Download Electricity Gone Wild

E ectricity
Gone Wi d!
By: Paulien Nuyts and Ivana Miljanic
Objectives for this chapter:
Learn what a parallel and series circuit is
Learn Kirchoff’s law and Ohms law
Learn teeming people analogies (for current) and
be able to explain how it represents current,
voltage, resistors, electrons, ammeters,
voltmeters, and batteries
Learn how to read ammeters and voltmeters in
different positions in series and parallel circuits
Learn what happens to brightness of bulbs when
more light bulbs are placed in a series or parallel
circuit
Learn all the schematic symbols for batteries,
wires, bulbs, ammeters, voltmeters, and resistors
Learn which path electrons are most likely to take
Learn how much voltage electrons use up at
different resistors
Know how to derive equations for power from
other equations
Learn Equivalent resistance
Know how adding more bulbs to a series or
parallel circuit changes the Equivalent Resistance
Know how to derive equations for equivalent
resistance from Kirchoff’s and Ohms law
Know what a 60 Watt bulb is
Learn all units for current, voltage, resistance, and
power
Learn how to do Kirchoff’s law, Ohms law,
Equivalent resistance, and power problems
Experiments
1. An experiment that you conduct to understand how the loop rule works which the
teeming people analogy explains is:
Materials: 2 identical light bulbs, 2 batteries, 5 wires, and a voltmeter
Precautions: Only do this experiment with an experienced science teacher, or a
responsible adult.
For this experiment you will only be making simple series circuits. First you will
connect the batteries and one light bulb in series. Measure the voltage of the battery
and then measure the voltage of the light bulb with the voltmeter, you will find that
they are the same, this is because the voltage charges on a closed path must equal
zero. This in simpler words means that the volts from the battery must be completely
used up by the resistors in each closed loop. Then add the second light bulb in series
to the circuit. Measure the voltage of the batteries with the voltmeter, and then
measure each individual light bulb’s voltage. You will find that both the light bulbs
each use up half of the batteries voltage. This is because Kirchoff’s loop rule tells us
that the voltage must be used up in every loop, and since the bulbs are equal in
resistance they use up the same amount of voltage, and because there are 2 light bulbs
that means they each use up half of the total voltage put into the circuit from the
batteries.
2.
Materials: yourself and food
Precautions: DO NOT starve yourself, eat once you are hungry
Throughout the day running around makes you tired and you lose energy and you
your stomach starts to growl. When this happens you usually get more food to eat.
You usually don’t eat when you are not hungry you first waste the food that has
already been eaten. This represents Kirchoff’s loop rule because you waste all of the
food that is in your stomach before you go and get more. You also burn more or less
calories depending on what you are doing; this represents the different amounts of
resistance in resistors.
The Environment
All around us we see examples of energy being wasted, and getting energy once you are
out of it, for example:



While running you waste energy that’s why at some point you’ll have to stop and
get more energy by eating.
Paulien concentrated so much on her science project that she had to take a 5minute break and get a snack to regain her energy.
Ivana always gets very tired during basketball practice, so right when she comes
home she eats a small snack
Series and Parallel Circuits
Series Circuit: In a series circuit an electron only has one choice, one path to follow
Parallel Circuit: In a parallel circuit an electron has more than one path to choose from
Kirchoff’s Laws
Junction Rule: At any junction, the sum of all currents entering the junction must be
equal to the sum of all currents leaving the junction
(Make example of junction rule)
Loop Rule: The sum of the charges in voltages around any closed path of a circuit must
equal zero.
(Make example of loop rule)
Ohms Law
Ohms law states that the current is equal to voltage divided by resistance. I=V/R
Analogies
There are a bunch of track runners running around a track, they start out at a Gatorade
stand gives them energy. This represents the battery. The track represents the wires. The
runners represent electrons and their urge to finish the race and get more Gatorade
represents the current. They burn all their energy by running and they cant run anymore
unless they get more Gatorade or “energy” which describes Kirchoff’s Loop and junction
rule. It shows the loop rule because all of the charges or energy given to the runners
through the Gatorade gets used up during the race. It shows the junction rule because all
the Gatorade that the runners drink gets used up by the end of the race. The voltage is the
energy given to the runners through the Gatorade. During the race the runners encounter
sandboxes, which make them use more energy, and decreases the current. The sandboxes
represent resistors. The ammeters are the fans watching the runners pass by. During
Gatorade commercials you can see the athletes sweating which represents how much
energy they are wasting at each sandbox. The sweat represents the voltmeter. This is a
good analogy that helps you understand how electrical circuits work. It explains the
experiments in the beginning because it shows the same thing, Kirchoff’s loop rule, and
how you go get energy when you run out of it. This analogy is helpful in helping you
remember how the resistors have to take up the exact amount of voltage that they are
given.
Ammeters and Voltmeters
A voltmeter measures how much voltage goes through between two points. A voltmeter
doesn’t change the current or resistance, it just measures the voltage. (Voltage is
measured in Volts). In a circuit a voltmeter is connected parallel to a circuit like in the
picture below. In the picture it’s not a full circuit but it shows how the voltmeter should
be placed. Whether it is placed over batteries or a light bulb it doesn’t matter.
If a voltmeter is placed in a series circuit over all of the resistors (connected on both sides
of the battery) then the voltmeter will equal the voltage of the batteries. If a voltmeter is
placed around separate resistors then it’s voltage consumption is going to be less than the
total voltage consumption but it might be less or more than the other resistor’s voltage
consumption depending on how many ohms the resistor has. In a parallel circuit all the of
the closed loops are going to read the same on the voltmeter because according to
Kirchoff’s loop rule the sum of the charges in voltage around a closed loop must equal
zero. If you put the voltmeter around one resistor on one side of the parallel circuit and
then take away the other side of the parallel circuit then voltage consumption wouldn’t
change.
An Ammeter measures how much current is going through a circuit. An ammeter doesn’t
change the current or resistance. (Current is measured in Amperes-Amps). In a circuit an
ammeter is placed within the circuit, as shown in the picture below. In the picture below
it shows some of the different ways an ammeter can be placed in a circuit.
= Ammeter
If an ammeter is placed in a series circuit no matter where, the current stays the same. If
an ammeter is placed in a parallel circuit the reading changes according to where it is put.
This is because the electrons choose the least resistant and shortest path.
The Brightness of the Bulbs
In a Series Circuit: Let’s say you have a circuit with two batteries and one light bulb.
Then that light bulb takes up all the voltage going through the circuit, which is proven in
Kirchoff’s loop rule. The voltage that the bulb takes up determines its brightness. If more
light bulbs are added in a series circuit then all the bulbs will become dimmer, because
more light bulbs are taking up that same voltage given by the battery.
In a Parallel Circuit: Let’s say you have a parallel circuit with two batteries and one
light bulb. When you add a light in parallel then the brightness of the bulbs isn’t going to
change. This is because it’s a closed loop, in a parallel circuit the loops are all separate. If
one of the lights were to go out it wouldn’t affect the other because the other is still in a
full circuit.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Schematic Diagrams
Schematic diagrams are used to express light bulbs, batteries, voltmeters, ammeters,
wires, a resistor, and many more, in an easier way.
Schematic Diagrams:
Light Bulbs
batteries
Wires
voltmeters
Ammeters
resistors
Which path electrons take?
In a parallel circuit electrons have to choose the path that they take. The higher
percentage of electrons is going to take the path that has fewer resistors; this is not
because they know which one has fewer or more resistors. For example let’s say there is
a room with 2 doors, one big and one very small. The room is very crowded and a fire
has just started in the back and people are panicking to get out. The people are being
shoved left and right, they certainly don’t know where the smaller or the bigger door is;
all they care about is getting out. Most of the people are still going to be shoved out of the
larger door, just like electrons choose the path with least resistance.
How much voltage electrons use up at different resistors
If it’s a series circuit and there in only one resistor then you add up the voltage of
the batteries, V, and divide it by the current, C. This is Ohms law. If there is more than
one resistor in a series circuit, you take the voltage, V, and divide it by the current, C, and
the you subtract the other resistors and you get the missing one. If it says that they are all
equal then you do V divided by C again, but this time you just divide it by the number of
resistors to get the value of one of the resistors. If the resistors are in parallel and there is
one resistor in each loop you take V and divide it by the current around that loop will be
given if you are asked to find the resistance. If there is more than one resistor in the loops
of a parallel circuit you take V and divide it by the given current around that loop and
then you subtract the rest of the resistors to find the one that you are looking for, or if the
are all equal you divide them by the number of resistors to get one of them.
Deriving the equation for power
We all know Power = work/time, but how do we connect this to current voltage
and resistance? Well, since the equations for voltage and current are as follows:
Voltage = Energy
Current = Charge
.
Charge AND
Time . the product of the two is energy/time
which is the same as work/ time, which is the equation for power. From this you can
conclude that Power = voltage*current. If you combine P=V*I (the equation we just
derived) and I=V/R (Ohms law) by substituting the value V/R for I in the first equation
you get P=V/R * V which is the same as POWER = V * V
R .
Equivalent Resistance
Equivalent Resistance is if you replace a bunch of resistors with one, the equivalent
resistance is the value or resistance of that one resistor.
Equivalent Resistance in a series circuit: To calculate how much resistance is in a
series circuit you have to use the formula below:
Req= R1 + R2 + R3 + R etc.
(Equivalent Resistance= Resistance 1 + Resistance 2 + Resistance 3 + Resistance etc.)
Equivalent Resistance in a parallel circuit: To calculate how much resistance there is
in a parallel circuit you have to use the formula below:
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
(One over Equivalent resistance= One over resistance one + One over
resistance two + One over resistance 3+ One over resistance etc
Adding more bulbs changes the Equivalent Resistance
In a series circuit the resistance changes in an exact pattern each time an identical light
bulbs are added. When one light bulb is added then the resistance is increased.
Data from an experiment done to show you how the equivalent resistance changes in a
series circuit:
# of bulbs
Voltage (Volts)
Current (Amps)
1
2
3
4
5
5
5
5
5
2.50
1.67
1.25
Equivalent
Resistance (ohms)
1
2
3
4
In a parallel circuit the resistance doesn’t change in a specific pattern even if the light
bulbs are identical, but it does change. When more light bulbs are added the equivalent
resistance goes down.
Data from an experiment done to show you how the equivalent resistance changes in a
parallel circuit:
# of bulbs
Voltage (volts)
Current (Amps)
1
2
3
4
5
5
5
5
5
10
15
20
Equivalent
Resistance (Ohms)
1
0.5
0.33
0.25
Deriving the equivalent resistance equations
Equivalent resistance is the value of all the resistors in the circuit combined. If
you replaced all of the resistors in the circuit with one that was equal to the resistance of
all of them, its resistance would be the equivalent resistance of the circuit. You can derive
the equations for finding the equivalent resistance of parallel and series circuits from
Kirchoff’s and Ohms law.
Series- Let’s say there are three resistors in a circuit, R1, R2, and R3, and one
battery whose voltage equals V. The potential differences of voltage across the resistors
will be V1, V2, and V3 (respective to resistors R1, R2, and R3). By Kirchoff’s loop rule
the sum of the charges in Voltage around this series circuit must equal 0, so V =
V1+V2+V3. Ohms law says that voltage=resistance*current, and the current is the same
throughout because it’s a series circuit, so V1=IR1, V2=IR2, and V3=IR3. If you
substitute these equations in you get V=IR1+IR2+IR3. If you divide both sides of this
equation by I (current), you get V/I=R1+R2+R3. V/I is the total voltage of the circuit and
I is the constant current throughout the whole circuit and resistance =V/I, since this is the
total voltage and the only current throughout the circuit it makes sense to say that this is
equal to the total resistance in the circuit, which is the definition of equivalent resistance,
therefore Equivalent resistance = R1+R2+R3.
Parallel- there are still those same three resistors R1, R2, and R3 by this time
they are in parallel. There is still one battery equal to V. The current through these loops
in the circuit are all different. The respective currents through these resistors are I1, I2,
and I3. According to Kirchoff’s junction rule the current flowing into each junction must
also come out of the junction and this equals the total current through the circuit, I.
According to this I = I1+I2+I3. Kirchoff’s loop rule says that the sum of the charges of
voltage in each of these loops must equal zero, since there is only on resistor in each loop
their voltage difference has to be equal to the voltage of the battery, V. According to
Ohms law I1=V/R1, I2=V/R2, and I3=V/R3. If you substitute these equations into the
first one you get I = V/R1 + V/R2 + V/R3. Since the total current is I, then
I=V/equivalent resistance. If you plug this into the equation you get V/equivalent
resistance = V/R1 + V/R2 + V/R3. Since the V is common on both sides you just divide it
out and get
.
1
= 1 + 1 + 1 .
Equivalent resistance
R1
R2 R3 .
A 60-Watt Bulb
What does it mean to be a 60-watt bulb?
In the United States the voltage going through an outlet is 120 Volts. When a 60-watt
bulb is plugged into 120 volts then it can be considered a 60-watt bulb. When 60 watts is
plugged into an outlet that is more or less than 120 volts then the bulb is no longer a 60watt bulb.
A 60-watt bulb is brighter in a parallel circuit than a series circuit because according to
Kirchoff’s loop rule all voltage must be used up in a closed loop. In a parallel circuit the
60-watt bulb gets more voltage than in a series circuit because in a series circuit the
voltage gets used up by more than one bulb and it’s the same voltage since it’s only one
loop.
Units for Current, Voltage, Resistance, and Power
Current: Amperes (Amps)
Voltage: Volts
Q ui ck Ti m e ™ an d a
T IFaFre( Unneco
p ret osss ed
pr rese. so r
edmed
ee ) dth ec
i s om
pi c tu
Resistance Ohms
Power: Watts
Summary
I hope that you have learned a lot from all of this. From this text
you should be able to know what a series an parallel circuit is, what happens to circuits
from more bulbs are added, what ammeters read when they are placed in different
positions, what voltmeters read when they are placed in different positions, learn what
schematic diagrams are, explain teeming people analogies, explain how electrons seem to
know which path to take in a circuit, explain what Kirchoff’s laws are, and be able to
predict current and voltage within a circuit, know Ohms law, know how to derive the
equation for power, know all the equations for power, know how adding more resistors to
a series or parallel circuit will change the equivalent resistance, know how to derive the
equation for equivalent resistance from Kirchoff’s and ohms laws, know what a 60 watt
light bulb really is and why it’s brighter when there are multiple in parallel rather than
series, know how to solve ohms law, Kirchoff’s law, equivalent resistance, and power
problems.
Kirchoff’s law, Ohms law, Equivalent resistance, and power
sample problems
1.
A book light with a resistance of 60 Ohms is put in series with a 24 -volt battery.
Find the current going through this circuit.
First you need to look for what equation you would use.
I=V/R
Secondly you have to plug in all your numbers.
I?= 24 volts/ 60 ohms
Solve.
I= 24/60 I= 0.4 Amps
2.
A
If each battery is 4 volts calculate the Resistance Voltage
R1
R1=R2
B
Vab=?
Vbc=?
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
R2
C
Ia= 0.5 A
R1=?
Steps
1. The voltage is split
between two resistors so
they would both be 2
ohms.
2. R=V/I (by Ohms law)
So you take the voltage
at each resistor and
divide it by the current.
R=2/0.5 R=4 Ohms
R2=?
3.
For this problem use the picture on the next page.
Find which bulb will use the most voltage and find the power of the circuit.
Steps:
First you have to find the current throughout the circuit, to be able to do that you have to
find the equivalent resistance first. The equivalent resistance would be 30 Ohms because
6+16+8=30. Then you can solve for current by using the I=V/R equation
I=60Volts/30Ohms. I=2Amps. Then you can solve for the voltage going through each
light bulb by using the same equation. I=V/R.
2Amps= Voltage/6 Ohms V= 12, 2Amps= Voltage/16 Ohms V= 32, 2Amps=Votlage/8
Ohms V= 16.
The bulb using the most voltage is the 16 Ohms bulb.
Now to find the power you have to use the equation P=IV
P=2Amps x 60 Volts
P=120 Watts
6 Ohms
60
Volts
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
16 Ohms
8 Ohms
4.
Current at A = 200ma
Current at B = 150ma
Current at C = ?
The current at point C would have to be 50ma. This is because all of the current that is
entering this circuit, 200ma, must leave it too. If 150ma of those 200ma goes in the first
loop and there is only one more loop then the 50ma that is left must go to the second
loop.
5.
If the battery is 2v and the resistors are identical then what will the voltmeter read from
points A to B?
(Use picture on next page)
The voltmeter would read 1v. This is because all of the charges in voltage, the 2v from
the battery, must be used up in the circuit because it is a closed path. Because there are
two identical resistors, to figure out how much voltage is used up by each you divide the
total voltage, 2v, by the number of resistors, 2, and you get 1v.
6.
Let’s start by adding the 6 and 8 ohms now it’s 14 ohms, then set up the equation for 14
and 2 in parallel: 1 = 1 + 1 from this you will get
Req 14
2
1.75 ohms. Now you take this and add it to 7 ohms because they are in connected in
series, and you get 8.75. Then you add the top part of this loop in the circuit(5+3) and
you get 8 ohms, so since the 8 ohms and 8.75 ohms are in parallel you have to set up the
equation again: 1 = 1 + 1
from
Req
8 8.75
this you will get about 4.179. Next you add the 2 and the 3 to it because they are
connected to it in series and you will get 9.179 ohms.
Problems
Let’s see how much you have learned.
Solve the following problems by using everything you have learned. (Hint: Draw pictures
or diagrams to help you visualize the situation)
1.
2.
3.
4.
5.
6.
7.
8.
If you have a parallel circuit with 4 batteries the first loop has one bulb and the
second loop has 2 bulbs. Find how much voltage each one of the bulbs has, if
each battery is 1.5 volts.
A motor with an operating resistance of 48 ohms is connected to a voltage
source. The current in the circuit is 5.6. What is the voltage of the source?
A transistor radio uses 0.6 Amps of current when a 7.0 Volt battery operates it.
What is the resistance?
What happens to the current through a 5 ohms resistor when a 10 Ohms resistor
is added in series. The voltage in this circuit is 4volts.
If you have a parallel circuit with 4 batteries and in the first loop there’s one
resistor, which is 14 ohms, in the second loop there’s two resistors one which is
10 ohms and the other is 4 ohms. Find the Equivalent resistance.
If you have a parallel circuit with one battery, which is 2volt, the first loop of the
circuit has one bulb, and the second loop has another bulb. The First bulb has a
resistance of 10 ohms and the second one has a resistance of 16. Find the power
of the circuit.
If you have a junction with 69 mA going in how many Amps should come out?
Find the equivalent resistance of this circuit:
9. Where is the most current going to go, loop 1, 2, or 3? (Use the picture for #8 to
answer)
10.
Is the current at B >, <, or = to the current at A?
11. If there is a 60v battery in a series circuit with 3 identical bulbs (3 ohms each), how
much power does each bulb have?
12. If the batteries are 2v each and the resistors are identical then what is the voltage
across one of the resistors?
13.
What is the voltage across points A and B and what is the current at point
A?
14. If you would increase the voltage of the last problem would the current at point A be
smaller or greater?
Answers
1. The first bulb in the first loop would be 6 volts; the two other bulbs in
the second loop are both 3 volts.
2. The Voltage= 268.8 Volts
3. Resistance = 4.2 Ohms
4. The current decreases
5. Equivalent Resistance= 7 Ohms
6. P= 0.65 Watts
7. 0.00069 Amps should come out
8. Req = 1.26 ohms
9. Loop 3
10. Current A = current B
11. 400 joules
12. 3 volts
13. Voltage across point A and B is 2v, the current at point A = 0.2ma
14. Greater