Help with Kirchoffs Circuit Law - Page 1
Forum section: electronics / beginners

author: acidblue
author_rank: Newbie
timestamp: September 22, 2014, 04:52:43 am
content: I am having a hard time trying to wrap this around my head.I am following this:http://www.electronics-tutorials.ws/dccircuits/dcp_4.htmlI am here about half way down the page when states this:We now have two “Simultaneous Equations” that can be reduced to give us the values of I1 and I2Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 AmpsSubstitution of I2 in terms of I1 gives us the value of I2 as +0.429 AmpsAs : I3 = I1 + I2I cannot for the life of me figure out how it gets theses values, -.143 and +.429What does it mean by "Substitution of I1 in terms of I2"I understand everything written above this part, but this piece i just dont get.

author: kolbep
author_rank: Frequent Contributor
timestamp: September 22, 2014, 05:33:43 am
content: The page you linked to is broken.

author: kolbep
author_rank: Frequent Contributor
timestamp: September 22, 2014, 05:35:44 am
content: http://www.electronics-tutorials.ws/dccircuits/dcp_4.html]http://www.electronics-tutorials.ws/dccircuits/dcp_4.htmlYou lost the l at the end of htm(l).

author: IanB
author_rank: Super Contributor
timestamp: September 22, 2014, 06:18:23 am
content: Quote from: acidblue on September 22, 2014, 04:52:43 amThere are two simultaneous equations. The first is:10 = 50 I1 + 40 I2The second is:20 = 40 I1 + 60 I2There are different ways to solve such a pair of equations, but the most straightforward is not the way they suggest, not substitution.Instead, multiply the first equation by 3/2:15 = 75 I1 + 60 I2Now subtract the second equation from this:(15-20) = (75-40) I1 + (60-60) I2Or-5 = 35 I1GivingI1 = -5/35 = -0.143 AThe value of I2 then follows by substituting back into either of the original equations.

author: Alex30
author_rank: Regular Contributor
timestamp: September 22, 2014, 12:02:31 pm
content: If you have 3 or more equations you can use matrices and find the determinant and then use cramers rule. Alternatively you can just use a graphics calculator or wolfram alpha. Do be careful if you use a graphics though as often if you are dealing with tens of milliamps so you will definitely lose accuracy if your calculator cannot handle lots of decimal places. I would have a go setting up these circuits in LTSpice so you can check your answers

author: acidblue
author_rank: Newbie
timestamp: September 22, 2014, 04:14:15 pm
content: Quote from: IanB on September 22, 2014, 06:18:23 amI don't understand what you mean by "substituting back". Substitute what?

author: zapta
author_rank: Super Contributor
timestamp: September 22, 2014, 05:37:11 pm
content: Quote from: acidblue on September 22, 2014, 04:14:15 pmIf you get one equation to look likeI2 = <some expression with only numbers and I1>Then you can replace each I2 in the second equation with that something on the right and you end up with one equation that has only I1. Then you solve it as you would normally do. Once you find the value of I1, put that number in the right hand of the something above and you will get the value (number) of I2.Edit: this is general algebra, not EE, and you can find many explanation on the net. For example this onehttp://www.wikihow.com/Solve-Simultaneous-Equations-Using-Substitution-Method.

author: klr5205
author_rank: Regular Contributor
timestamp: September 22, 2014, 06:07:59 pm
content: For additional math explanations, I found this website to be incredibly helpful when I was in school:http://patrickjmt.com/The sheer volume of content is staggering, and the videos that I did watch were very clear and informative.

author: acidblue
author_rank: Newbie
timestamp: September 22, 2014, 06:40:55 pm
content: If I understand you correctly i can do :20=40 I1 + 60 I2 to20=40(-.143) + 60 I2Since I1 = -.14320=-5.72 = 60 I2-20 -5.72 = 60 I2-25.72 = 60 I2 -27.52/60I2 = 0.42867rounded up gives 0.429 which is correct.Ok, I got it now.Thanks for the help

author: jlmoon
author_rank: Unknown
timestamp: September 23, 2014, 03:44:57 pm
content: Quote from: acidblue on September 22, 2014, 06:40:55 pmI think you got your signs messed up here. But you're pretty closeSecond eq: 20 = 40(-.142) + 60I2> 20 = -5.68 + 60 I2.. changing sides > 20 + 5.68 = -5.68 +5.68 (cancel) + 60 I2resolves ... 25.68 = 60I2divide to get rid of 60 on right side > 25.68/60 = I2I2 = .428 A

author: Rick Law
author_rank: Super Contributor
timestamp: September 24, 2014, 12:36:54 am
content: Quote from: acidblue on September 22, 2014, 06:40:55 pmCongrats. You got the idea of substitution! The (bold) part that you did is it! The rest is just arithmetic (Like keeping the sign right, it is important too).To understand what you did and how you can repeat your success:In a nut shell, when you have one unknown, you need just one equation. When you have two unknowns you need two equation, so forth.So, the idea is to use one equation to express as:first unknown=(expression ofother constantsandsecond unknown)Then with the second equation, you can replace each occurrence offirst unknownby the stuff on the right of the equal. The stuff on the right equalsfirst unknown, so the equation is still valid. But by doing so you eliminatedfirst unknownfrom the picture. That process is substitution. You substitutedfirst unknownwith something that doesn't contain thefirst unknown.Now your second equation has only thesecond unknownandconstantssince all occurance offirst unknownhave been replaced withsecond unknownandconstants. Juggle it a bit and you can get it to a point:second unknown=(expression of constants only)