hello
im now stuck on a two loop mesh that I have to solve with simultaneous equations.
I get one correct current which I think is I1 = 0.857ma
then I get lost at current two which I get 0.476ma, instead of 0.856ma

equation 1 is 4v 6kX + 4kY
equation 2 is 2v 4kX + 5kY
equation 3 is 8v 12kX + 8kY
equation 4 is 6v 12kX + 15kY
4-3 gives -2v 12-12=0 15-8= 7kY

-2v/7k = -0.286mA I2=-0.285mA

putting this into equation 1 is
4v 6kX + 4k x -0.285mA = -1.144v

4+-1.144/6= 0.476

I think the problem is the -1.144, if it wasn't minus I would get the right answer but I don't know what to do with that minus.
any help would be appreciated!!!
many thanks
simon

 

Your circuit shows specific values for the resistances and then you comment at the bottom states that you "simplified" the circuit by changing three of the values. So are the other two values unchanged?

Where are these four equations coming from?

What are X and Y?

We are NOT mind readers!

 

there were two resistors that I combined to make R1, two resistors that I combined to make R3. R2 was on its own anyway.
Y is 0.285ma
X came out at 0.457ma

equation 1 is the first loop
equation 2 is the second loop
equation 3 is the first loop multiplied by 4
equation 4 is the second loop multiplied by six
this part is for the simultaneous equations.
X is 0.856ma or there abouts. I checked on multisim and that's what came up. the answer for Y is correct.
I don't know where im going wrong.

 

Again, what are X and Y?

Is X the current of the Thames river at high tide on the summer solstice?

It might as well be for all that you've specified what it means?

And what do you mean by

4v 6kX + 4kY

That is NOT an equation.

What do you mean by "equation 1 is the first loop"?

You need to stop be sloppy with your descriptions, because that results in sloppiness in your results. In this case, your sloppiness is leading you to ignore a couple of signs that you aren't including in your work.

So take a step back and be explicit in your descriptions and your developement. Treat it as though you are explaining it to someone that is a class behind you.

 

You are confusing poor old WBahn, who really is very good at helping folks, because you are confusing yourself by mixing up three different circuit analysis methods.

It is always a good idea to state your justification at the end or beginning of each line of calculation.
It is actually mandatory in formal engineering calculations in real life industry.

Start again and decide for each equation are you using KVL or KCL and state this. You may mix equations from these two, but do not add then in stuff from other methods.

It is a good idea, though not mandatory, to mark on your circuit the currents, assumed directions and node labels to assist. You can then identify the loops uniquely. There are three loops, but you can only pick two from three to use.

Get into good habits early and you will help yourself immensely.

 

Oh, I'm pretty sure I know what he means to say, but I'm trying to get him to put in the effort to be better at saying it -- and I really do think that there is a good chance that, if he does, that he will spot his mistakes.

I'll also through out that it really is mandatory that you either mark your currents, including directions, and such on the circuit OR that you otherwise clearly define them. To solve for X and not give any indication of what the hell X means is pretty bad. It is almost as bad to indicate that X is the current in such-and-such resistor and not give any indication of what the reference direction for that current is.

@ninjaman: Take a step back and look at your original post as though you are reading something that someone else posted and all you know is what is in that post.

Based just on what is in that post, what is I1? Meaning, which current in that circuit is it? The post says that the person got this current correct with a value of 0.857mA, yet nowhere in the rest of the post is there any evidence that this was a value that was ever arrived at for anything. The post then talks about "the second current" and how the answer is supposed to be 0.856mA but the poster got 0.456mA. Again, the poster gives not clue as to which current in the circuit is being referred to as "the second current". A bit later, the poster does talk about I2, which you would think might be the second current previously referred to, but the poster says that, for this current, he got an answer of -0.285mA.

Similarly, what is being referred to by X and Y?

How would YOU like to grade someone's work that looked like this? How much time would YOU be willing to spend trying to figure out what the person meant and what they were trying to do? How quickly would you adopt the attitude that it if the person submitting the homework for grading didn't care enough to make their work easy to follow, that they must not care too much about receiving a good grade?

To drive home the importance of the proper care and feeding of homework graders, consider that they have a very finite amount of time to spend grading any particular person's paper. For example, my Circuits II grader has 2.5 hours per week that they are contracted for. In that time, they have to grade 40 papers with three problems each and have to enter the grades in the database. Even if we ignore the housekeeping overhead, that gives him 150 minutes to grade 120 problems. So he basically has just one minute to spend grading a given problem from start to finish. How much time do you think he is going to spend deciphering what someone meant if that person couldn't be bothered to make it clear from what they presented?

Perhaps this will make it a bit clearer what I mean. Let's say that I ask you to find where I am going wrong in the following:

6 4M + 3*7.62 = 22.86

6 + 22.86/4 = 11.755

Pretty cryptic, huh?

For what it's worth -- you are very, very close to having the right answer. Just take the time to present your work carefully and you will find it.

 

Oh, I'm pretty sure I know what he means to say,

Aw, I was just being nice.

 

Aw, I was just being nice.

Ah... so sweet of you!

 

(1) 4v = I1 x 6k + I2 x 4k x5
(2) 2v = I1 x 4k + I2 x 5k x4
(3) 20v = I1 x 30k + I2 x 20k
(4) 8v = I1 x 16k + I2 x 20k

(4) - (3) = -12v I1 x -14k

-12/-14 = 0.857mA = I1 (I think)

this is what I have so far, I don't know where to go from here. any help would be appreciated.
thankyou

 

Use Superposition Theorem.
http://en.wikipedia.org/wiki/Superposition_theorem

 

hey thanks shteii01, I didn't think about that. but I have to show it using simultaneous equations, and show will kvl kcl.
I was getting confused with the direction and which voltage I had to use and which resistance. I have the answer now but confirming it with superposition is a good idea.
all the best
simon

 

[rant]
I see that you haven't taken my advice.

Following this advice not only helps the user immensely, it also helps those who come after trying to sort things out.

Throughout industry there is what is known as the 'independent check'.

That is someone makes a check of the calculations before the design or whatever is implemented.

Would you like to walk out onto a platform above Niagara falls that had not been independently checked for structural integrity?

Or perhaps would you like to receive an injection that had been drawn up without an independent check on the quantity of drug in the syringe?

Make things easy for yourself and others.

[/rant]

Do your last two posts mean you need further help or not?

Getting KCL and KVL right is a bit of an art.

 

please could you explain the art as it sounds interesting.
also, why the [rant]
if you have an issue then please don't answer any of my posts. unless your paid to. in which case please let me speak to your manager!

 

Happy to explain, but first can we agree sign conventions?

This site uses the convention of electron current, that was popular a few years ago.

I prefer to use what is known as conventional current. That is the positive direction of current flow is from postive to negative.

Which are you studying in your course?

 

i think we are using positive to negative.
conventional current flow, though i cant remember the lecturer saying much about it for this question. he told me that if the answer was a minus then the direction i thought was correct is wrong and current flow is in the other direction. i read about direction being arbitrary as the result would show the proper direction with a minus or plus. i had a difficult time finding the answers as i couldn't tell which resistance i was supposed to use with what current to find the right voltage drop.

 

i had a difficult time finding the answers as i couldn't tell which resistance i was supposed to use with what current to find the right voltage drop.

Do you think not labelling any of the currents may have had anything to do with this?

Have you identified the three loops in your original diagram?

 

I know there are two loops, one inside R1 and R3, one inside R2 and R3. then there is a loop around the outside. I think it is the process that I get confused with when trying to find voltage drops. I had to look are what I was taking away. it was a guess. it worked out, though doesn't make sense.

 

Try read this pdf file.

 

(1) 4v = I1 x 6k + I2 x 4k x5
(2) 2v = I1 x 4k + I2 x 5k x4
(3) 20v = I1 x 30k + I2 x 20k
(4) 8v = I1 x 16k + I2 x 20k

(4) - (3) = -12v I1 x -14k

-12/-14 = 0.857mA = I1 (I think)

this is what I have so far, I don't know where to go from here. any help would be appreciated.
thankyou

You STILL haven't defined what I1 and I2 are!

Don't make the people reading your work guess what you mean or have to reverse engineer it. A grader will give you a low grade, a customer will go elsewhere, and an employer will tell you to go elsewhere.

Since it doesn't seem likely that YOU are going to do this soon, I will do it for you.

I1 is the clockwise mesh current in the left mesh.
I2 is the counterclockwise mesh current in the right mesh.

Next, pay attention to what you write -- the math means what you write, not what you want it to mean.

(1) 4v = I1 x 6k + I2 x 4k x5

means

(1) 4v = (I1 x 6k) + (I2 x 4k x5)

Which is not what you mean. A suitable shorthand for what you mean is

(1) [4v = I1 x 6k + I2 x 4k] x5

And then there's still

(4) - (3) = -12v I1 x -14k

What is it that you have against equal signs? When you say "-12V I1" that can only reasonably mean that -12V is multiplied by I1.

So try writing it like this:

(#1) 4v = 6kΩ(I1) + 4kΩ(I2) // Mesh #1
(#2) 2v = 4kΩ(I1) + 5kΩ(I2) // Mesh #2

(#3) -12V = -14kΩ(I1) // 4(#2) - 5(#1)
(#4) I1 = (-12V)/(-14kΩ) = 0.857mA

A few stylistic notes:

1) I set off the first two equations with a blank line from the rest. The reason is that the first two equations ARE the circuit analysis! They are the rosetta stone for this problem. Everything else is algebra.

2) I put the variables into parens in order to set them off and get rid of the 'x' multiplication signs since these are easily mistaken for variables. Generally, it's better to use '*' for multiplication in text if you need a multiplication sign.

3) I used the '#' in the equation numbers just to make it easier to identify them in the notes off to the side and to make it obvious they aren't just numbers so that 4(2) isn't mistaken as being merely 8.

As for where to go from here. How about solving for the other current?

If you will just be careful with the math, you will see where you made your mistake. Your basic circuit analysis is not the problem, it's sloppy math notation leading to sloppy math.

 

please could you explain the art as it sounds interesting.
also, why the [rant]
if you have an issue then please don't answer any of my posts. unless your paid to. in which case please let me speak to your manager!

That [rant] is for YOUR benefit. The idea is to point out why we are harping on these things that don't seem to be getting across. Remember, you are not only learning to play with some equations to get some numbers, but you are training and being educated to enter a profession -- be it as an engineer or a technician or whatever -- and that it is a profession in which you will affect people's lives. Incompetent doctors kill, but usually one at a time; incompetent engineers kill in job lots.

Also, telling people that are trying to help you for free to cater to your sensitivies or go away is a good way of saying that you really don't want any help at all -- or at least a good way of ending up with the same result.