Electric Circuits

AP Physics Rapid Learning Series - 16
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El t i Ci itElectric Circuits
Rapid Learning Core Tutorial Series
www.RapidLearningCenter.com/
© Rapid Learning Inc. All rights reserved.
Wayne Huang, Ph.D.
Keith Duda, M.Ed.
Peddi Prasad, Ph.D.
Gary Zhou, Ph.D.
Michelle Wedemeyer, Ph.D.
Sarah Hedges, Ph.D.

Objectives
Understand and utilize
Ohms law
By completing this tutorial, you will:
Ohms law.
Calculate electric power.
Describe the characteristics
of series and parallel
circuits.
Calculate various electric
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Calculate various electric
quantities in circuits.
Learn the basic concepts
pertaining to Kirchoff’s
laws.
Concept Map
Physics
Studies
Previous content
New content Alternating
Current
Electrical
Forces
Electric Charge
Direct
Current
Described by
or
4/83
Moves in
Circuits
Ohm’s
Law
Current
Described by
Series
Circuit
and
Parallel
Circuit

Ohm’s Law
Ohm’s law describes the basic quantities
present in any electrical circuit
5/83
present in any electrical circuit.
Basic Electric Circuit
+
-
-
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In a circuit, the electrons are flowing or moving.
They are not stationary or static.

Sign Conventions
Electrons will obviously move from the negative
to the positive terminal of the battery.
However, long ago it was thought that positive
+ Conventional current
, g g g p
charge flowed. This convention remains.
7/83
Most of the time the actual direction of flow is very
unimportant.
-
actual current
Common Schematic Symbols
Conducting Wire: Switch: Light Bulb:
Battery or
Voltage Source: AC Voltage: Capacitor:
8/83
A
Ammeter:
V
Voltmeter: Resistor:

Current
Current describes the number of electrons flowing
in a circuit.
It’s very analogous to water flowing in a hose orIt s very analogous to water flowing in a hose or
pipe.
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It doesn’t count the actual number of electrons, that
would be too cumbersome.
Amperes
Current is typically measured in Amperes. (Amps for
short or A)
1 Amp = 1 Coulomb / second.
1 Amp is a relatively large amount of current, often
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p y g ,
milliamps, mA are used.
1000mA = 1 A

Direct Current
DC, Direct Current: the charge flows in one
direction only.
Examples: batteries
Conducting wire
-
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Alternating Current
AC, Alternating Current: electrons in the circuit
move in one direction, then switch and then flow in
the opposite direction.
Conducting
wire
-- -- -
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Examples: wall outlets

AC Frequency
Since AC current changes direction, or oscillates,
you can describe how often it changes direction.
Almost all US outlets use a frequency of 60 Hz.
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Adapters/Transformers
Often it is useful to convert AC into DC, or vice versa.
To save batteries,
you’ve probably
plugged a radio or
CD player into the
AC wall outlet with
the help of an AC to
14/83
p
DC adapter.

Changing Current Types
Diodes allow electrons to flow in one direction only.
They help change AC into DC. They are a one way
path for electrons.
Notice the
input/output
voltages, and
frequency.
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A diode inside
this transformer
helps change AC
into DC.
Voltage Sources
No current will flow unless there is a voltage
source. This is also known as a potential
difference.
Sometimes, this potential difference is supplied by
batteries!
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Electric Potential Analogy
Imagine a rock laying on level ground. It will not
move anywhere (no current) since there is no
difference in elevation (voltage).
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Electron Flow
However, an object at the top of a hill could roll
down (current flowing) because of the elevation
difference (voltage).
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Shocking Experience
To receive a shock, there must be a voltage
difference applied to you. (electrons must “roll”
downhill)
This is often referred to
as an electric potential
difference.
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A Bird on a Wire
A bird could sit on a high voltage wire and receive
no shock at all. Its entire body is at the same high
voltage. No voltage difference.
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However, if it touches the ground, tower, or wire
with a different voltage then there would be a large
voltage difference, and the current would flow!!!

Grounding
To prevent electric shock, most cords have a third
prong that is used to ground the cord.
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If there are any extra electrons, they are immediately
sent to the ground, not you.
Electrical Resistance
In almost all circuits, the electrons flow with some
opposition or resistance.
Resistance is measured in units called Ohms.
The symbol is the Greek letter omega: Ω
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Resistors on a
circuit board

Analogy
A hose is like wire
A pump is like voltage
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As mentioned earlier, there are many similarities
between electrical circuits, and “water” circuits.
A valve is like a switch
Ohm’s Law
Mr. Ohm discovered an extremely useful
relationship: Voltage = Current x Resistance
IRV =
Voltage, V
Resistance,
Ω
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Current,
Amperes, A

Ohm’s Law Example
A light bulb operates on a 110 volt circuit. The
bulb draws a current of .91 amps. What is the
resistance of the light bulb?
V = IR
R = V/I
R = 110V/.91A
I = 120.8 Ohms, Ω
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Internal Resistance
You may measure the potential difference of a battery
at 9V:
9V
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9 V

Voltage Difference
However, if you insert that battery into a circuit,
measure its voltage again, you may find its somewhat
lower!?!
8.5V
8.5 V
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Internal Resistance Diagram
This difference is due to the internal resistance of the
battery itself.
+-
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Once the battery is hooked into a circuit, there is
current flowing.
battery
When this current encounters even minor
resistance in the battery, there is some voltage
drop when compared to earlier.

EMF
When in a circuit, the reduced voltage is referred to
as the terminal voltage.
29/83
The potential difference when nothing is connected is
referred to as the EMF, electromotive force. This term
has nothing to do with the traditional notion of force.
Lost Voltage
When you take away the voltage lost due to the
internal resistance from the EMF, the remaining
voltage is the terminal voltage:
EMF - IRbattery = Vterminal
voltage is the terminal voltage:
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EMF - “lost voltage” = terminal voltage

Energy Conservation
Energy, or voltage, is still conserved. Some is simply
used up by the internal resistance of the battery itself.
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The actual internal resistance of most batteries is
very small, less than 1 Ohm.
Internal Resistance Example
As in the previous discussion, the 9 V battery has
a terminal voltage of 8.5 V when connected to a
100 Ω resistor. What is the internal resistance of
the battery?
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?

Internal Resistance Solution
First, it may be useful to find the current:
.085A
8.5VV
I ===
Then, use the idea of internal resistance:
EMF - IRbattery = Vterminal
.085A
100ΩR
I
8 5Vr085A9V
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8.5Vr.085A9V battery =−
5.88Ωrbattery =
.5Vr.085A battery −=−
Electric Power
Just as power was used to discuss
mechanical work per unit of time
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mechanical work per unit of time,
electrical power works the same way.

Electric Power
Just like mechanical power, electrical power
describes work done per unit of time.
1 Watt = 1 Joule / 1 second
1000 Watts = 1 kilowatt
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Power Formula
In terms of electrical quantities, power can be
calculated by multiplying current x voltage.
IVP =
Electric
Potential, V
Electric
Power
Watts, W.
36/83
Current,
Amperes, A

Alternate Power Formula
Another formula for power can be found.
Since V = IR
And P = IV
Substitute V into the power equation and obtain:
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P = I2R
Power Problem
How much current flows through a 100W light
bulb if its connected to a 120V fixture?
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Example Solution
The bulb is 100 W when it has a potential difference
of 120 V applied.
P = IV
I = P/V
I = 100W/120V
I = 0.83 Amps
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Electric Bill
Look at an electric bill.
You’ll notice that you pay $ for every
kilowatt-hour used.kilowatt hour used.
( ~ $.10 per kW hr )
Thus, for every hour you use 1000
Watts, you would pay $ 0.10.
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A kilowatt-hour is a unit of energy,
not power.

Another Power Problem
Maybe your parents are always telling you to turn
off the lights and save electricity/money? How
much would it cost to run a 100 W bulb for 4 hours
if it were connected to a 120 V fixture? ($0 10 perif it were connected to a 120 V fixture? ($0.10 per
kW hr)
100 W = .1 kW
0.1 kW x 4 hrs = 0.4 kW hrs
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0.4 kW hrs x $0.10 = $0.04!
Series and Parallel Circuits
Electrical components can be connected
in various ways This drastically changes
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in various ways. This drastically changes
the properties of the circuit.

Series Circuits
One simple way to arrange components in an
electrical circuit is to create one large continuous
loop with the components:loop with the components:
Two
Batteries
Switch
Light
Bulb
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Bulb
Resistor
This is similar to a TV series where one episode
follows another.
Series Circuit Characteristics
1. The current is constant throughout the circuit.
f2. Individual components may use varying amounts of
voltage.
3. The total voltage use is equal to the voltage of the
battery/power supply.
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4. A break in the circuit interrupts the entire circuit.

Adding Resistors in Series
When resistors are added in series, the total
resistance of the circuit is the sum of those
individual resistors.
...RRRR 321S ++=
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Series or
total
combined
resistance, Ω
Individual
resistors, Ω
Parallel Circuits
Another way to connect a circuit is in parallel. In
this arrangement, each component is connected
separately in its own “loop”.
Two
Batteries
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Three Resistors in Parallel

Parallel Circuit Characteristics
1. The current in the different branches can vary.
f f2. The total current of the circuit is the sum of the
individual branches.
3. All branches of the circuit receive the same
voltage of the battery/power supply.
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4. A break in one loop doesn’t affect the others.
Resistors in Parallel
To find the equivalent resistance of resistors
added in parallel:
...
R
1
R
1
R
1
R
1
321P
++=
48/83
Parallel or total
combined
resistance, Ω
Individual
resistors, Ω

Parallel Resistor Example
Calculate the total effective resistance of two 10
Ohm resistors connected in parallel.
111
+
21P RRR
+=
10Ω
1
10Ω
1
R
1
P
+=
21
Combine, use
common
denominator if
needed
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10Ω
2
R
1
P
=
5ΩRP =
Take the
reciprocal of
each side of
equation
Parallel Resistor Observations
5 Ohms. Notice that the total overall resistance is
lower than either one of them individually!
This occurs because there are multiple paths for
the electrons to take, lowering their resistance.
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Too Many Resistors
Adding more and more
devices in paralleldevices in parallel
decreases the total or
overall resistance.
This allows too much
current to flow! Obviously
this can be dangerous!
51/83
Combination Question
Often, when one Christmas tree light goes out, they
all go out!!! How are these type of lights wired
together?
52/83
These are wired in series. A parallel arrangement
would eliminate this…

Fuses and Circuit Breakers
To prevent too much current from causing a fire,
fuses are designed to melt and break the circuit
before that happens.
53/83
Today, most homes have circuit breakers. These
don’t melt, but are switched off to interrupt the
circuit.
MetersMeters
A multimeter can measure voltage, current and
resistance.
Meters can provide either digital or analog
readouts.
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Digital Analog

Measuring Current
Current can be measured using an ammeter or
a multimeter.
A meter that only
measures current is
called an ammeter.
55/83
Using Meters to Measure Current:
When measuring current, the meter leads
are connected in-line with the load or
voltage source.
A
56/83

Connecting a Meter for Current
Notice that the
meter leads or
probes are placed
in line with the
load. This is called
a series
ti
57/83
connection.
Using Meters to Measure Voltage
When measuring voltage, the meter leads
(probes) are placed across the load or voltage
source
V
source.
58/83

Connecting a Meter for Voltage
Notice that the meter
leads or probes are
placed across the
load. This is called a
parallel connection.
59/83
Circuit Problems
This section will detail how to calculate
the various electrical quantities in a
60/83
the various electrical quantities in a
circuit.

Series Circuit
Calculate the current and electric potential
difference for each component of the circuit
shown.
R2=10Ω
5V
61/83
R1 = 5Ω
A good first step is to simplify the circuit.
Circuit Simplification
Because this is a series circuit, to combine the
resistors and simplify the circuit, they are merely
added together.
R2=10Ω
5V 5V
Rs=15Ω
62/83
R1=5Ω
...RRRR 321S ++=
15Ω10Ω5ΩRS =+=

Current in a Series Circuit
5V
=15Ω
IRV =
V
I =5V
Rs=
Use the voltage of the power supply and the total
resistance of the circuit to find the total current
R
.33A
15Ω
5V
I ==
63/83
flowing through the circuit.
Because the electron flow has no where else to go,
this amount is also the current flowing through
both resistors. I1 and I2 is that same .33 Amperes.
Voltage in a Series Circuit
10Ω
5V
Since we know the current
flowing through each
resistor, we can use Ohm’s
R2=
R1=5Ω
5V
law to find the potential
difference for each of those
resistors.
IRV1= IRV2=
64/83
)(.33A)(5ΩV1=
1.67VV1 =
Notice how the sum of the two voltages adds up
to the power supply for the circuit.
)(.33A)(10ΩV2=
3.33VV2 =

Parallel Circuit
Calculate the current and electric potential
difference for each component of the circuit
shown.
5V
R1=5Ω R2=10Ω
65/83
Notice how this parallel circuit contains the
exact same components as the series circuit,
they are just arranged differently.
Again, a good first step is to simplify the circuit.
Circuit Simplification
5V
21P R
1
R
1
R
1
+=
R1=5Ω R2=10Ω
111
10Ω
3
R
1
P
=
10
66/83
10Ω5ΩRP
+=
10Ω
1
10Ω
2
R
1
P
+=
3.3ΩΩ
3
10
RP ==
Notice this parallel
resistance is less than
either one individually.

Shortcut Formula
An equivalent formula can be used for two
resistors, R1 and R2, connected in parallel.
Sometimes this formula is easier to manipulate.
21
21
P
RR
RR
R
+
=
67/83
It may be easier to remember this formula as the
product over the sum for the two resistors.
Voltage in a Parallel Circuit
5V
R1=5Ω R2=10Ω
The easy part about any parallel circuit is the
voltage applied to each item.
Si h it h it i d d t
68/83
Since each item has its own independent
connection to the battery or power supply, each
item receives that voltage.
In this case, V1 and V2 are each 5V.

Current in a Parallel Circuit
5V
R1=5Ω R2=10Ω
Once you realize that the electric potential for each
resistor is 5V, finding the current is easy using
Ohm’s law, V=IR.
V 2V
I
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1
1
1
R
V
I =
2
2
2
R
V
I =
1A
5Ω
5V
I1 == .5A
10Ω
5V
I2 ==
Current Observations
5V
R1=5Ω R2=10Ω
Notice the two currents add up to the same value as
the total current in the circuit. This is a good way to
check your work.
5VV
70/83
1.5A
3.3Ω
5V
R
V
I
T
T ===
1.5A.5A1AIII 21T =+=+=

Kirchhoff’s Laws
Kirchhoff’s laws describe more complex
circuits The concept is relatively simple
71/83
circuits. The concept is relatively simple,
but the application to a circuit can be a
bit complex.
Multiple Voltage Sources
Typically, Kirchhoff’s laws are used when there are
multiple voltage sources and/or loops in the circuit.
+
_
72/83
The multiple voltage sources may even
“oppose” each other.

Junction Rule
The current going into a junction (intersection) is
equal to the current leaving the junction.
8 A
4 A4 A
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This junction rule can be considered a restatement of
conservation of charge.
4 A4 A
Conducting wire
Loop Rule
The sum of the voltage changes for all elements in
a loop must equal zero.
VResistor1V
4V
74/83
The loop rule can be considered a restatement of
conservation of energy.
03V1V4V =−−+
Resistor 3V

Using the Loop Rule
• Guess (educated?) at a direction of conventional
current flow. Draw it into your diagram.
• Label ends of battery. Long end +.
current flow. Draw it into your diagram.
• Resistor voltage is – if you move in the same
direction as the current.
• Battery voltage is + if you go from the negative to
positive battery terminal.
75/83
• Finally, create an equation using the loop rule and the
path of the current.
• If the current is - , your original direction was wrong.
p y
Example
+
_10V 5 Ω
CurrentOhm units
have been
+
_
6V
10 Ω
guess
05I10V10I6V =−−−−
IRV =
have been
omitted for
clarity
Starting
point
76/83
05I10V10I6V =
015I16V- =−
1.07A
15Ω
16V
I −=
−
+
=
Current
goes
opposite
direction

Multiple Loops
When there are multiple loops, both the loop and
junction rule may be required.
You may have multiple unknowns.
You must have as many equations as unknowns in
order to solve.
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Once one variable is found, substitute it to find the
others.
Kirchhoff’s Laws Example
I1
I2
312 III +=
22 Ω+
_
9V
I2
I3
6V
15 Ω
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09V22I-15I 12 =+−Top loop:
Bottom loop: 015I6V 2 =−−
6V

Problem Solving - 2
Work with the bottom loop first since its simplest:
Amps
15
6
I2 −=015I6V 2 =−− 15
Thus, our original guess for the direction of I2 was
wrong!
Put I2 back in and find I1.
09V22I-15I 12 =+−
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12
09V22I-15(-.4) 1 =+−
0922I6 1 =+− .68AI1 =
Problem Solving - 3
With our revised current direction, we now know I1 and I2.
22 Ω+ _
I1
6V
15 Ω
I2
I3
80/83
.4A +.68A = 1.08A in the opposite direction to what
we originally thought.
I3
We can finally get I3:

Parallel
connections have
Parallel
connections have
Current is
charge per
unit of time
Current is
charge per
unit of time
Kirchhoff’s loop
and junction rule
Kirchhoff’s loop
and junction rule
Learning Summary
Series connectionsSeries connections
connections have
separate loops.
1/RP=1/R1+1/R2+…
connections have
separate loops.
1/RP=1/R1+1/R2+…
unit of time.
1Ampere =
1C/1s
unit of time.
1Ampere =
1C/1s
j
describe more
complex circuits.
j
describe more
complex circuits.
Ohm’s law:Ohm’s law:
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Series connections
are in one
continuous loop.
Rs = R1+R2+R3…
Series connections
are in one
continuous loop.
Rs = R1+R2+R3…
Ohm s law:
V = IR
Power formula:
P = IV = I2R
Ohm s law:
V = IR
Power formula:
P = IV = I2R
Congratulations
You have successfully completed
the tutorial
Electric Circuits
82/83

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