Equation

Hi Alan yeah i see what you mean about a bug in the calculator. I was practicing finding matrix inverses today and put in a matrix from the book i'm reading. Put in all 0's on the B matrix and i got the same answer as the book but two of the numbers were signed as being - instead of +. I don't know why that is apart from there's something wrong there.

Anyway when i corrected the signs and took the negatives off and multiplied with the constants matrix i got the correct answer.

It's all fine but the problem would be if you didn't recognize that there was an issue there. You'd be getting the answers all wrong probably and racking your brain to work out why!!

 Gary

Thanks Alan for taking the time to write all that out. It pretty much explains it all for me!!

I think there is an error in one of the equations there in the book. I think he's made one negative when it shouldn't be or switched something around by mistake. It happened before with the other problem you helped me solve where he had reversed two figures and put them in the wrong sequence.

Easy done when dealing with all that data.

Thanks for all the help i appreciate it.

 Gary

Tried it out but not getting the answer from the book. Answer to the problem for the three unknowns is 326, 34, 53. All ma measurements. 0.326, 0.034, 0.053 Amps converted from the ma's. That's the results on page 61 of 'Electronics A Systems Approach'

Probably error on my part, doing something wrong! But the figures i'm getting are nowhere near the answers.

Thanks for the tips Alan. For the 3 x 1 vector i've never worked with those yet or studied anything about them so i didn't know i could use that option for doing the final multiplication of the inverse and constant matrices. I thought i always had to have matrix selected for both sides!!

I'll look into the SMath tip and look it up via google as it sounds like what i'm looking for.

Thanks again for all your help.

Cheers Alan. I just noticed you can't do a 3 x 1 for the constants matrix. Unless i'm missing something (which is very possible!) 2 is the lowest i see in the options.

Not too much of a problem multiplying the inverse with the constants and adding everything up to get the answers just by hand. But it would be nice to just have that constant matrix with the 3 rows 1 column option and multiply it straight with the SC.

What do you think is the best route for calculating these systems of equations say if you have more than 3 equations? There are so many options, sometimes as a relative novice you're in a bit of a quandary as to what to use.

I was reading about Cramer's rule but the Algebra book i'm looking at says that get's more involved and is a bit more work. I think it looks quite a good way to calculate though.

For me i think after you know the steps to take to set up the equations in matrix form, using a calculator like the SC would be best as it saves you a major headache!!!

Thanks CPhill for taking the time out to try to help. Thanks for the link as well, i'll check it out.

Alan: You came through for me again mate! I didn't know if it was possible to do it through the SC but i had a hunch it was but just didn't have a clue what to do or how to go about it!

You've helped me so much with that answer.

The book we talked about before 'Electronics: A Systems Approach' page 62. The coefficient matrix on the left there is the one i'm talking about calculating. From the Algebra book i'm studying it says get the inverse matrix and then multiply it with the constants matrix to get the solution to the systems of equations problem but using a matrix instead of doing it by hand.

Thanks Alan you're a great help. 

 Gary

Ha! You must have read my mind Alan! That's exactly what happened at the end bit where he talks about

getting the results through multiplying the different matrices.

I tried to muck around with it using the online calculator but not really knowing what i was doing i didn't get any

results!!  It said something about 'No inverse' in red warning letters and i was obviously doing it all wrong. Too

advanced for me at this point so i'll stick to just doing the equations.

Yeah the bit in the beginning about Kirchoff's law confused me as well because he was talking about the current

flowing into and out of the node which is what the law is about so i kinda scratched my head at that point.

Thanks for the link i'll be sure to study up on it.

Gary

Hi Alan,

Yeah i'm wanting to expand my knowledge enough that i can get around these calculations for circuits for working out voltages and currents. 

I also need to learn how to work with matrices. The problem like we discussed on here i think may be able to be solved by a matrix. Maybe your way is the best though for the type of equation we've talked about. The matrices seem like a lot more work. Maybe that's just because i don't know about them yet though.

I'm going to include a link to a video i've been watching on YT on nodal analysis. The guy get's two equations, re-arranges them and then instead of solving them by elimination or substitution he puts them through a matrix and gets the result.

Again i'm a total beginner again so don't have a clue whether that's the best way to approach these things or not. I'm just at the flailing around in the dark stage with anything complicated!

Have a look and see what you think. I know Neil Storey's book go's into matrices a bit further in than where i am at the moment. That's a bit complex for me at the moment though.

Alan you're the best! That answers my questions, as i've forgotten so much about maths!

I bought one of those Algebra for dummies books to go right back to basics again and do the remedial work i need.

Really i only need to know enough to be able to do the math for circuits like the ones in the book we have by

Neil Storey and stuff related to that.

I got back into doing this because i'm a guitar player and love working on amps replacing capacitors and

resistors and the likes and stuff so am always studying amp schematics. Most of those things can

be worked out by using Ohm's Law which is pretty simple. But when you get to more advanced

problems/circuits like

we're discussing then that's where my lack of maths knowledge really gets me!

I have two amp textbooks by a guy called Merlin Blencowe. He has a Preamp one and a Power Amp one. In

those he also has quite a lot of maths in places. Some very easy involving Ohm's law and it's variants. But

others quite complex. So that has been another reason for me to try to learn more about all this. It becomes

very interesting to me after a while.

You'd understand everything that Merlin writes in the books regarding equations. But some of it i struggle

with. I learnt a lot of good stuff from chapter 2 in Neil Storey's book about phase angles and sine waves that

helped me deciphering stuff from Merlin's books 

Thanks very much Alan this has helped me a lot.

 Gary

Trying to follow all the steps Alan you wrote out in the second answer you gave. Forgetting about the mistake

in the book in the printing in the book for a second.

In the first rearrangement of the first equation: where are you getting the 5 from? are you just dividing the 50

 by the 10 in the denominator? I've been off maths so long and i'm so rusty! I thought that in that case that

would

cancel out the denominator and you couldn't use it further on.

Same with the bit where  we have 5 - 13/60 where is the 13/60 coming from?

Man i wish i was good at Maths!

Thanks Alan.

I'll send him a message saying he's got a mistake in the book for future editions and see what he says.

You're right sometimes there are mistakes when so much data is being put on paper.

I'm principally interested in this because i'm into amplifiers and there's a lot of calculations for gain stages,

rectification and all the rest of it. Anyway in reading a couple of my textbooks i came across some sums that i

looked at from every angle and they just didn't make sense! So i contacted the author and he said i was right

that the book had a typo in there! So it might very well be that.

Thanks for helping me.

Thanks Alan yeah the way you've written them at the top is the way the problem appears in the book.

This is from a book called 'Electronics: A Systems Approach' by a guy called Neil Storey who works at Warwick University here in the UK. I emailed him to help me, but he's on holiday!!

I don't know why the numbers don't match.

Thanks for taking the time to write all that out! Appreciate it.

Thanks Alan, Confession time i made a mistake in what is posted. The first equation at the end should also have

0 - V2/15. Those denominators are all resistors in the circuit.

The actual answers in the book are V2 = 32.34v  V3 = 40.14v though in my novice state with combining equations i can't get those answers.

The first way you wrote out the second equation is exactly the way the author has written it in the book. The second one is out because i forgot to include that part of the equation in my rush to post it!

Thanks for replying.