Basic Circuits Math - Using Substitution and Matrices to Solve Circuits Equations

Summary

In this video, the presenter discusses two methods for solving circuit math problems: the traditional substitution method and a matrix calculation approach using a linear equations calculator. They share their personal struggles with math and demonstrate how to apply Kirchhoff's Current Law (KCL) to set up equations for a sample circuit with resistors and voltages. The presenter simplifies the equations step by step, showing the process of substitution, and then contrasts it with the matrix method, which is faster and less error-prone. They emphasize the importance of having an intuitive understanding of the circuit's behavior, regardless of the method used.

Takeaways

• 😀 The speaker initially avoids math tutorials due to a lack of confidence in their math skills but decides to discuss circuit math.
• 🔍 The video focuses on two methods for solving circuit equations: the substitution method and the matrix method.
• 📚 The substitution method is demonstrated first, showing a step-by-step approach to solving for voltages in a circuit without a calculator.
• 🧩 The matrix method is introduced as a more efficient way to solve circuit equations, especially when using tools like a linear equations calculator.
• 🤔 The speaker emphasizes the importance of understanding the intuitive feel of the math involved in circuits, ensuring the results make sense.
• 🔢 The substitution method involves simplifying equations and substituting values to isolate variables, which can be prone to errors.
• 📉 The matrix method simplifies the process by inputting the equations into a calculator, reducing the chance of manual calculation errors.
• 🛠️ The video provides an example circuit with resistors and voltages to illustrate the math involved in solving for unknowns.
• 📝 The speaker discusses the process of using Kirchhoff's Current Law (KCL) to set up the equations for the circuit.
• 📉 The matrix calculator on circuitbred.com is recommended for those who prefer a tool-assisted approach to solving circuit equations.
• 💡 The video concludes by encouraging viewers to practice both methods, understanding the foundational concepts and the practical application of circuit math.

Q & A

What was the initial stance of the speaker on creating math tutorials for the Circuit Bread channel?

The speaker initially stated that they would not create math tutorials for the Circuit Bread channel due to their self-acknowledged lack of strength in math and the embarrassment of making mistakes on camera.

What is KCL in the context of the script?

KCL stands for Kirchhoff's Current Law, which is a fundamental principle used in circuit analysis that states the sum of currents entering a junction or node is equal to the sum of currents leaving the node.

What is the purpose of the example circuit provided by the speaker?

The example circuit is used to illustrate the process of solving circuit equations using two different methods: substitution and matrix calculation, to help viewers understand how to deal with the math that comes from circuit problems.

How many resistors are in the example circuit, and what are their values?

There are four resistors in the example circuit with values of 100 ohms, 200 ohms, 300 ohms, and 400 ohms, labeled as R1, R2, R3, and R4 respectively.

What are the two methods discussed by the speaker for solving circuit equations?

The two methods discussed are the substitution method and the matrix calculation method, with the latter involving the use of a linear equations calculator.

Why might the substitution method be preferred in some educational settings?

The substitution method might be preferred in educational settings because it provides a hands-on approach to understanding the process of solving equations, which is often a requirement in school curriculums.

What is the main advantage of using a matrix calculator for solving circuit equations according to the speaker?

The main advantage of using a matrix calculator is that it is faster, less prone to manual calculation errors, and can handle more complex equations with multiple unknowns more efficiently.

What is the speaker's recommendation for dealing with complex algebraic errors during manual calculations?

The speaker recommends taking one's time, being careful, and learning to double-check work to avoid algebraic errors during manual calculations.

What is the significance of having an intuitive feel for the circuit equations being solved?

Having an intuitive feel for the circuit equations is important to ensure that the results make sense in the context of the circuit's behavior and to quickly identify any potential errors in the calculations.

How can viewers get help or ask questions about the content of the video?

Viewers can ask questions or seek help by leaving comments on the YouTube video, visiting the CircuitBred.com website, or joining the new Discord channel mentioned by the speaker.

What is the speaker's final encouragement for viewers who found the tutorial helpful?

The speaker encourages viewers who found the tutorial helpful to like the video, subscribe to the channel, and share it with friends.

Outlines

00:00

📚 Introduction to Circuit Math Tutorial

The speaker begins by admitting their initial reluctance to create math tutorials due to past online mockery and self-acknowledged weaknesses in mathematics. Despite this, they decide to tackle circuit math, explaining that while they are comfortable with math, they tend to make simple mistakes. The video aims to discuss common math problems encountered in circuit analysis, focusing on two primary methods for solving them: substitution and using a linear equations calculator. The speaker sets up an example circuit with a 10-volt source and various resistors, labeling currents and nodes to demonstrate Kirchhoff's Current Law (KCL) and establish equations for solving the circuit.

05:00

🔍 The Substitution Method for Circuit Math

The speaker delves into the substitution method for solving circuit equations, starting with simplifying the equations by isolating variables. They multiply both sides of the equation by common denominators to consolidate terms involving the same variable. The process involves moving terms from one side of the equation to the other, aiming to express one variable in terms of another. The speaker emphasizes the importance of taking one's time to avoid common algebraic errors, acknowledging their own struggles with this aspect of math. They guide the viewer through the steps of substitution, highlighting the complexity and potential for mistakes, especially when dealing with multiple equations and variables.

10:02

🧩 Transitioning to Matrix Calculations for Circuit Analysis

After demonstrating the substitution method, the speaker acknowledges its complexity and potential for error, especially when not using a calculator. They introduce the second method, which involves using a matrix calculator to solve the circuit equations. The process involves setting up the equations in a matrix form, with coefficients and constants properly aligned, and then inputting these into a calculator to find the solution. The speaker simplifies the equations before inputting them into the calculator, emphasizing the importance of being consistent and accurate in this step. They show how this method is faster and less prone to manual errors, making it a preferable choice when tools are available.

15:03

🛠️ Conclusion and Encouragement for Intuitive Understanding

In the final paragraph, the speaker concludes the tutorial by reflecting on the two methods presented: the traditional substitution method and the modern matrix calculation method. They stress the importance of understanding the intuitive feel of the math involved in circuit analysis, regardless of the method used. The speaker encourages viewers to ask questions if anything was unclear and invites them to engage with the community through comments, the website, or the new Discord channel. They end with a call to action for likes, subscriptions, and shares, expressing hope that the tutorial was helpful and informative for the viewers.

Keywords

💡Circuit Bread

Circuit Bread is a term used metaphorically in the script to refer to a journey or venture into the world of electronics and circuits. It is not a literal bread but a channel or platform where the speaker discusses various aspects of circuitry. The term is used to set the context for the video's theme, which revolves around the educational content related to electrical circuits.

💡Math Tutorials

Math Tutorials refer to instructional content designed to teach mathematical concepts and problem-solving techniques. In the script, the speaker mentions an initial reluctance to create math tutorials due to personal challenges with math, which adds a layer of humility and relatability to the narrative. The video, however, goes on to discuss methods for solving mathematical problems that arise in circuit analysis.

💡Ohms

Ohms is a unit of electrical resistance. In the context of the video, resistors with specific ohm values are part of a circuit that the speaker is analyzing. The script mentions 100 ohms, 200 ohms, 300 ohms, and 400 ohms resistors, which are integral to the circuit's function and the mathematical equations derived from it.

💡KCL (Kirchhoff's Current Law)

Kirchhoff's Current Law (KCL) is a fundamental principle in electrical circuit analysis which states that the total current entering a junction or node is equal to the total current leaving it. The script uses KCL to derive equations for the currents at different nodes in the circuit, which is a key step in solving for unknowns in the circuit's behavior.

💡Substitution Method

The Substitution Method is a technique used in algebra to solve systems of equations by expressing one variable in terms of another and then substituting this expression into the other equations. The script describes using this method to solve for voltages (v1 and v2) in the circuit, which is portrayed as a more traditional and sometimes cumbersome approach.

💡Matrix Calculator

A Matrix Calculator is a tool used to perform various mathematical operations on matrices, including solving systems of linear equations. The script contrasts the manual substitution method with using a matrix calculator, highlighting it as a more efficient and less error-prone method for solving circuit equations when tools are available.

💡Linear Algebra

Linear Algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. In the script, linear algebra is implicitly used when the speaker discusses solving the circuit equations using a matrix calculator, which is a practical application of this mathematical field.

💡Voltage (V1, V2)

Voltage, denoted as V1 and V2 in the script, is the electrical potential difference between two points in a circuit. The speaker is trying to solve for these voltages across different components in the circuit using mathematical equations derived from KCL and Ohm's Law, which are central to understanding the circuit's operation.

💡Current (I1, I2, I3, I4)

Current, represented by I1, I2, I3, and I4 in the script, is the flow of electric charge in a circuit. The speaker uses Kirchhoff's Current Law to establish relationships between these currents at different nodes, which is essential for setting up the equations that are solved to analyze the circuit.

💡Educational Content

Educational Content refers to material that is intended to teach or instruct, often on a specific subject. The script is part of a broader category of educational content aimed at helping viewers understand the mathematical aspects of circuit analysis, making complex concepts more accessible.

Highlights

Creator initially avoided math tutorials due to self-acknowledged weaknesses in math.

Introduction of two best methods for handling circuit math.

Explanation of how to deal with math that comes from circuits without a calculator.

Demonstration of using a linear equations calculator on circuitbred.com for circuit math.

The complexity of circuit equations increases with more unknowns and equations.

Detailed walkthrough of setting up a simple circuit example with resistors and currents.

Application of Kirchhoff's Current Law (KCL) to derive equations for the circuit.

Simplification of equations to isolate variables for substitution method.

Challenges and common mistakes in manual algebraic manipulation.

Advantages of using a matrix calculator for solving circuit equations.

Step-by-step guide on how to input equations into a matrix calculator.

Comparison of the substitution method and matrix calculation method for solving circuit equations.

Importance of understanding the intuitive feel of circuit math for error checking.

Acknowledgment of the difficulty of manual math and the benefits of using tools.

Recommendation to practice both methods for a comprehensive understanding of circuit math.

Invitation for viewers to ask questions and engage with the community for further clarification.

Encouragement to like, subscribe, and share the tutorial for others to benefit.

Transcripts

[Music]

when we started the circuit bread

channel i

very explicitly said that i wasn't going

to do any math tutorials when we were

talking about it because

i as i have been mocked on the internet

by some i am not great at math it's not

my strength i'm i'm okay

at it i'm comfortable with it i enjoy it

but i make dumb mistakes and i

am very embarrassed to do math on camera

so

i am breaking that rule because we are

talking about

the way that you deal with the math that

often comes from your circuits one

problems so a lot of the times when you

solve for a circuit your

your equations end up in a somewhat

similar form they are basically

the same in their component parts but

then you just have different values and

different levels of complexity

so i wanted to talk about how i have

found

the for me the two best ways of doing

basically

circuit math and kind of going through

the steps and showing one

the what i find to be the much harder

yet

much more common way of doing things

because it's something that you can do

without a calculator

and then also i want to show you the way

to do it

when you have access to tools like

our linear equations calculator on the

circuitbred.com website

so with that i set this example up which

i gotta

admit gave me a greater appreciation for

people that write textbooks and teachers

that come up with example problems

because

i struggle coming up with good problems

that aren't just the same things over

and over again

so i wanted to create this circuit and

i just want to state that if you are

solving for i

in any of this it's pretty obvious so

don't

don't think about it too deeply like the

actual circuit itself we're more focused

on the math

but in this case i have basically a

simple circuit with 10 volts right here

and then i have a resistor 100 ohms 200

ohms 300 ohms

400 ohms and that's r1 r2 r3 r4

and that's i1 going through that i2 i3

i4 just again

for me it makes sense to make the eyes

match the r's

so r1 is got the current i1 going

through it

if possible and then finally i've

identified this node as v1 and

this node as v2 and then i just wrote

those out because i got a little bit

tiny and i'm not sure how well you can

see it

and then going from there i said using

kcl

v1 i have i1 minus i2 minus i3 equals 0

because i1 is going in

i2 is leaving i3 is leaving and then on

v2

i have i2 plus i3 minus i4 equals 0

right there and if i'm going too fast

that probably means you just need to

spend a little bit more time with kcl

but again all i'm doing here is

identifying the currents in and out of

those two nodes that i called v1 and v2

and i probably should have called n1 and

n2 so

sorry about that so now

using kcl we continue on and again this

is all just setting up for the math so

give me a minute i1 we've established

that that is

10 minus v1 over r1 i2 is the same thing

v1 minus v2 over r2 and so on so i've

written out what i1 i2 i3 i4

are and then i've replaced those down

here so v1 equals

10 minus v1 over 100 minus v1 minus v2

over 200

minus i3 which is v1 minus v2 over 300

equals zero

so all of that setup was just to get to

it's just i wanted to give you some

context

but that's just to get to these two

equations

so now we have two unknowns and two

equations so now i'm going to take it

down not

slow it down

okay so we have two equations

and two unknowns now this is where i

talked about

this is a very similar setup you see in

circuits and the complexity goes up

because sometimes you get

three equations and three unknowns four

equations and four unknowns

i've never seen above four frankly but

i'm sure you can

and even at four it just turns into a

huge mess

so if you have four bless you

good luck but right now we'd like to

focus on these two equations

and focus on first the substitution

method

of solving for v1 and v2 in these

and then we will go through the method

of

using a matrix calculator to solve for

it and kind of go over the pros and cons

of the two

and also um just exactly how you do it

so let

us take this first equation and for

substitution

we take this first equation and we get

it

so basically we get either v1 or v2

depending on what we want to do

over to the same side so using this

first one

let's start by simplifying this i'm just

going to multiply both sides by 100

so i'm now going to have 10 minus v1

minus v1 over 2

plus v2 over 2

minus v1 over 3

plus v2 over three

equals zero and this is this is where i

make those

math mistakes calculus whatever uh the

complex algebra whatever

it's me accidentally putting pluses

where i shouldn't and minuses where i

shouldn't that's my

weakness and if that's your weakness

take your time

learn to take your time so you don't be

like me

okay and then we can take this and we

can simplify this so that we have all

the

v1s and all the v2s together so i'm

basically going to just do

10 um that's going to come out to be

plus oh

common denominator what is that 5

v two over six

um and then let's move the v one over to

the other side

to make that positive and that is going

to be so v1

another common denominator is going to

be

6 again so i've got 11

v1 over 6.

and this is kind of the fun thing about

randomly throwing in numbers you

sometimes get some crazy math

where your numbers are getting out of

control okay and then from here i'm just

going to

multiply both sides by 6 11 to get that

to be v1 all by itself

so that's just going to be 6 over 11

times 10 plus 5 v2

over 6 equals

v1 okay so again stick with me this is

the first step

with substitution so what we have done

here now is we've taken that first

equation and we have pulled out

where uh pulled out everything so we

just have everything equals

one variable now we

keep this in our minds we kind of let's

say hey

we're going to need you in just a second

and let's go to another piece of paper

where we will do the second equation

okay so this is the second equation

again

this is what we got right here

and so now we're basically going to do

the same simplification on this

second equation that we did on the first

one so i'm going to multiply everything

by

so you get v1 minus

v2 plus 2v1

over 3 minus 2v2

over 3. then

minus v2 over 2

equals 0.

okay and then we now need to well i mean

we could even at this point

do the substitution but it would be more

complicated

and just the whole point of this is

now that we know what v1 equals we can

take

v1 and stick it in here

and then now we only have v2 because

if we take this and stick it in there we

no longer have v1 and all it is is this

and then it all comes out but

putting this in every place that we see

v1 here which in this case is actually

only twice

it's more complicated than it needs to

be so let's simplify this

before we do that replacement so

let's see v1 minus oh v1 plus

two-thirds v1 is going to give us

five-thirds v1

5v1 over 3.

um and then let's throw v2 onto the

other side just for the fun of it

and let's see so it's v2 two-thirds and

then one-half

and that is going to give us 13 halves

v2 did i do that right nope six six

i skipped a step in my brain and again i

don't know how many times i have to say

this

don't be me be good at math okay so now

to further simplify this multiply both

sides by three so now we just have five

v one equals thirteen v

2 over 2. well i mean we can even

simplify that farther and just get

v1 equals 13 v2

over 10. okay so now

we take this and we replace

oops replace this v1 with that

and we have let's see 6

over 11 times 10 plus 5

v2 over 6

equals 13

v 2 over 10.

okay i think right now if you're still

with me

congratulations you're probably saying

okay this is getting complicated i feel

like we've got

multiple pieces of paper here

everything's kind of going out of crate

going crazy and that's one of the

challenges with this method

you see i haven't had to use a

calculator yet i i think i'm going to

switch over to decimal pretty quick

because it does

the fractions are going to get crazy but

this is this is

how come this version

this method is a bit of a struggle is

because there's

just a lot of simple math that you can

get messed up so

be very careful take your time because

if you're on a test you're most likely

going to have to do

the math like this and again good luck

if that's the case

but let's let's take this and we'll just

turn that into

60 11

plus okay how does that 30 60

30 30

over 66 v2

okay i'm just gonna get this into

decimal really quick so let me put this

uh let's get this into decimal

um oh man once this all simplifies out

if i multiply

both sides by 10 and divide by 13

i basically get this turns into

four point two plus

oh 0.35

v2 equals v2

which then gives us 4.2

equals 0.65 v2

which then gives us

6.46 equals v2

okay that was it

now that we know what v2 is we can go

back into the original equation and plug

that number in and we'll be able to

solve for

v1 but even with me writing

big on these that was that was a bit

cumbersome

again very simple math but cumbersome

and this is with only

two equations and two unknowns so that

is a substitution method that is where

you

solve for one in in respect to the other

one

and then go back substitute it and then

you only have one equation

and you actually can figure out what

what you need to do

now that this is a process

that it's really good to practice it

it's good to know because

this helps kind of give a nuts and bolts

feel for what's going on

but if you're ever in a situation where

you have access to a computer or have

access to tools

yet not necessarily to a spice program i

would highly recommend

doing this second method which basically

just uses

linear algebra so let me see let's go

now we're on to the second method so

first method substitution that was

exciting

i never want to do that again so second

method

you basically can just look at the first

method and

not go quite as far before punching it

into the calculator okay

so the second method the the way you do

it is you basically just have to

match the the v1 and the v2

and then the actual number that you have

and

put it into a matrix properly so what i

mean by that is

let's go back this is solving for this

was the first equation that we had

and we can look right here we can take

that

and move it around a little bit so for

the matrix calculation we want again it

makes sense and it doesn't matter as

long as you are consistent

but it makes sense to put this as

negative

11v1 over six

plus five v2 over six

equals negative ten

so that is just negative eleven points

11 over 6 times v1

plus 5 6 times

v2 and then that is equal to negative

10.

so that's that's important to note

because now we are going to the second

one and we are our second equation

which we figured out um we just

can go and look right here making sure

that this is still visible

we can take this and put it up here and

we can say

5 v 1 over

3 minus 13

v 2 over 6 equals

0. and now we have done all of the

manual math that we need because we have

v1 v1

v2 v2 and then our number right there so

now we actually jump to our matrix

calculator

which you're going to see this is super

fast we are almost done

hallelujah so now we are on this linear

equations calculator and again if you

had

more um more unknowns and more equations

you can just go in and change it again

if you get over five

good luck but in this case for a

we will look and see oh it's 11 negative

11

divided by six then we have

five divided by six then we've got

negative 10

and then this one we've got 5 thirds

and then negative 13 over 6

and then 0. and

there we go 6.451 6.46 so obviously i've

now made some sort of rounding error

doing all of this that's not too

surprising

but instead of having to do all these

calculations and putting in the

substitution

and basically just having a heck of a

time i was able to

get it to this point which was really

quite straightforward

and then put it into the calculator and

have it do it for me and

not only is it faster it's less likely

to have any mistakes because

you aren't doing as much manual math and

again if you're a mathematician and this

stuff is easy for you why are you

watching this video

i don't know but if you're like me and

math is you know

okay but not the best thing in the world

for you

this is very helpful now the reason we

went over the substitution is again

in most classes your teacher's going to

make you do it

by hand and maybe you'll have like a ti

ti-83 if you're lucky maybe a ti-89

which you can do

the matrix calculation on a ti-89 is

just a little bit more complicated

i mean i'm not going to get into that

but

these are the two methods the painful

yet good to learn and good to know for

school

and then the not quite so painful faster

to do

method if you just need to figure

something out really quick

now on both of these it goes back to

what i think i

i at least i hope i always stress on any

of this is having an intuitive feel

as you're looking at this um both of

these

do these make sense uh is it makes sense

that

the first voltage here v1 is a higher

voltage than here yes does it make

sense that if they're both less than 10

yes does it make sense that

this is actually still over half of the

total voltage yes because there's

400 ohms there so to have the same

current going through here as going

through all of these you need a pretty

high voltage here so

all of this stuff makes sense no matter

what

tool you are using it is super important

that you

understand intuitively what's going on

and what makes sense because if you've

gotten v1 as 13 volts and v2 is negative

73

that should have sent a red flag like

that's not at all what it should be

but that's it okay that was um i i fear

maybe why i should never do math

tutorials that was kind of painful

painful for me hopefully it wasn't

painful for you

i really hope that helped you out i

really hope that lays a foundation for

you to understand

how you can do all the circuits math

using both substitution and if you have

the ability

doing it on the linear equations

calculator if you have any questions

please

drop it in the comments below on youtube

go to circuitbred.com put it there

you can go to our new discord channel

and i don't know when you see this maybe

it won't be a new discord channel by

then

and ask any questions to see if we can

help clarify anything that i wasn't as

clear as i'd like to be in this video

if this was useful and this was helpful

give it a like subscribe to our channel

share with your friends all that good

jazz thank you very much and we will see

you in the next one

you