Engineering [Circuit Analysis] Am I on the right track with this problem?

[AndrewEE]

Homework Statement

Find the value of V1, V2, V3, V4

Relevant Equations

None

*A is the node all the way to the left above the 30V but I didn’t end up using it because I don’t think you need it since the same as 30V, right? *

I know I messed up somewhere because shouldn’t V1-4 = 30? After finding my V1 I used that value to find the other Vs and it ended up being close to -1000V

 

[BvU]

Hello @AndrewEE,

!​

Relevant Equations:: None

Seems strange to me. List the equations you use.

No wonder I can not understand

If I have to guess, I would say you try to use Kirchoff current law. Where does the 30 A come from ?

I agree with your @V₄, @V₁, but not @V₂

And this one

is probably KVL, but I don't see it. Why 30 ?

I also find it confusing to see the black V1 in a different location than the red V1. I suppose the black one can be ignored.You may also want to apply KCL to a point B -- the minus side of the 30 V on the left

$\$

 

[DaveE]

It's a nice try, thanks for showing us your work. However, there are several problems with the equations you've set up.

@A: (30-V₁)/20 and (V4-30)/15 are currents, through R1 and R4. But 30 is the voltage at node A. It doesn't make sense to add up voltages and currents. Similar to the way in beginning physics problems you won't see Kg and meters summed in a formula. You can really only add up the same types of things.

@V2: You've correctly summed two of the three currents that can flow to that node. Or perhaps you assumed that the current through R3 is zero, without showing any justification for that.

I didn't review much more than this yet. I think you may be making too many evaluation steps at the beginning. Label the currents and make KCL equations at each node. In this network every node will have three currents, except V3 and the ground node which has four. Then in the next step describe each of those currents using the voltages and ohms law and substitute those into your KCL equations. Some currents will remain as variables, like the current through V1. Be careful to get the polarities correct. This will give you a set of equations to solve.

 

[AndrewEE]

It's a nice try, thanks for showing us your work. However, there are several problems with the equations you've set up.

@A: (30-V₁)/20 and (V4-30)/15 are currents, through R1 and R4. But 30 is the voltage at node A. It doesn't make sense to add up voltages and currents. Similar to the way in beginning physics problems you won't see Kg and meters summed in a formula. You can really only add up the same types of things.

@V2: You've correctly summed two of the three currents that can flow to that node. Or perhaps you assumed that the current through R3 is zero, without showing any justification for that.

I didn't review much more than this yet. I think you may be making too many evaluation steps at the beginning. Label the currents and make KCL equations at each node. In this network every node will have three currents, except V3 and the ground node which has four. Then in the next step describe each of those currents using the voltages and ohms law and substitute those into your KCL equations. Some currents will remain as variables, like the current through V1. Be careful to get the polarities correct. This will give you a set of equations to solve.

Thank you for your reply! Working through your reply... I'm a little confused on how to tell which direction the current is flowing through a resistor when you don't know the values of the voltages before and after. I know it will flow from high to low but when you have an unknown voltage, how do you tell? As in, (30-V1)/20 versus (V1-30)/20.

As well, I also am not 100% sure what to do when there's only a voltage. I know I = V/R but how do you find I when there's no R? That's why I added the 30, I figured that if there's no resistor, I = V?

For your other point at V2, again, I didn't know what to do with the 10V between V2 and 3. I sort of figured that since I'm looking at currents I should ignore the voltage source if there's no resistor? I went at it both ways, at node A I treated the V as I, at V2 I treated V as 0. From your response, I think both ways are incorrect.

I apologize if my questions are dumb and I'm wasting your time. I have a learning disability that makes learning new concepts challenging. Once I get it, I get it. It just takes me longer than others but I've always wanted to be an engineer so I'm pushing myself.

 

[AndrewEE]

Hello @AndrewEE,

!​

Seems strange to me. List the equations you use.

No wonder I can not understand

View attachment 279598​

If I have to guess, I would say you try to use Kirchoff current law. Where does the 30 A come from ?

I agree with your @V₄, @V₁, but not @V₂

And this one

View attachment 279599​

is probably KVL, but I don't see it. Why 30 ?

I also find it confusing to see the black V1 in a different location than the red V1. I suppose the black one can be ignored.You may also want to apply KCL to a point B -- the minus side of the 30 V on the left

$\$

Thank you for your reply! Hopefully I can answer you're response. The 30A is actually the 30V supply. When not given resistance, I'm not exactly sure how to handle voltage to translate it to current. I know I = V/R but when there's no R, I don't know what to do.

Also, I thought the sum of all voltages across a closed circuit have to = 0. So, wouldn't
V₁ + V₂ + V₃ + V₄ + 30V + 10V = 0? I wrote the equation wrong, it was supposed to be =40 not =30.

As far as the black and red, my professor did that. I think it's to show where the voltage source is versus what node you look at to find it, I'm not entirely sure though.

Everything on the bottom wire = 0 because of the ground, right? So, 3A + 1A - 2A - I_{across 30V}?

I'm going to copy and paste this in here too from the other reply on this thread because I feel bad for not knowing more about this like my classmates do. I apologize if my questions are dumb and I'm wasting your time. I have a learning disability that makes learning new concepts challenging. Once I get it, I get it. It just takes me longer than others but I've always wanted to be an engineer so I'm pushing myself.

 

[DaveE]

I'm a little confused on how to tell which direction the current is flowing through a resistor when you don't know the values of the voltages before and after.

You just need to define a current direction up front. Don't worry about guessing the correct direction. Then you need to subtract the voltages across the resistor with the correct polarity to match the defined current direction. If you guess the current direction wrong, you will just get a negative value for that current in your solution. The important thing is to make all of your definitions and equations consistent. This is why it is good to add the current definitions explicitly as a first step. Screwing up the polarities is by far the most common mistakes people make because it easy to do and the initial definitions are somewhat meaningless.

As well, I also am not 100% sure what to do when there's only a voltage. I know I = V/R but how do you find I when there's no R?

You can't substitute for the current through a voltage source. You would just define it as an unknown current and solve for it in your equations. The same thing is true for the voltage across a current source.

I apologize if my questions are dumb and I'm wasting your time.

These are not dumb questions, in fact they are really normal questions for this sort of problem. You don't need to apologize for learning, we are all (hopefully) learning new stuff and asking questions.

One of the nice things about PF is that everyone that answers is doing it because we want to. If I thought you were wasting my time, I just wouldn't answer. I do that all the time for posts that don't interest me.

 

[BvU]

because I feel bad for not knowing ... I apologize ..

I wholeheartedly agree with @DaveE : they are good questions and of course we are learning (Dave seems to be the expert here, I try to learn as I go along, so I have to look up things that I have little real experience with, e.g. KCL and KVL ).

Everything on the bottom wire = 0 because of the ground, right? So, 3A + 1A - 2A - Iacross 30V?

Not because of the ground but because of KCL plus the fact that there is no current to ground (nothing else is connected to ground). So this way you have the current into the 30 V source -- and thereby the current into A !

Thank you for your reply! Hopefully I can answer you're response. The 30A is actually the 30V supply. When not given resistance, I'm not exactly sure how to handle voltage to translate it to current. I know I = V/R but when there's no R, I don't know what to do.

Now you know: the current comes from elsewhere (the bottom end of the 30 V). And in = out otherwise charge would accumulate ad infinitum.
Re

So the 30 is now 2 Amps and you have given the two others a direction (away from A) so they both have a minus sign.

Then:

Also, I thought the sum of all voltages across a closed circuit have to = 0. So, wouldn't
V1 + V2 + V3 + V4 + 30V + 10V = 0?

KVL is about the sum of voltages around a closed loop. I.e. a single path that returns to the point where it starts. For example from A to A via R1, R2, V2, R3, R4:

V1 $-$ 30 V $\quad+\quad$ V2 $-$ V1 $\quad+\quad$ 10 V $\quad+\quad$ V4 $-$ V3 $\quad+\quad$ 30 V $-$ V4 $\quad=\quad$ 0​

​

Which leaves you with $\$ V3 $-$ V2 $\quad=\quad$ 10 V, so I don't think it's very helpful, but there are several more loops ! As Dave says: define a name and a direction for the currents in these loops and thus build a set of equations. If all is well you end up with a set of N equations and N unknowns ... fingers crossed .

I haven't done the whole exercise (and I'm not even sure I can bring it to a good end) so I depend on you to do the work -- keep us posted !

$\$

 

[Jody]

Hey!

This is a great question. I remember feeling super stuck on this too. Someone showed me a "trick" to solve all those negative or positive sign problems and this meme felt like it became real.

I'm going to echo what Dave said in #6 although here's an example of the setup I first drew in another thread (here). You assign it up front. The easiest thing for me to do is just to assign them all the same direction... all inward, or all outward... whichever makes me feel more comfortable. If you do that, then you never have to bookkeep the sign! Nice! I always mess up on the sign and so getting rid of that bookkeeping step gives me a lot of confidence in my answers.

This becomes even easier when you solve your equations symbolically first, then plugging in the numbers later. Let's say for instance I've solved this problem for what I want and my final answer needs $i_2$... my professor gave me the current going to the right like $i_4$ we can say $10mA$ :(... No problem! $i_2 = -i_4 = 10 mA$. I just plug and chug into my symbolic solution and the work is done.

What if you pick the "wrong" direction? I show with a classic voltage divider that you can do whatever you want. I solve one going all inward, then another all outward. You'll get the same answer. Try it out!