Welcome to my video on circuit analysis. In this video we're going to talk about how to
find the current through and the voltage across every resistor in this circuit. Before I get
started let me introduce myself. My name is Chris. I am the creator of all the videos
here at math meeting. And if you need any extra help with your homework send me pictures
to homework solutions at Math meeting. My email is on the right hand side of the screen.
Send pictures and I will get back to you immediately. Or if you want tutoring send me an email to
live tutoring at math meeting. And we can set something up. But let's get started right
away with this example. In this example we need to find the current through and the voltage
across each resistor. To do this our first step is to find an equivalent circuit. So
we need to simplify this circuit so there is an equivalent circuit with only resistor.
So if we want to simplify this circuit with only one resistor we need to use these two
formulas that I wrote for you at the top of the screen. If there is two or more resistors
that are in series then we use this formula on the left side of the screen. And the formula
is the total resistance is equal to R1 + R2 + R3...In other words we just add all the
resistors together. But if two or more resistors are in parallel. Then we use this formula
on the right. And that's equal 1/Rt is equal to 1/R1 + 1/R2 + 1/R3 etc...depending on how
many resistors we have. So we are going to use these two formulas to simplify this circuit.
So there is only one resistor left. What I like to do is start with the resistors that
are the farthest away from the battery source. So we have a battery source over here of 24V.
And the resistors that are the farthest away from this battery source are these two resistors
over on the right hand side of the screen. These two resistors are in series. Because
they are in series we need to add them together. We're going to use this formula here on the
left hand side of the screen. So if we add them together we're going to have 3 ohms plus
12 ohms. 3 + 12 = 15. So we can erase these two resistors on the right hand side of our
circuit. And after we erase these two resistors. We can place them with one equivalent resistor
of 15 ohms. All right so the 12 ohm and the 3 ohm resistor in series made this 15 ohm
resistor. And now once again we want to simplify the resistors that are farthest to the right
of our battery source. So I am going to simplify these two resistors that are in parallel with
each other. Because they are in parallel with each other we have to use this formula on
the right hand side of the screen. To give myself a little more space I am going to erase
these formulas on the top. So once again we are going to simplify these two resistors
that are in parallel with each other. So we're going to use our parallel resistor formula.
And the equivalent resistance or the total resistance for two resistors that are in parallel
is equal to 1/R1. (R1 is equal to 10 ohms.) Pluse 1/R2. (R2 is equal to 15 ohms.) So we
have 1/10 plus 1/15 which is equal to 5/30. 5/30 simplifies to 1/6. So 1/Rt is equal to
1/6. So we know that the total resistance is equal to 6 ohms. So I can erase both of
these resistors on the right hand side of our circuit and replace them one equivalent
resistor of 6 ohms. Ok so we're almost finished simplifying this so there is only one resistor
left. The only thing we have to do now is simplify these three circuits that are in
series with each other. All three of these circuits are in series with each other. So
the only thing we need to do is add them together to get our equivalent or total resistance.
So the total resistance for three resistors that are in series is R1 plus R2 plus R3.
So we have 2 + 6 which is equal to 8. And 8 + 4 is equal to 12. So if we add all of
the resistors together we have a total or equivalent resistance of 12 ohms. So I am
going to erase all of these resistors that are in series and replace them with one equivalent
resistor of 12 ohms. So know that we have simplified this circuit so there is only one
resistor. Now we can find the total current running through the circuit. And we can use
the formula the V = I * R. So the voltage we know is 24 volts. and the resistance. The
equivalent or total resistance is 12 ohms. So we can plug that in. And we know that the
current is going to be equal to the voltage which is 24V divided by the resistance which
is 12 ohms. 24/12 is equal to 2. So the current is going to be equal to 2 amps. So we know
that the total current running through the circuit is 2 amps. So we know that the current
leaving the battery is equal to 2 amps. And we can use this information to find the current
through and the voltage across every resistor in this entire circuit. So what I like to
do keep going back to each diagram that we simplified. So if we go back one more diagram
that we simplified the circuit is going to look something like this. So we know once
again the total current is 2 amps. So the current leaving the battery is 2 amps. And
we know from my last video that the current never changes for any resistor that's in series.
So when this current goes through this resistor that's in series we know it's not going to
change. So we know that the current going through this first resistor is also equal
to 2 amps. So the current I1 is going to be equal to 2 amps. And we also know that the
current going through the second resistor which is also in series isn't going to change.
So the current going through this second resistor I2 is also equal to 2 amps. And I think you
get the idea by now. This current going through this third resistor is not going to change
because it's in series. It's also going to be equal to 2 amps. So now that we found the
current through each resistor now let's find the voltage across each resistor. And we can
use the same formula as before. To find the voltage across resistor number 1. The voltage
across resistor number is equal to the current of resistor number 1 times the resistance.
So we know that the voltage across 1 is equal to the current through 1 which is 2 amps.
So we have 2 amps for the current. We know that the resistance is equal to 2 ohms. So
we got to multiply it by 2 ohms. 2 times 2 is equal to 4. So the voltage across resistor
number 1 is equal to 4 volts. So now let's find the voltage across resistor number 2.
The voltage across 2 is going to be equal the current of 2 amps. So we have a current
of 2 amps. Multiplied by the resistance of 6 ohms. Multiplied by 6 ohms. 2 times 6 is
equal to 12. So the voltage across the second resistor is going to be equal to 12 volts.
Now let's do the same exact thing for our third resistor. We know that the voltage across
the third resistor is going to be the current which is 2 amps. So we have 2 amps being multiplied
by the resistance. We know the resistance is equal to 4 ohms. So 2 times 4 ohms is equal
to 8. So the voltage across the last resistor (resistor number 3). The voltage across resistor
number 3 is going to be equal to 8 volts. So now that we found current through and the
voltage across every one of these resistors let's go back one more diagram when we were
simplifying. Let's recap what we already know. We know that the current going through resistor
number 1 is equal to 2 amps. We know that the voltage across resistor number 1 is equal
to 4 volts. And we also know that the current going through resistor number 3 is equal to
2 amps. And we know the voltage across resistor number 3 is equal to 8 volts. All of this
we already know from the work we did before. So now let's try and solve for these two resistors
on the right side of the circuit. So how do we do this? Well we know the voltage leaving
the battery is equal 24 volts. And we know once we go across this resistor we have a
4 volt drop. So 24 - 4 is equal to 20. So the voltage after the first resistor is equal
to 20 volts. Which means that the voltage over here is also equal to 20 volts. Because
we haven't crossed any resistors. And we can do the same thing for the end of the circuit.
We know the voltage at the end of the circuit is equal to 0 volts because the potential
always goes to 0. And we know there is an 8 volt drop before we crossed the last resistor.
so 0 + 8 is equal to 8. So we know that we have a voltage of 8 volts before we crossed
the third resistor. So we know the voltage over here is also equal to 8 volts because
we have not crossed any resistor. If you don't cross any resistor the voltage is going to
stay the same. Now look what we have. We know before we crossed this resistor we have a
voltage of 20 volts. And then once we cross that resistor it goes from 20 to 8. So we
know 20 to 8 is a difference of 12. So we know the voltage across this resistor is equal
to 12. We'll call this resistor 4. The voltage across resistor 4 is equal to 12 volts. And
the same thing is going to happen for the resistor on the right. we know that we are
starting with 20 volts. Once we cross the resistor it goes to 8 volts. 20 - 8 is equal
to 12. We'll call this resistor number 5. So the voltage across resistor 5 is going
to be equal to 12 volts as well. So now let's find the current going through these resistors.
We can use the same formula. If we rearrange the formula we know the current going through
resistor number 4 is going to be equal to the voltage across 4 divided by resistance
of 4. We know the voltage across 4 is equal to 12 volts. So we have 12 volts being divided
by the resistance which we know is 10 ohms. So 12 divided by 10 ohms. 12/10 is equal to
1.2. So we know the current going through resistor 4 is equal to 1.2 amps. And now let's
do the same thing for resistor number 5. Once again the current through resistor number
5 is equal to the voltage across 5 divided by the resistance of 5. And we know the voltage
across 5 is equal to 12 volts. So we have 12 volts being divided by the resistance of
5 which we know is equal to 15. So we have 15 ohms as the resistance. 12 divided by 15
is equal to .8. So we know the current through resistor number 5 is equal to .8 amps. And
this totally makes sense because we started with a current of 2 maps. And when it splits
off into these two parallel resistors we have one current of 1.2 and one current of .8.
(1.2 plus .8 is equal to 2.) So the total current has not changed. It splits and some
of it goes down the left side and some of it goes down the right side. So now let;s
go back one more diagram so we can completely solve the original circuit. So now the only
thing we have left are these two resistors. So once we solve those two resistors we have
completely solved our entire circuit. So we know the current going through the right side
of these resistors in parallel is equal to .8 amps. So we'll call this resistor number
6. So the current through resistor 6 is equal .8 amps. And once again these resistors are
in series with each other. So the current doesn't change. So we'll call this resistor
number 7. So the current through resistor number 7 is also equal to .8 amps. So now
the only thing we have left to do is use our formula. So we know the voltage across resistor
number 6 is going to be equal to to the current through 6 multiplied by the resistance of
resistor number 6. So this is going to be equal to the current which is .8 amps. So
the current is equal to .8 amps multiplied by the resistance. The resistance we know
is 3 ohms. 0.8 time 3 is equal to 2.4. So we know the voltage across resistor 6 is equal
to 2.4 volts. So now let's move on to resistor number 7. We know the voltage across resistor
7 is equal to the current of 7 multiplied by the resistance of 7. This is equal to the
current which is 0.8. So we have 0.8 amps multiplied by the resistance. The resistance
we know is 12 ohms. Multiply by 12 ohms. 0.8 times 12 is equal to 9.6. So we know the voltage
across resistor number 7 is equal to 9.6 volts which makes sense because 9.6 + 2.4 is equal
to 12. So the voltage these two resistors are exactly the same as the voltage across
this resistor. And now we have officially found the current through and the voltage
across every single resistor in this original circuit. So I hope this gave you a better
idea on circuit analysis and how to solve circuits. Check my next video if you want
to keep on learning. Thank you so much for watching. Don't forget to subscribe. I really
appreciate all of you guys who are supporting me making these videos. I will see you in
my next one.