[music]
>> On this video,
we're going to take
a Closer Look at Circuits.
Any time moving charges
find a closed loop to travel in,
that's a circuit.
There are three basic types
of circuits.
Series, parallel and complex,
which is a combination
of series and parallel.
Exercises with these
three types usually involves
solving for the voltage,
current or resistance
in the circuit.
To organize
all the information involved
in complicated circuits,
it's helpful to use a chart
called a VIR chart.
A VIR chart is a table
with blanks for the voltage,
current and resistance
in our circuit.
We use Ohm's Law
and our knowledge of
how circuits work to fill in
the blanks.
Once all the blanks
are filled in,
we know everything there is
to know about voltage,
current and resistance
in our circuit.
Before we get practice using
VIR charts to solve circuits,
let's remember some of
the rules about series
and parallel circuits.
In a series circuit,
the total current in the circuit
is the same
across every resistor.
And the total voltage
across the circuit equals
the sum of the voltages
across each resistor.
Also in a series circuit,
the total resistance equals
the sum of the individual
resistors.
In a parallel circuit,
the voltage in each branch
of the circuit is the same.
While the total current
equals the sum of the current
in the branches.
Resistors add inversely
so that one over the total
resistance equals the sum
of the inverse
of the individual resistors.
This VIR chart is for a circuit
with three resistors.
We're calling them
R1, R2 and R3.
We want to fill in what
the voltage is across them,
what the current is
through them,
and what is the resistance
of each one.
In the bottom row,
we'll figure out those values
for the entire circuit.
The first step is to fill in
any blanks in the table we can.
We know all three
resistor values
and the overall voltage.
Remember that the tools we have
to work with are Ohm's Law
and our knowledge
of how circuits work.
In this case,
with the information we have,
there's only one thing
we can do.
Add up all the individual
resistors to figure out
the total circuit resistance.
Since we're working
with a series circuit,
we find the total current
resistance by adding up
each individual resistor.
7 ohms, plus 38 ohms,
plus 20 ohms, equals 65 ohms.
Great.
So we fill in the blank
for the total resistance
with 65 ohms.
What can we solve for next?
If you said total current,
you're right.
We can rearrange Ohm's Law
so that it says current equals
voltage divided by resistance.
45 volts divided by 65 ohms
gives us 0.69 amps of current.
We fill that blank in
on the chart.
We don't have enough information
to use Ohm's Law again,
but we do know that we're
working with a series circuit,
and in a series circuit,
current only has one path
it can travel.
It doesn't branch off
or get depleted anywhere.
That means, in this circuit,
current is the same everywhere
through each resistor.
The total circuit current
is the same as
the current running
through R1, R2 and R3.
So we fill in all the blanks
for the current with 0.69 amps.
The last thing left
is to solve for the voltage
across each of
the three resistors.
For this, Ohm's Law will get us
where we want to go.
Any time you know two out of
the three variables within
a row, you can solve for
the third using Ohm's Law.
The voltage across R1
equals the current, 0.69 amps,
times 7 ohms,
which equals 4.8 volts.
The voltage across R2 equals
0.69 amps times 38 ohms,
which equals 26 volts.
The voltage across R3
equals 0.69 amps,
times 20 ohms,
which equals 14 volts.
Let's fill in these last blanks
in the VIR chart.
To check if our work is correct,
we can apply our knowledge
of how the numbers
should add up.
The voltages across
the three resistors
should add up to
a circuit total of 45 volts.
Adding the voltage across R1,
R2 and R3, we get 44.8 volts.
Within rounding error,
it matches what we expect it
to match.
Current is the same everywhere.
A series circuit.
Looking at our I column,
we see it is.
And for resistors in series,
the total resistance
equals the sum of
the individual resistance.
7 ohms, plus 38 ohms,
plus 20 ohms, equals 65 ohms.
So our resistance values
add up as expected.
With the chart complete,
we can answer any questions
about voltage,
current or resistance
in this series circuit.
So that's how to solve a series
circuit using a VIR chart.
Now let's try the same method,
but with a circuit
built-in parallel.
Just like before,
our first step is to fill in
all the blanks
in the VIR chart we can,
based on the circuit
information given.
Here's what that would
look like.
Unlike with the series circuit
we just looked at, here we have
a couple of options.
We could add up
the resistor values
to solve for overall
circuit resistance,
like before, but there's also
another possibility.
Remember, that
in a parallel circuit,
the voltage used up across
each branch of the circuit
is the same.
And with the circuit
in our example,
since there are only
three resistors that are
all wired in parallel,
the voltage used up in each one
is the same as the voltage
provided by our battery.
That means we fill in
all the blanks in the V column.
They're all the same.
14 volts.
Great.
So looking at the chart now,
you may see an opportunity.
Remember, when we know
two out of three numbers
in a row,
we can solve for the third.
We can use Ohm's Law
to discover the current
running through each of
the three resistors.
We'll call those currents
I1, I2 and I3,
and solve for them using
a rearranged form of Ohm's Law.
I1 equals the voltage
across R1, divided by R1,
which equals 14 volts,
divided by 10 ohms,
which equals 1.4 amps.
I2 equals the voltage
across R2, divided by R2,
which is 14 volts
divided by 25 ohms,
which equals 0.56 amps.
I3 equals the voltage
across R3, divided by R3,
which equals 14 volts
divided by 4 ohms,
which is 3.5 amps.
Let's fill in these
current values in our table.
In parallel circuits,
if we know the currents moving
through each of the branches,
we can add them together
to figure out the total
circuit current.
This means we can go further
in filling out the VIR chart.
We can add the current
in each of the three resistors
to solve for the total
circuit current.
I total equals 1.4 amps
plus 0.56 amps, plus 3.5 amps,
which equals 5.5. amps.
The last step in filling out
our table is to use Ohm's Law
once more to solve for
overall circuit resistance.
The total resistance
in the circuit equals
the total voltage
divided by the total current,
which equals 14 volts
divided by 5.5 amps,
which is 2.5 ohms.
With that, we have
a completed VIR chart
for this parallel circuit.
Now we can move on to
solving a complex circuit.
A combination of the series
and parallel circuits.
If you look at this circuit,
you can see that there is
a parallel piece to it,
as well as, a series aspect.
So we have our complex circuit
and now we'll solve for
current voltage and resistance,
step by step.
Once again, a VIR chart
will help us organize
all our information.
Take a minute to draw
a VIR chart for this circuit.
Did you draw something
like this?
If so, great. If not,
let's draw one together.
Remember, a VIR chart
includes one column for voltage,
current and resistance.
A row for each resistor
and one last row
for the total V, I, and R
in the circuit.
Now, what's the first thing
we do?
Right!
We fill in all the given
information.
In this case,
we know all the resistance
values and the total voltage
for the circuit.
As we look over this chart,
what could we do next
to fill in boxes.
We don't know the voltage
across any of the resistors,
and we don't know the overall
individual current running
in our complex circuit.
That leaves us with resistance.
In general,
when you have a complex circuit,
you'll approach it by solving
for the equivalent resistance
in the circuit.
This is easier said than done,
because remember,
our resistors are set up
in a parallel and series
arrangement and don't easily
combine together.
We'll have to add them
in a step by step process,
being careful to follow
the parallel and series rules
of resistor addition.
Our goal is to figure out
the total resistance of our
four resistors in this circuit.
We can begin by dividing
our resistors into which ones
are arranged in series,
and which are in parallel.
R1 and R2 are in parallel,
and R3 and R4 are in series.
Since R1 and R2 are in parallel,
that means their equivalent
resistance is calculated
using the parallel resistor
addition rule.
For two resistors in parallel,
one over R1 total
equals 1 over R1
plus 1 over R2.
Let's find a common denominator
for our values on the right.
That will be R1 times R2.
This makes our value
on the right of the equation
R2 plus R1, divided by
R1 times R2.
To get our total
in the numerator
so we can solve for it,
we take the reciprocal
of both sides and we are left
with R total equals R1 times R2,
divided by R2 plus R1.
Plugging in 3 ohms for R1
and 8 ohms for R2,
we find the total resistance
of the parallel resistors
in our circuit is 2.2 ohms.
Now we move on to R3 and R4.
The resistors that are
arranged in series.
Remember the series addition
rule says that
the total resistance
equals the sum of
the individual resistors.
In our case, our total
equals R3 plus R4.
Plugging in 5 ohms for R3
and 10 ohms for R4,
our total series resistance
is 15 ohms.
At this point,
it will be helpful to redraw
our circuit to reflect
the progress we've made.
We've solved for the equivalent
resistance of both the parallel
and the series resistors.
That means we can redraw
our complex circuit to show
these equivalent values
in place of the original
four resistors.
Here's what that looks like.
Since our goal is to solve
for the total resistance
in the circuit,
we have one last step,
to add these equivalent
resistance together.
Since they are in series,
R total, the overall
resistance of our circuit
equals 2.2 ohms, plus 15 ohms,
which rounds to 17 ohms.
We solve for the equivalent
resistance in this circuit.
At its simplest,
the circuit's resistance
can be represented like this.
Now we can return to our
VIR chart and fill in
the box for
total circuit resistance.
It is 17 ohms.
What do you think
we can do next?
If you said solve for
the total current
in the circuit, you're right.
Now that we know total
voltage and resistance,
we know two out of the three
variables in a row,
and we can use Ohm's Law
to figure out
the total current.
Ohm's law says the voltage
equals current times resistance,
which is the same as saying
that current equals voltage
divided by resistance.
I total equals V total,
divided by R total,
which is 12 volts
divided by 17 ohms,
which equals 0.71 amps.
As we fill in
the total current box
in our table,
we can also see
that the total circuit current
runs through R3 and R4.
After all,
they are arranged in series.
We know two out of three
variables for resistors 3 and 4,
so let's solve for the voltage
across this resistors
using Ohm's Law.
The voltage across R3
equals I3 times R3,
which equals 0.71 amps
times 5 ohms,
which is 3.6 volts.
The voltage across R4
equals I4, times R4,
which equals 0.71 amps
times 10 ohms,
which equals 7.1 volts.
Now let's see what we should
look at next
to finish our chart.
We still need to find
the voltage and current
for resistors 1 and 2.
What value is the same
for resistors in parallel?
If you said voltage,
you're right.
So how can we find
what the voltage is for
these two resistors?
Remember, we calculated
the total resistance
for these two in parallel
to be 2.2 ohms.
The total voltage across
this equivalent resistor
would be equal to 0.71 amps
times 2.2 ohms,
which gives us a voltage
of 1.6 volts.
Because this equivalent resistor
is made of two resistors
in parallel,
the voltage across each one
will be 1.6 volts.
We can now put that
in our chart.
We now know two out of
the three variables in our rows
for resistor 1 and 2,
so we can solve for the current
through these resistors.
I1 equals V1, divided by R1,
which equals 1.6 volts
divided by 3 ohms.
Which is 0.53 amps.
I2 equals V2, divided by R2,
which is 1.6 volts,
divided by 8 ohms,
which is 0.2 amps.
With that, we can complete
our VIR chart
for this complex circuit.
We can now specify
the voltage across
any resistor in the circuit.
We have all the resistors values
and we know how much current
runs through any and all
parts of the circuit.
The last thing is to check
that our solutions make sense.
Check number 1, if we add up
our individual voltages
for any given current path,
do we get the total 1.6 volts,
plus 3.6 volts,
plus 7.1 volts is 12 volts.
Within rounding error,
it matches what we expect it
to match, 12 volts.
Check number two,
does our total current running
across the parallel branches
equal the current running
through the rest of the circuit?
0.53 amps, plus 0.20 amps,
equals about 0.71 amps.
Great.
So our results hold up
as we test them against
what we know to be true
of our series
in parallel circuits.
That's a Closer Look
at Circuits.
For more "Physics in Motion",
go to our homepage
where the entire series
is available to you.
[music]