

(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.)

May I know, what do you mean by this?

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.

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

 

[galaxy_twirl]

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.

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

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.

I see. I didn't know that reference nodes are chosen because of common connections. What I learned 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 learned 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? :)

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.

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

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

 

[gneill]

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

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.

I see. I didn't know that reference nodes are chosen because of common connections. What I learned 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? :)

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.

I also learned 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? :)

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.

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

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

Sure. Isn't that better than writing a supernode equation for the reference node? :)

 

[galaxy_twirl]

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.

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.

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.

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? :)

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.

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

Sure. Isn't that better than writing a supernode equation for the reference node?

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. :)

 

[gneill]

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 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.

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. :)

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

 

[galaxy_twirl]

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

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

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.

 

[gneill]

Control systems sound like.. robots! Am I right?

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.