### Video Transcript

In this video, weโll be learning
about the characteristics of combination circuits, which are circuits that contain
resistors in both series and parallel. We will begin by learning how to
determine the equivalent resistance of a combination circuit. But first, we need to refresh our
memory on how to find the equivalent resistance of resistors only in series or only
in parallel.

Resistors in series are connected
along a single conductive path, as shown in the diagram below. Resistors in parallel are connected
along multiple conductive paths such that the current is split amongst them, as
shown in the diagram below.

To determine the equivalent
resistance of a series circuit, the value of one resistor that could replace the
entire circuit, also known as the total resistance, we add each of the resistors
together. For the diagram that we drew, the
total resistance will be equal to ๐
one plus ๐
two plus ๐
three. If we had more resistors in series,
we would need to take those values into account and add them in our equation. In a series circuit, the total
resistance will always be larger than any of the individual resistors.

For a parallel circuit, finding the
total resistance is a bit trickier. One over the total resistance is
equal to one over ๐
one plus one over ๐
two plus one over ๐
three. The value of any additional
resistor added in parallel would need to be taken into account. The total resistance for a parallel
circuit will always be smaller than the smallest resistor. Letโs apply the rules for finding
equivalent resistance for series and parallel circuits to a combination circuit.

To determine the equivalent
resistance of a combination circuit, we need to simplify our circuit down to either
a simple series or a simple parallel circuit. Looking at our combination circuit,
we can see that it would be a simple series circuit if not for these two resistors
which are in parallel. This means that if we replace those
two resistors with one resistor of equivalent resistance, we would have a simple
series circuit.

Letโs draw out the simplified
circuit. In our simplified circuit, we still
have ๐
one of five ohms and ๐
four of four ohms, just as we did in the circuit up
above. But this time we have only one
resistor where before we had two resistors in parallel. Now, we need to find the equivalent
resistance of those two resistors in parallel using the equation for finding
resistance of resistors in parallel.

Remember that one over the
equivalent resistance is equal to one over the first resistance plus one over the
second resistance plus one over the third resistance so on and so forth for however
many resistors that are in parallel. Plugging in our values, the left
side of the equation stays the same with one over the equivalent resistance. And the right side of the equation,
since resistor two has a value of six ohms, we have one divided by six ohms plus,
resistor three has a value of three ohms, one divided by three ohms.

When adding fractions, we need to
find the least common denominator. In this case, it would be six
ohms. We have to multiply one over three
ohms by two over two so that it becomes two over six ohms. One over six ohms plus two over six
ohms equals three over six ohms, which can simplify down to one over two ohms. To solve for the equivalent
resistance, we need to multiply both sides of our equation by two ohms and
equivalent resistance. On the left side of the equation,
๐
equivalent will cancel out. And on the right side of the
equation, two ohms will cancel out, leaving us with an equivalent resistance of two
ohms.

We can, therefore, say that our
single resistor ๐
is equal to two ohms, which is the equivalent resistance of the
three-ohm and the six-ohm resistors in parallel. We use the letter ๐
to represent
our resistor so that we donโt get confused when we go to find the equivalent
resistance for the whole circuit. Now, we have a simple series
circuit with three resistors: ๐
one, ๐
, and ๐
four. We can use the equation for
resistors in series to determine the total resistance of the circuit.

Remember that the equivalent
resistance for a series circuit is equal to the first resistance plus the second
resistance plus the third resistance and so on and so forth for however many
resistors there are in series. Plugging in our values, the total
resistance for the circuit is equal to ๐
one, five ohms, plus ๐
, two ohms, plus ๐
four, four ohms. When we add those three values
together, we get an equivalent resistance of 11 ohms. This is the same as having a
circuit with only one resistor of value 11 ohms attached to the battery.

Next, weโll take a look at a
combination circuit that can be simplified down to a simple parallel circuit.

Looking at our circuit, we can see
that it would be a simple parallel circuit if not for this branch where we have two
resistors in series. This means that if we replace those
two resistors with one resistor of equivalent resistance, we would have a simple
parallel circuit.

Letโs draw out the simplified
circuit. In our simplified circuit, we still
have resistor one of two ohms and resistor four of four ohms just as we did up
above. But this time we have one resistor
on the branch where we had resistor ๐
two and ๐
three in series. This means that our one resistor
has an equivalent resistance of the two resistors in series.

To determine the equivalent
resistance of those two resistors in series, we need to use the equation to find
total resistance in a series circuit. On the left side of the equation,
we keep ๐
equivalent for the total resistance of our branch. On the right side equation, we put
one ohm in for the value of ๐
two and three ohms in for the value of ๐
three. When we add one ohm plus three ohm,
we get four ohms. This means that resistor ๐
has a
value of four ohms, which is the equivalent resistance of ๐
two and ๐
three in
series. We use the variable ๐
so that we
donโt get confused when we go to solve for the equivalent resistance of the entire
circuit.

Now, we have a simple parallel
circuit of three resistors: ๐
one, ๐
, and ๐
four. We can use the equation for a
parallel circuit to determine the total resistance. The left side of the equation stays
the same with one over ๐
equivalent. On the right side of the equation, we have
one over two ohms for our first resistor, one over four ohms for our second
resistor, and one over four ohms for our third resistor.

When weโre adding fractions, we
need to use the least common denominator. In this case, it would be four
ohms. This means that we need to multiply
one over two ohms by two over two. We can now add two over four ohms
plus one over four ohms plus one over four ohms, which gives us a sum of four over
four ohms and can be simplified down to one over one ohm. We multiply both sides of the
equation by one ohm and ๐
equivalent so that ๐
equivalent cancels out on the left
side of the equation and one ohm cancels out on the right side of the equation. Leaving us with an equivalent
resistance for the circuit of one ohm. This is the same thing as having a
circuit with only one resistor of value one ohm attached to the battery.

To determine any other
characteristics of a combination circuit, we need to use Ohmโs law. Ohmโs law is an equation that
relates the potential difference across the resistor ๐ to the current through the
resistor ๐ผ and the value of the resistance ๐
. If we wanna find the total current
through each of our combination circuits, we need to use the potential difference of
each of the batteries as well as the equivalent resistance of each of the
circuits.

To find the current using Ohmโs
law, we need to divide both sides of the equation by ๐
. On the right side of the equation,
the ๐
s would cancel out. And the left side of the equation,
weโd be left with ๐ divided by ๐
. We can apply this new variation of
the equation to both circuits to determine the total current in each.

For our top circuit, the potential
of the battery is 12 volts and the equivalent resistance of the circuit is one
ohm. 12 volts divided by one ohms is 12
amps. This means that the total current
in our circuit is 12 amps. In our bottom circuit, the
potential difference of the battery is 22 volts and the equivalent resistance of the
circuit is 11 ohms. 22 volts divided by 11 ohms is two
amps. The total current through this
circuit is two amps.

The current that we found in both
of these circuits is the total current not the current through each resistor. We can also use Ohmโs law to
determine the potential difference across one of our resistors. If we wanna determine the potential
difference across the four-ohm resistor, we need to know the current going through
it as well as the value of its resistance.

The current going through the
four-ohm resistor would be two amps. We know this because the four-ohm
resistor is in series with the other components of the circuit. In a series circuit, the current
doesnโt split. Therefore, the total current goes
through each of the individual components that are in series. We found that the total current was
two amps. Therefore, the current through the
four-ohm resistor is also two amps. The resistance value is four
ohms. When we multiply two amps by four
ohms, we get a potential difference across our four-ohm resistor of eight volts.

Another approach to solve any of
these problems when equivalent resistance wonโt work is to apply Kirchoffโs two
laws. Letโs refresh our memory on these
two laws before we apply them to our previous circuits.

Kirchoffโs first law states that
the current into a junction equals the current out of a junction. Applying this law to a parallel
circuit when the current splits across multiple paths so that the current in one
path is ๐ผ one, the current in the second path is ๐ผ two, the current in the third
path is ๐ผ three, and so on. Then adding all those currents
together gives us the total current ๐ผ ๐. Applying this law to a series
circuit where the current doesnโt split, the total current will be equal to the
current through each of the resistors: ๐ผ one for resistor one, ๐ผ two for resistor
two, ๐ผ three for resistor three, and so on.

Kirchoffโs second law states that
the sum of all voltages around any closed loop in a circuit must equal zero. We can apply this law to a series
circuit because the potential difference is shared or split across components in
series. This tells us that for a single
battery circuit, the potential difference across all the resistors added together โ
๐ one for resistor one, ๐ two for resistor two, ๐ three for resistor three, and
so on โ is equal to the potential difference of the battery, ๐ ๐.

In a simple parallel circuit, this
is why the resistor on each branch has the same potential difference as the
battery. If we wanna know the current
through the three-ohm resistor, we need to apply Kirchoffโs second law to determine
the potential difference across the branch. If we follow the path that comes
from the battery to the five-ohm resistor up to the three-ohm resistor down to the
four-ohm resistor and back to the battery, we can then apply Kirchoffโs second law
to this path.

The total potential difference is
the potential difference of the battery, 22 volts. ๐ one is the potential difference
across the five-ohm resistor. ๐ two is the potential difference
across the three-ohm resistor. And ๐ three is the potential
difference across the four-ohm resistor. Since we do not know the potential
difference across any of the resistors, we will need to plug in for each one Ohmโs
law.

For each potential difference, we
plug in ๐ผ๐
. For the five-ohm resistor, we have
an ๐ผ of two amps and an ๐
of five ohms. For the three-ohm resistor, we
donโt know the current, so we leave it as ๐ผ. And the resistance is three
ohms. For the four-ohm resistor, we plug
in two amps for the current and four ohms for the resistance.

Simplifying through distribution,
two times five is 10, and two times four is eight. To isolate the current, we need to
subtract eight volts from both sides and 10 volts from both sides. This will cancel out the eight
volts and the 10 volts on the right side of the equation. Subtracting 18 volts from 22 volts
leaves us with the left side of the equation of four volts. Our last step is divide both sides
of the equation by three ohms, canceling out the three ohms on the right side of the
equation. This leaves us with a current of
four-thirds amps.

If we apply Kirchoffโs first law,
we could find the current through the six-ohm resistor. In this case, the total current
that comes into the junction, ๐ผ ๐, must be equal to the current that goes through
the three-ohm resistor plus the current that split to go through the six-ohm
resistor. To determine our unknown current,
we need to subtract the current through the three-ohm resistor from both sides of
the equation. This will cancel out the current
for the three-ohm resistor on the right side of the equation.

When we subtract the current
through the three-ohm resistor from the total current, we will get the current that
goes through the six-ohm resistor. We plug in two amps for the total
current and four-thirds amps for the current through the three-ohm resistor as we
just found. When subtracting fractions, we need
to make sure that we have the least common denominator. In this case, that would be
three. So, we multiply two by three over
three, which is the same thing as six-thirds. When we subtract four-thirds amps
from six-thirds amps, we get two-thirds amps.

Letโs apply what we just learned
about combination circuits to an example.

In the circuit shown, the current
takes multiple paths from the positive battery terminal to the negative battery
terminal. Find the total resistance of the
circuit.

That given diagram shows a
combination circuit as resistors are both in parallel as well as in series. Before we simplify our circuit to
solve for the total resistance, letโs remind ourselves how to find total resistance
in both series and parallel circuits. In a series circuit, the total
resistance or equivalent resistance is equal to the sum of each individual resistor,
๐
one, ๐
two, ๐
three, and so on, until all the resistors are accounted for. In a parallel circuit, one over the
total resistance or equivalent resistance is equal to one over ๐
one plus one over
๐
two plus one over ๐
three and so on and so forth until all the resistors are
accounted for.

Looking at our diagram, we can see
that if we could replace our two resistors in parallel with an equivalent resistor,
then we would have a circuit that is a simple series circuit. We have drawn a simplified version
of the circuit below. We need to determine what value of
๐
will be equivalent to the 12-ohm and 18-ohm resistors in parallel. To do this, we need to use the
equation to find the total resistance of a parallel circuit.

We used ๐
for equivalent
resistance, so the left side of the equation becomes one over ๐
. The right side of the equation are
our two resistors, one over 12 ohms plus one over 18 ohms. When adding fractions, we need to
find the least common denominator. The least common denominator for 12
ohms and 18 ohms would be 36 ohms. In order for each of our fractions
to have a denominator of 36 ohms, our one-over-12-ohm fraction needs to be
multiplied by three over three and our one-over-18-ohm fraction has to be multiplied
by two over two. This makes our fractions three over
36 ohms plus two over 36 ohms. When we add three over 36 ohms plus
two over 36 ohms, we get five over 36 ohms.

To isolate ๐
, we multiply both
sides of the equation by 36 ohms and ๐
. This cancels out ๐
on the left
side of the equation and 36 ohms on the right side of the equation, giving us the
equation 36 ohms is equal to five ๐
. Then, we divide both sides by five,
canceling out the five on the right side of the equation, giving us an equivalent
resistance of 36 divided by five ohms. In decimal form, this would be 7.2
ohms.

Looking at our simplified circuit,
we can see that our three resistors are in series with each other. Therefore, we need to use the
series equation to find total resistance. The total resistance of our
circuit, ๐
equivalent, is equal to 14 ohms, the value of the first resistor, plus
10 ohms, the value of the second resistor, plus 7.2 ohms, the value of the
equivalent resistance of the 12- and 18-ohm resistors that were in parallel. When we add our three resistors
together, we get a value of 31.2 ohms. All the values in our problem have
been given to two significant figures. Therefore, we need to round our
total resistance to two significant figures, which gives us a total resistance for
our circuit of 31 ohms.

Key Points

Characteristics of circuits
containing resistors with both series and parallel combinations can be determined by
calculating the equivalent resistance of circuit branches. Kirchoffโs laws can be used to
calculate currents in circuit branches where currents cannot be calculated using
equivalent resistance methods.