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Electric Circuits
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Do Light Bulb Demo
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Electric Circuits
There are two different types of electrical circuits.
Series
Parallel
and
More than 1 path
One Path
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Series Circuit:
Circuit in which a current flows through each
component, one after another. There is only one
path for the current to follow.
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Current
Because there is only one path for electrons to
travel in a series circuit. The current at any
one point in the circuit would be equal to the
current at any other point in the circuit.
IT  I1  I 2  I 3  ....
The current through each resistor is the same.
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Resistance
Once again because there is only one path in a
series circuit, for the case of resistance this
means that each electron will have to go
through every resister with in the circuit.
Therefore the total resistance is the sum of all
the resistors found in the circuit.
RT  R1  R2  R3  .....
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Potential Difference (Voltage)
In the picture above, the skate boarder has a drop in gravitational
potential energy at each stage during the trip.
This is very similar the drop in electric potential energy (voltage)
in a series circuit as an electron passes through a resister.
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In a series circuit the total voltage drop across the entire circuit
is equal to the sum of the individual voltage drops across each
resister.
2V
4V
6V
2  4  6  12
12V
VT  V1  V2  V3  .....
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Example:
What would the potential difference be across the second resister?
4 Volt Drop
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http://www.stmary.ws/highschool/physics/home/animations3/electricity/circuits2_bigger.swf
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Example:
Find the total current in the circuit below.
V  IR
V
I
R
RT  R1  R2  R3  ...
RT  2.0  4.0  6.0
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I
 3.3 A
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Example:
Four loads (3.0 , 5.0 , 7.0 , and 9.0 ) are connected in
series to a 12 volt battery.
a) Find the equivalent resistance of the circuit.
24 
b) Find the total current in the circuit.
0.50 A
c) Find the potential difference across the 7.0  load.
3.5 V
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Parallel Circuit:
Circuit in which there are two or more paths for the current to flow.
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Here’s A Thought
Picture walking through a crowded building, and there are that
many people that you can hardly move. Then the hallway splits into
two hallways of the same size. Half the people go one way and the
other go the other way.
Now you have much more room because there are only half as
many people in the hallway.
What would happen if the hallway was to split again? And again?
Would you have more and more room to walk each time?
If you have more room would it be easer for you to walk down the
hallway? Could you say that there is less resistance to your motion?
This is what happens within a parallel circuit, the more paths that
electrons have to chose from, the less over all resistance there is to
their motion.
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In a parallel circuit the total resistance can be found by the
following formula
1
1
1
1
 
  ......
RT R1 R2 R3
If each resister is 6 Ω, then we get
1 1 1 1
  
RT 6 6 6
1 1

RT 2
R  2
Because the current had 3 equivalent paths to chose from the
over all resistance is 1/3 of what it would have been otherwise.15
Find the total resistance of the given parallel circuit
1
1
1
1
 
  ......
RT R1 R2 R3
50Ω
1
1
1
1



RT 50 100 250
100Ω
1
5
2.5
1



RT 250 250 250
250Ω
250
RT 
 29.411
8.5
RT  30
Not how the total resistance is smaller
than any of the individual resistors.
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In a parallel circuit the total current is equal to the sum of all of
the individual currents through each branch of the circuit
3A
2A
5A
3A
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In a parallel circuit the total current is equal to
the sum of all of the individual currents through
each branch of the circuit
IT  I1  I 2  I3  .....
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12V
12V
What
In would
a parallel
the circuit
voltage
be across
the total
a resister
voltagethat
is
wasthe
hooker
sameup
asatthe
any
location
voltage
across
each
two
Hook
up between
a wire
tothe
Measure
the
voltage
wires?
individual
resistor.
each
terminal
anywhere
between
the
12V
two wires
VT  V1  V2  V3  .....
12V
Take a 12 volt battery.
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http://www.stmary.ws/highschool/physics/home/animations3/electricity/circuits2_bigger.swf
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Find the total current of the given parallel circuit
IT  I1  I 2  I3  .....
V1 10
I1 

 0.2
R1 50
V2 10
I2 

 0.25
R2 40
V3 10
I3 

 0.5
R3 20
IT  0.2  0.25  0.5
IT  0.95 A
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Solve the following circuit.
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VT  V1  V2  V3  20volts
IT  I1  I 2  I3  .....
18  4  4  I3
V1 20
I1 

 4A
R1 5
V2 20
I2 

 4A
R2
5
I 3  10 A
V3
R3 
I3
20
R3 
 2
10
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Example:
A 60 V battery is connected to four loads (3.0 , 5.0 , 12.0 ,
and 15.0 ) in parallel.
a) Find the equivalent resistance of the circuit.
1.46 
b) Find the total current leaving the battery.
41.0 A
c) Find the current through the 12.0  load.
5A
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DO
Pg. 719 #’s 27 – 31 (pdf 89)
Pg. 724 #’s 32 –35 (pdf 89)
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Complex Circuit:
A complex circuit is one that
has both series and parallel
components at the same time.
Despite their name, the method
for working with complex
circuits is quit easy. We
simplify the circuit into a
smaller less complicated circuit,
one section at a time.
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The first thing we see here is 3 different series circuits (which
can be simplify) embedded within the complex circuit.
Now we have a
plane parallel
circuit that we can
easily work with.
Example:
Find the equivalent resistance of the circuit below.
Step 1: Combine the 2 resisters that are in parallel.
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1
1
1
2
1




RT 40 40 40 20
RT  20
Step 2: Combine the 3 resisters that are in series.
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RT  8  20  12  40
Step 3: Combine the 2 resisters that are in parallel.
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1
1
1
2
1




RT 40 40 40 20
RT  20
Step 4: Combine the 2 resisters that are in series.
RT  20  20  40
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DO
Pg. 728 #’s 36 & 37 (pdf 89)
Pg. 733 #’s 1-7 (pdf 89)
RRHS Handout
Do Your Ohm Work
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