Download Grade 6 Math Circles Circuits Electricity

Faculty of Mathematics
Waterloo, Ontario N2L 3G1
Centre for Education in
Mathematics and Computing
Grade 6 Math Circles
March 18/19, 2014
Circuits
Electricity
Electricity is a type of energy that deals with the flow of electric charge. The two main
types of electricity are Static Electricity and Current Electricity. We will be focusing
on current electricity due to the mathematics involved in electric circuits.
Manipulating Equation Pyramids
Before we get into calculating and solving problems concerning current electricity, let’s learn
how to use an Equation Pyramid.
When we have three values that are related to each other, more often than not we can set
them up in an equation pyramid. While not primarily used in math as a subject, it is an
extremely useful tool for science relationships you may come across. Let’s look at one using
the values A, B and C:
How to use:
• Cover the letter representing the value you are looking for
• If the remaining letters are beside each other, you multiply the two values:
Example: If we cover A, we get B and C beside each other. Therefore A = B × C
• If the remaining letters are on top of each other, you divide the two values:
Example: If we cover B, we get A on top of C. Therefore B =
A
C
Example: If we cover C, we get A on top of B. Therefore C =
A
B
1
Ohm’s Law
The main formula we use in circuit calculations is Ohm’s Law. This law states:
V =I ×R
Where the letters stand for the following:
• V represents Voltage, which is the amount of electrical force that pushes electricity
through the circuit. It is measured in volts (V).
• I represents Current, which is the flow of electrons that moves within the circuit. It
is measured in amperes (A).
• R represents Resistance, which is a property which impedes a current. Resistance
causes electric energy to be turned into heat. It is measured in ohms (Ω).
We will be working with the following Equation Pyramid when doing our calculations today:
Exercises I
1. By rearranging Ohm’s Law or using the Equation Pyramid, write the equations for:
V=
I=
R=
2. What is the voltage given off by a circuit where there is a current of 40 amperes, and
a resistance of 2 ohms.
3. A circuit has a total resistance of 3 ohms and ends up giving a voltage of 18 volts.
What is the current in the circuit?
4. If a circuit gives off 120 volts of energy and it is found that there is a current of 40
amperes, what is the total amount of resistance in the circuit?
2
Circuits
A circuit is a closed path on which electric current travels. A circuit must always have an
object which produces a current and one that consumes the current. In this case we will be
looking at a circuit with batteries and light bulbs:
+
battery
-
Since we will be working with many circuit diagrams, drawing detailed batteries and light
bulbs (OR all having different representations) results in inconsistency when comparing
answers. We need some uniformity or notation:
+
-
Battery
Light Bulb
Using the above symbols we can redraw the diagram that we found at the top of the page.
This is the notation we must use in this lesson and allows us to make sure that we are all
talking about the same circuit.
Complex Circuits
Not all circuits follow a simple track. Circuits found in real life come in many forms and
have their own advantages and disadvantages. The way the current flows also affects how
much of the current is used on each light bulb or device. Let’s investigate the two main ways
that a circuit can be set up: Series and Parallel.
3
Circuit in Series
One way that we can connect light bulbs in a circuit is in series, which means that the
current flows through the light bulbs one after another.
The following circuits represent light bulbs that are in series. That is to say that if the
current flows through the circuit, it will not be broken into several paths in order to charge
every light bulb.
-
+
+
Circuits in Series have the following three properties:
1. The total current in the circuit (ie. the total current leaving the battery) is equal to
the current flowing through each of the light bulbs in the circuit.
Itotal = I1 = I2 = I3 = ...
2. The total voltage in the circuit (ie. the total voltage leaving the battery) is equal to
the sum of voltage at each of the light bulbs.
Vtotal = V1 + V2 + V3 + ...
3. The total resistance in the circuit (ie. the total resistance at the battery) is equal to
the sum of resistance at each of the light bulbs.
Rtotal = R1 + R2 + R3 + ...
Circuit in Parallel
One way that we can connect light bulbs in a circuit is in parallel, which means that the
current flows through the light bulbs in a branched path.
4
-
-
The following circuits represent light bulbs that are in parallel. That is to say that if the
current flows through the circuit, it will be broken into several paths in order to charge every
light bulb. The point at which the circuit branches is called a node.
+
+
Circuits in Series have the following three properties:
1. The total current in the circuit (ie. the total current leaving the battery) is equal to
the sum of the current flowing through each of the light bulbs in the circuit.
Itotal = I1 + I2 + I3 + ...
2. The total voltage in the circuit (ie. the total voltage leaving the battery) is equal to
the voltage at each of the light bulbs in the circuit.
Vtotal = V1 = V2 = V3 = ...
3. The inverse of the total resistance in the circuit (ie. the total resistance at the battery)
is equal to the sum of inverses of the resistance at each of the light bulbs.
1
Rtotal
=
1
1
1
+
+
+ ...
R1 R2 R3
Exercises II
-
1. For the following circuits, state if the circuits are parallel or series.
-
-
+
+
+
5
Total Calculations
For each of the diagrams below, determine if the light bulbs are connected in series with
each other, or in parallel with each other. Then, determine the total voltage, current and
resistance leaving the battery in each circuit.
+
a) Example:
V=4V
-
I = 2 Amps
V=6V
Solution:
The light bulbs are connected in series.
Let’s investigate each part:
– Voltage: We know that that since the light bulbs are in series that we must use
the following formula:
Vtotal = V1 + V2
The values for these are given as follows: V1 = 6V and V2 = 4V . Subbing into
the formula we see:
Vtotal = 6V + 4V = 10V
– Current: We know that that since the light bulbs are in series that we must use
the following formula:
Itotal = I1 = I2
The values for these are given as follows:
I1 = I2 = IT otal = 2Amps.
– Resistance: We know that that since the light bulbs are in series that we must
use the following formula:
Rtotal = R1 + R2
We must use V = I × R to find the resistance at each light bulb:
R1 =
6V
V1
=
= 3Ω
I1
2A
and
R2 =
V2
4V
=
= 2Ω
I2
2A
Now we can find the total resistance;
RT otal = R1 + R2 = 3Ω + 2Ω = 5Ω
6
b) Example:
I = 4 Amps
R = 2 Ohms
-
I = 2 Amps
R = 4 Ohms
+
Solution:
The light bulbs are connected in parallel.
Let’s investigate each part:
– Voltage: We know that that since the light bulbs are in parallel that we must
use the following formula:
Vtotal = V1 = V2
We must use V = I × R to find the resistance at each light bulb:
V1 = I1 × R1 = 2A × 4Ω = 8V
and
V2 = I2 × R2 = 4A × 2Ω = 8V
Now we can find the total voltage:
V1 = V2 = Vtotal = 8V
– Current: We know that that since the light bulbs are in parallel that we must
use the following formula:
Itotal = I1 + I2
The values for these are given as follows:
IT otal = I1 + I2 = 2A + 4A = 6AAmps.
– Resistance: We know that that since the light bulbs are in parallel that we must
1
1
1
use the following formula:
=
+
Rtotal
R1 R2
Now we can find the total resistance;
1
RT otal
=
1
1
1 1
1 2
3
+
= + = + = Ω
R1 R2
4 2
4 4
4
But we must find the reciprocal of our answer (or flip the fraction:
1
RT otal
3
4
= Ω which means Rtotal = Ω
4
3
Note: Once you have found your Vtotal , Itotal and Rtotal for the entire circuit,
all your total values should follow: Vtotal = Itotal × Rtotal . You may use this
to check if you are right.
7
Problem Set
-
1. For each of the diagrams below, determine if the light bulbs are in series with each
other, or in parallel with each other.
a)
b)
-
+
+
c)
-
d)
e)
f)
-
+
-
-
+
+
+
8
2. For each of the diagrams and corresponding tables below, fill in the missing values in
the table using Ohm’s Law and circuit properties.
3
4
2
a)
1
-
5
+
Vtotal =
V1 =
V2 =
V3 =
V4 =
Itotal = 5A
I1 =
I2 =
I3 =
I4 =
Rtotal =
R1 = 1Ω
R2 = 3Ω
R3 = 2Ω
R4 =
V5 =
I5 =
3
Ω
2
R5 = 12 Ω
-
b)
+
1
2
Vtotal =
V1 =
V2 = 4V
Itotal =
I1 = 2A
I2 =
Rtotal =
R1 = 2Ω
R2 = 2Ω
c)
2
-
1
+
Vtotal = 18V
V1 =
V2 =
Itotal =
I1 = 3A
I2 =
Rtotal =
R1 =
R2 = 4Ω
-
d)
1
2
+
3
4
Vtotal =
V1 =
V2 =
V3 = 12V
V4 =
Itotal =
I1 =
I2 =
I3 =
I4 =
Rtotal =
R1 = 4Ω
R2 = 2Ω
R3 = 3Ω
R4 = 1Ω
9
-
e)
1
+
2
3
Vtotal =
V1 =
V2 =
V3 =
Itotal =
I1 =
I2 = 2A
I3 =
Rtotal = 2Ω
R1 = 4Ω
R2 = 8Ω
R3 =
1
2
-
f)
+
4
3
5
Vtotal =
V1 =
V2 = 16V
V3 =
V4 =
V5 = 28V
Itotal =
I1 =
I2 = 4A
I3 =
I4 =
I5 =
R2 =
R3 = 5Ω
R4 =
R5 =
Rtotal
25Ω
= R1 = 3Ω
3. * For each of the diagrams and corresponding tables below, fill in the missing values
in the table using Ohm’s Law and circuit properties.
Note: All of the circuits below contain both parallel and series components. To solve
for the missing values, you will need to solve for the total values in the parallel component. Then you can simplify the circuit by treating the parallel component as a single
light bulb in series with the rest of the circuit. The values for current, voltage, and
resistance of the simplified light bulb will be the total values of the parallel component.
1
a)
-
2
+
3
Vtotal =
V1 = 6V
V2 =
V3 =
Itotal =
I1 =
I2 =
I3 = 2A
Rtotal =
R1 = 12Ω
R2 = 4Ω
R3 = 4Ω
10
1
2
-
b)
Vtotal =
Itotal =
V1 =
3
A
2
Rtotal =
4
3
5
+
V2 = 3V
V3 =
V4 = 6V
V5 =
1
A
2
I5 =
I1 =
I2 =
I3 = 1A
I4 =
R1 = 4Ω
R2 =
R3 =
R4 =
R5 = 2Ω
c)
1
3
-
2
4
+
5
Vtotal =
V1 =
V2 =
V3 =
V4 =
V5 =
Itotal = 5A
I1 =
I2 =
I3 =
I4 =
I5 =
Rtotal =
R1 = 5Ω
R2 = 2Ω
R3 = 3Ω
R4 = 4Ω
R5 =
d)
1
Ω
13
1
-
2
3
+
4
5
Vtotal
25V
= V1 =
V2 =
V3 =
V4 =
V5 =
Itotal =
I1 =
I2 =
I3 =
I4 =
I5 =
Rtotal =
R1 = 10Ω
R2 = 4Ω
R3 = 3Ω
R4 = 8Ω
R5 = 1Ω
11
4. ** a) Each light bulb in the diagram has resistance R. Determine the total resistance
of the circuit in terms of R.
4
-
1
2
7
5
8
+
3
6
9
b) Given that RT = 15Ω and IT = 2A, use the total resistance found in part (a) to fill
in all the missing values in the table below.
Vtotal =
V1 =
V2 =
V3 =
V4 =
V5 =
Itotal =
I1 =
I2 =
I3 =
I4 =
I5 =
Rtotal =
R1 =
R2 =
R3 =
R4 =
R5 =
V6 =
V7 =
V8 =
V9 =
I6 =
I7 =
I8 =
I9 =
R6 =
R7 =
R8 =
R9 =
12