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Unit 12
Electricity and RC Circuits
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Conduction Electrons
In organic compounds, electrons are bound to specific atoms.
In metallic compounds, some of the electrons are not bound to a specific
atom.
They are free to move throughout the metal.
These electrons are called conduction electrons.
If a potential difference is placed across the wire (like when you connect
the wire to a battery), then the electrons will move.
As they move, the electrons collide with the metallic atoms.
Depending upon the number of collisions an electron has, it may move
faster or slower through the metallic structure.
Remember, moving electrons in a wire are known as current.
12-1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Batteries
Batteries come in many shapes and sizes that have a variety of
voltage and currents.
Identify the voltages of the four common battery types shown below.
12-1A
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Batteries
This slide will explain how a battery provides electricity for use in your small
electrical appliances.
Free electrons and “holes,” which are the absences of electrons, are produced
within the battery due to electrochemical reactions.
12-2
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Capacitors
Like batteries, capacitors come in a wide assortment
of shapes.
A capacitor acts as a buffer to temporally store excess
charges.
At other times a capacitor acts as a short lived battery
in order to provide additional current (electrons) when
the need is larger than a battery can provide.
Basically, a capacitor consists of two parallel plates.
One of these has a net positive charge while the other
has a net negative charge.
In the next slide we will take a closer look at the
operation of a capacitor.
12-3
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Capacitors
Observe the capacitor acting in the capacity of a short lived battery.
12-4
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Resistors
Resistors impede the flow of electrons
(current).
They impede this flow because certain
electrical items have maximum limitations on
the current they can handle.
Consider the Light Emitting Diode (LED) in
the figure to the right.
The battery supplies too high a current to the
LED.
As a result, the LED is damaged by the
current.
When a resistor is placed into the circuit, the
current is reduced to a level appropriate for use
with the LED, and the LED is not damaged.
12-5
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Batteries and Current
A complete circuit is one that connects a battery to an electrical component
back to a battery.
Remember, current flows from the negative end of a battery through the
light bulb (resistor) and back into the positive end of the battery.
The symbol “I” is used to denote current.
Current is the number of electrons passing a certain point in a circuit per
unit of time.
We draw an arrow in the direction that the current would flow through the
circuit.
12-6
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Multiple Batteries and Current
Depending on how they are placed,
multiple batteries in a circuit can
enhance or impede the current in a
circuit.
Consider the circuit below. The
current flowing through both batteries
travels in the same direction.
As a result, the net current is enhanced
in the circuit.
Now, let us flip one of the batteries.
When we flip the battery, its current
now acts in the opposite direction.
As a result, it impedes the current.
Since both batteries are the same, no
net current flows, and the bulb goes out.
I1
I2
12-7
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Multiple Batteries and Current
In this figure the batteries are of
different voltages
As in the previous figure, when the
batteries are connected in the same
way, the net current is enhanced.
When one battery is flipped, the net
current is impeded.
Determine the net current direction
and the net voltage of the circuit
below.
The key to determining the net
current direction lies in considering the
voltages of the batteries.
The net current will flow in the
direction of the current belonging to
the battery with the highest voltage.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
I2
I1
12-8
WS ???? #2
Determine the voltage and the net current direction in the circuits
below.
- +
- +
+ -
+ 6 Volt
6 Volt
6 Volt
6 Volt
12-9
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Circuit Configurations: Series Resistors
Look at the
drawing animation
below.
The skateboarders
leaving the school
only have one place
to go.
They also only
have one place to
return to.
Their round trip
occurs in a series
circuit.
12-10
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Circuit Configurations: Series Resistors
Now lets replace the
school building with a
battery.
Turn the house and
restaurant into a light
bulb (resistor).
Change the roads into
wires.
Now let some
electrons flow through
the circuit.
They have only one
destination: through
the resistor and back
the battery.
12-11
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Circuit Configurations: Multiple Series Resistors
The lights below are in series with each other because the same current flows
through all three of them.
Notice how the first bulb is brighter than each consecutive bulb thereafter.
This is because the electric potential (voltage) drops as it passes through the
resistor.
As a result, there is less voltage available for the next resistor.
I1
12-12
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Circuit Configurations: Parallel Resistors
In a parallel circuit,
the skateboarders have
a choice of going home
or to the Big Burger for
lunch.
They make this
choice at the
intersection.
When there is an
intersection resulting in
a choice between two
or more directions, the
circuit is a parallel
circuit.
12-13
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Circuit Configurations: Parallel Resistors
Now let us make the
appropriate replacements in
this circuit and turn it into an
electrical circuit.
Now allow the current to
flow.
Notice how more electrons
flow through the small bulb.
This action happens
because the small bulb has a
lower resistance than the
large bulb.
We will look at resistance
more in the next slide.
12-14
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Resistance
Lets consider the circuit to
now have two ramps.
Most unskilled
skateboarders would choose
the lower ramp.
As a result, we could say
that the lower ramp has a
lower resistance.
Now let us make the
appropriate replacements.
In this replacement, we
replaced the ramps with
actual resistors instead of
with light bulbs.
As you can see, more
electrons flow through the
orange resistor because it has
a lower resistance.
12-15
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Circuit Configurations: Multiple Parallel Resistors
The circuit to the right shows
three parallel resistors (bulbs) that
are in series with a fourth resistor.
When we close the top switch, the
two lighted bulbs are in series, and
the top bulb is slightly brighter than
the bottom bulb.
They share equal currents.
What do you think will happen
when we close the middle switch?
Notice that the top two bulbs were
equally dimmed while the bottom
bulb remained the same.
The currents through the top bulbs
are equal but half that through the
bottom bulb.
A similar result is observed when
the bottom switch is closed.
I1
- +
12-16
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A Closer Look – Series Resistors
When the switch to the right is
closed, which bulb will be brighter?
Why?
The voltage drops across R1
making R2 less bright than R1.
If we remove R2, then what will
happen to R1?
Why?
The same current flows through
both light bulbs (resistors).
If you remove one of these series
resistors, then the current can not
flow, and the bulbs go out.
R1
- +
R2
12-17
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A Closer Look – Parallel Resistors
Note: R2 = 2.0  and R1 = 1.0 
When the switch to the right is
closed, which bulb will be brighter?
Why?
R1 will be brighter because more
current will flow through it due to the
fact that it has less resistance.
If we remove R1, then what will
happen to R2?
Why?
R2 will get brighter because all of
the current now passes through R2
instead of being split between R1 and
R2.
R1
- +
R2
6 Volt
12-18
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
A Closer Look – Combination Resistors
Note: R4 = 4.0 , R3 = 3.0 , R2
= 2.0 , and R1= 1.0 .
When the switch to the right is
closed, which bulb will be brighter?
Why?
All of the current flows through R1
before it splits after R1.
What will happen to R4 if we
remove R1?
Why?
All of the bulbs go out because the
electrons can not flow through R1.
What would happen to R1, R3, and
R4 if we removed R2?
Why?
What would happen to R2 and R1 if
we removed R4?
Why?
R4
R1
R2
- +
R3
12-19
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Schematic Symbols
When engineers design
electrical circuits, they
replace actual pictures of
electrical components
with schematic symbols.
Here are five electrical
components we will use
frequently.
The schematic symbol
for each of these
electrical components is
as follows.
Here are some other
symbols you must be
familiar with.
V
A
12-20
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Parallel and Series Electrical Configurations
There are two basic electrical configurations: series and parallel.
In a series connection, all electrical components share the same current.
In a parallel connection, the current through each component varies depending
upon the components resistance.
Let us take a look at the schematic diagrams for the circuits below.
In the lower left picture, the two resistors are series.
Note that we can move one of the resistors any where in the circuit while
maintaining the same current through each.
In the lower right picture the resistors are parallel.
Again, we may move one resistor and still have a parallel circuit.
The resistors do not need to be geometrically parallel in order to be electrically
parallel.
12-21
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Equivalent Circuits
It is often desirable to reduce numerous electrical resistors in a circuit to an
equivalent circuit with fewer resistors.
In a series circuit, the series resistors may be replaced with a single resistor with
the equivalent resistance to that of the ones it replaced.
On paper, simply redraw the circuit with only one resistor in the place of the two
(or more) you are replacing.
The same concept holds true for parallel resistors.
The exact same procedure is followed when doing equivalent circuits with
capacitors instead of resistors.
R1Req R2
R1eq
R2
12-22
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Series Test
Series Component Test – must be
able to go from only one side of a
component to only one side of an
equivalent component without
passing an intersection or a
nonequivalent component.
An intersection kills a series
possibility.
Are R4 and R2 Series?
Are R5 and R3 Series?
Are R4 and R5 Series?
Are C2 and C1 Series?
Are C3 and C4 Series?
R3
R5
C4
C3
C2
R1
R2
R4
C1
12-23
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Parallel Test
Parallel Component Test – must be able to go
from both sides of a component to both sides of
an equivalent component without passing through
a nonequivalent component.
An intersection has no impact on a parallel
circuit possibility.
Are R3 and R1 Parallel?
Are R4 and R2 Parallel?
Are C2 and C1 Parallel?
R3
R2
C3
C2
R1
R4
C1
12-24
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Series & Parallel Test
The circuit to the right is a very
complicated RC (Resistor-Capacitor) circuit.
Let us apply the tests for series and parallel
circuits in order to reduce the circuit to its
simplest form.
Always replace series components with
their equivalent series component before
attempting to replace parallel components.
Do you see any series components?
Once all series components are replaced,
proceed to replace any parallel components
that may remain.
Are the remaining resistors series?
Why or why not?
Once there are no longer any series or
parallel components, the circuit is reduced as
far as possible.
12-25
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Equivalent Resistance for Series Circuits
When you replace series resistors with an equivalent resistance, you must calculate
the value of the new resistor.
When replacing two series resistors with an equivalent resistance, use the
following formula in calculating the equivalent resistance.
Req  R1  R2
If you are replacing many resistors, use the following formula in order to calculate
the equivalent resistance.
Req  R1  R2  ...
R1eq
R2
12-26
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Equivalent Resistance for Parallel Circuits
When you replace parallel resistors with an equivalent resistance, you must
calculate the value of the new resistor.
When replacing two parallel resistors with an equivalent resistance, use the
following formula in calculating the equivalent resistance.
1
1
1


Req R1 R2
If you are replacing many resistors, use the following formula in order to calculate
the equivalent resistance.
1
1
1
R1


 ...
Req R1 R2
Req
R
2
12-27
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Equivalent Capacitance for Series Circuits
When you replace series capacitors with an equivalent capacitance, you must
calculate the value of the new capacitor.
When replacing two series capacitors with an equivalent capacitance, use the
following formula in calculating the equivalent capacitance.
1
1
1
 
Ceq C1 C2
If you are replacing many capacitors, use the following formula in order to
calculate the equivalent capacitance.
1
1
1


 ...
Ceq C1 C2
C1eq
C2
12-28
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Equivalent Capacitance for Parallel Circuits
When you replace parallel capacitors with an equivalent capacitance, you must
calculate the value of the new capacitor.
When replacing two parallel capacitors with an equivalent capacitance, use the
following formula in calculating the equivalent capacitance.
Ceq  C1  C2
If you are replacing many capacitors, use the following formula in order to
calculate the equivalent capacitance.
C2
Ceq  C1  C2  ...
C1eq
12-29
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
WS ??? # 1 and WS ??? #1
Find the equivalent capacitance or resistance for the circuits in the following
problems.
WS ??? # 1. If C1 = 5.0 F, C2 = 25.0 F, and C3 = 9.5 F, then Ceq = ____?
WS ??? # 1. If R1 = 5.0 , R2 = 25.0 , and R3 = 9.5 , then Req = ____?
C1
R1
R3
C2
R2
C3
Ceq  C1  C2
1
1
1
 
Ceq C1 C2
Req  R1  R2
1
1
1


Req R1 R2
12-30
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
WS ??? # 1 and WS ??? #1
Find the equivalent capacitance or resistance for the circuits in the following
problems.
WS ??? # 1. If C1 = 5.0 F, C2 = 25.0 F, and C3 = 9.5 F, then Ceq = ____?
WS ??? # 1. If R1 = 5.0 , R2 = 25.0 , and R3 = 9.5 , then Req = ____?
Ceq  C1  C2
C1
1
1
1
 
Ceq C1 C2
R1
R2
Req  R1  R2
C3
1
1
1


Req R1 R2
C2
R2
12-31
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
WS ??? # 4
Find the equivalent capacitance or resistance for the circuits in the following
problems.
If R1 = 5.0 , R2 = 25.0 , and R3 = 9.5 , then Req = ____?
R1
R2
Req  R1  R2
Ceq  C1  C2
1
1
1


Req R1 R2
1
1
1
 
Ceq C1 C2
R1
R2
R3
12-32
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
WS ??? # 4
Find the equivalent capacitance or resistance for the circuits in the following
problems.
If C1 = 12.0 F, C2 = 25.0 F, C3 = 5.0 F, and C4 = 1.5 F, then Ceq = ____?
C1
C2
Req  R1  R2
Ceq  C1  C2
C3
1
1
1


Req R1 R2
1
1
1
 
Ceq C1 C2
C4
C4
12-33
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Resistor/Capacitor Circuits WS ??? # 2
Reduce the circuit below as far as possible
by finding the equivalent capacitance and
resistance. Draw the final circuit for each
problem. C1 = 22.8 F, C2 = 2.3 F, C3 = 5.9 F, C4
= 5.0 F, R1 = 2.2 , R2 = 14.8 , R3 = 9.5 ,
and R4 = 12.0 .
R1
R2
C1
R3
C3
Req  R1  R2
Ceq  C1  C2
1
1
1


Req R1 R2
1
1
1
 
Ceq C1 C2
C2
12-34
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Ohm’s Law
Ohm’s Law gives us a mathematical expression relating the voltage (V), Current
(I), and Equivalent Resistance (R) of a circuit.
V  I Req
Previously, we reduced the circuit below to its equivalent resistance.
If we do not know the voltage but we do know the current and the resistance, then
we can use the equation to find the voltage.
V  I Req
If we know the voltage of the battery and the resistance, then R
we can find the
1
current flowing through the resistor.
V
I
Req
Req
R
2
12-35
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Ohm’s Law
You can use Ohm’s Law to calculate the current through a resistor if you know the
voltage across the resistor and the resistance of the resistor.
Consider the parallel circuit below.
Suppose the voltage (V) of the battery in both circuits below is 10.0 V.
Since both sides of the battery are connected to both sides of both resistors, the
voltage across both resistors would be 10 volts.
However, they would not have the same current because the current splits before it
reaches the resistors.
You can use Ohm’s law in order to find the current through these resistors.
I  I1  I 2
V
I1 
R1
I1
R1
I1
I2
R2
I2
I
V
I
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
V
I2 
R2
12-36
Ohm’s Law
Both sides of the resistors below are not connected to both sides of the
battery.
As a result, they do not have the same voltage across them.
However, as these resistors are in series, they share the same current.
The voltage drop across R1 (from A to B) is given by the equation below
right (Ohm’s Law).
The voltage would also drop across R2 and can be calculated with the
equation below left.
VBC  I R 2
C
R2
B
V
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
R1
I
VAB  I R1
A
12-37
Ohm’s Law Example 1
R1 = 10.0 , R2 = 20.0 , and R3 = 30 .
What is the current through the circuit below?
What are the voltage drops through R1, R2, and R3?
I1
R1
R2
R3
12-38
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Ohm’s Law Example 2
R1 = 10.0 , R2 = 20.0 ,
and R3 = 30 .
What is the current through
each resistor in the circuit
below?
What are the voltage drops
through R1, R2, and R3?
R1
R2
- +
R3
12-39
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Power
 Once we know both the current and the voltage across a resistor, we can
determine the power consumed by that resistor.
 The power consumed may be determined using the following equation.
P  VI
 The power consumed in R1 in both circuits below may be determined as follows.
P1  VI1
V
I1 
R1
V 
P1  V  
 R1 
V2
P1 
R1
VAB  I R1
P1  VAB I
I1
R1
I1
I2
R2
I2
I
V
I
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
P1   IR1  I
P1  I 2 R1
C
R2
B
V
R1
I
A
12-40
Batteries and emf
So far we have considered batteries as perfect sources of electrons meaning that all the
electrons produced by the electrochemical reactions inside the battery are delivered to the
electrical circuit.
However, this statement is not true because of the fact that the materials the battery is
made of resist the flow of electrons produced by the battery itself.
We call this resistance the internal resistance (r) of the battery.
This internal resistance, when multiplied by the current flowing through the battery,
reduces the electric potential (voltage) the battery can deliver to the circuit.
Consider the D-Cell battery below that provides a voltage (V) of 1.5 V to a circuit.
The actual electrical potential produced by the battery known as the electromotive force
(emf or ) is larger than the voltage V delivered.
The relationship between the emf and the battery voltage is given in the below equation
where I is the current produced by the battery.
The schematic symbols for a battery with and without internal resistance is given
below.
V    Ir
V

V
r
12-41
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Electro Motive Force (emf)
The V below is the ideal value of the voltage across the battery (i.e. 9 Volts).
Vab  IR
Although the potential difference across the terminals of a battery is V when no
current is flowing, the actual potential difference is reduced when a current is
flowing through the battery due to internal resistance (r)in the battery.
The actual potential difference is known as the electromotive force, .
When we consider the internal battery resistance, the figure below would change to
look as follows.
The electromotive force may be found by using the following equation.
Vab    Ir  IR
R
The internal battery resistance
is treated like a regular resistor,
R, when doing calculations.
A
B
 V r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
12-42
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