Basic Circuits Math - Using Substitution and Matrices to Solve Circuits Equations
Summary
TLDRIn 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
📚 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.
🔍 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.
🧩 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.
🛠️ 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.
Mindmap
Keywords
💡Circuit Bread
💡Math Tutorials
💡Ohms
💡KCL (Kirchhoff's Current Law)
💡Substitution Method
💡Matrix Calculator
💡Linear Algebra
💡Voltage (V1, V2)
💡Current (I1, I2, I3, I4)
💡Educational Content
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
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