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Electric Circuits
Electric Circuit
•
•
1.
2.
Is a closed path for the flow of electrons.
Consists of:
Source of electricity
Wires to conduct the flow of electrons
(electric current)
3. Objects (resistors or loads) along the
path that require electricity to operate
(ex. Lamps)
Current
• Current: the movement of negative
charge (flow of electrons). The
amount of charge that ‘flows” past a
certain point in a conducting wire every
second.
• If you were to describe a current of
water you might state the amount of
litters of water that flows past a point in
a pipe in one minute. (watch)
•Current/Water analogy
Watch
Current
•
•
•
•
•
•
One coulomb of charge is equal to the
amount of charge in 6.25x1018 electrons
(6.2 billion billion electrons).
The symbol for current is I
Current is measured in amperes A
The symbol for charge is Q
Charge is measured in coulombs C
Time (t) is usually measured in seconds s
Current
•
The mathematical relationship among these
variables is:
Q
I
t
Amount of current=amount of charge moving
past a point
time
Current
•
A current of 2.0 A means that 2.0 C of
charge is moving past a point in the
circuit every second.
I=Q
t
2.0 A= 2.0 C
1s
Calculations
• Example 1
If 350 Coulombs (C) of charge pass a point
in a conductor in 20 seconds, what is the
electric current (I) through that point?
Formula: I=Q/t
I= ?(unknown)
Q= 350C
t=20 s
I=Q / t
I=350C / 20s
I=17.5 A
There are 17.5 Amperes of
current passing though that
point.
Calculations
• Example 2
If 200 C of charge pass a point in 3 minutes,
what is the electric current through that
point?
• What is your first step?
• Covert minutes to seconds
Calculations
------------- x 60 s = 180s
• 3 minutes
1 --------min
I= ?
Q= 200 C
t= 180 s
that
I=Q/t
I= 200/180s
I= 1.1A
There are 1.1 amps of
current passing though
point.
Review
• Read p.324-325; 328-329
• Questions p.329 (1-5)
Electricity on the Move
Source of Electricity:
• You can think of a battery as an object
that has energy to make electrons move
around a circuit.
Energy
Electrical Potential Energy
•
•
The purpose of a battery in an electric
circuit is to provide energy to move the
negative charge (electrons) through
the conductors in the circuit.
All Energy is measured in Joules (J)
Batteries
•
•
•
Chemical reactions occur in a battery that result in buildup of
electrons in the negative terminal
Electrons in the battery move to the other terminal making it
positive.
These “energized” electrons now have the ability to do work
on something.
Potential Energy
•This electrical energy stored
in a battery is called Potential
Energy. (This is the driving
force responsible for the
moving charges in a circuit.)
•In order to actually do work
the battery must be connected
to an object (load or resistor)
and the circuit must be
complete. (See page 330,
figure 10.3 a & b)
Potential Energy
The units for potential energy per unit of charge
are:
J (Joules)
C (Coulomb)
or
Volt (V)
Another name is given to J/C. It is called a Volt (V)
Formula
• Voltage can be looked at as how much energy is
carried per unit charge
• Voltage (potential energy) = Energy / charge
E
V
Q
Potential difference = energy/charge
Potential Difference
• When describing energy in a circuit we speak of
potential difference.
• Potential Difference: the difference in potential
energy per coulomb charge at one point in a circuit
compared to another. This is measured in Volts (V).
• For example: If one coulomb of charge at one point
in a circuit has one more joule, of potential energy,
than at another point in the circuit the potential
difference is one volt.
1C+ 1J
vs.
1C + 2J = Potential Difference of 1 V
• Using a voltmeter, you can measure voltage by
looking at the difference in energy between two
points on the circuit
Example
• In a battery, 45J of chemical energy are converted into
electrical energy by separating positive and negative
charges. This energy places 15C of charge at the
negative terminal, leaving a deficit at the positive
terminal. What is the potential difference between the
two terminals of the battery?
E
Given
V=
Q
• E= 45 Joules
V = 45J = 3.0 V
• Q= 15 Coulombs
15C
A battery that uses 45J of
• V=?
chemical energy to separate
15C of charge generates a
potential difference of 3V.
Example 2
• Within a battery, 180 J of chemical energy are converted
into electrical potential energy. This amount of energy
produces 30 C of negative charge (electrons) at the
negative terminal, and a deficit of electrons at the
positive terminal. What is the potential difference
between the negative and positive terminal of the
battery?
Given
• E= 180 Joules
• Q= 30 Coulombs
• V=?
E
V= Q
V=
180 J
30C
= 6.0 V
A battery that uses 180J of
chemical energy to separate
30C of charge generates a
potential difference of 6.0V.
Review
• Read pages 330 to 336
• Answer questions 1 to 5 on page 336.
Resistance
• When electrons move through a conductor
the atoms resist the flow of electrons.
• Resistance is the property of a substance
which indicates how much that substance will
interfere with the flow of electrons
• Example: The resistance of Tungsten
(filament in light bulbs) is 400 times greater
than copper wire.
Resistance
• There is a mathematical relationship between
resistance potential difference and current.
V
R
I
resistance = potential difference / current
•The unit for resistance is V/A or Ohm (Ω)
Ohm’s Law
• The scientist Ohm did experiments that found
resistance was always the same no matter how
much voltage (potential difference) was placed
on them.
V  I R
Potential difference = current x resistance
•An ohmic resistor has constant resistance. Many
electrical appliances are NOT ohmic
Example
• What is the resistance of a heating coil if a current
of 10.0 A goes through it when connected to a wall
outlet providing a potential difference or voltage of
120 V?
V = 120V
V
R
I
I = 10.0 A
V 120V
V
R 
 12.0  12.0
I 10.0 A
A
Example 2
• What is the current running through a light bulb
having a resistance of 5.0 Ω (V/A) when the
potential difference across the bulb is 12.0 V?
V = 12.0 V
V
R
I
R = 5.0 Ω (V/A)
V 12.0V
I 
 2.4 A
R 5.0
Power
• Power is defined as energy per unit time.
• Electrical Power is defined as the amount of
electrical energy that is converted (into light,
sound, heat or motion) every second.
• The mathematical equation for power (P) is:
E
P
t
Power = energy / time
Power
• The unit for power, Joules/second, is called a
watt W.
• Example: when sixty joules of electrical energy is
converted into light and heat by a light bulb every
second.
• The bulb has a power of 60 watts (W)
• It’s not convenient to talk about power in electric
circuit this way so another equation has been
developed. (page 344)
Power
• The new mathematical equation for power (P) is:
P  I V
Power = current x potential difference (voltage)
•Power gives us information about how many joules of
electrical energy are being converted from electrical energy
into another form of energy every second.
Example
• A current of 0.83 A passes through a light bulb
which is connected to a 120 V wall outlet. What is
the power of the light bulb??
V = 120V
I = 0.83 A
P  IV
P  IV  0.83A120V  100W
Power Rating
• It’s useful to know how much energy
an electrical device would use in a
certain amount of time.
• To calculate this we use a Power
Rating
• Many electrical devices, like a light
bulb have this stamped on them.
This stamp tells us that the light bulb has a power rating of
60 W, 14 W, or 12.5 W when connected to a 120 V wall
outlet
Power Rating
• If we want to calculate how much energy an
electrical device uses, we would multiply the
power (watts) by time (seconds).
E  Pt
Energy (joules) = power (watts) x time (seconds)
Energy Efficiency
•There is no electrical device that converts all of the electrical
energy going into it , into the energy the electrical device is
producing.
•Not all electrical energy going into a light bulb is converted to light.
•The efficiency of an electrical device can be calculated but using
the following equation:
Percent efficiency of electrical device = (useful energy
output/total electrical energy input) x 100%
•The useful energy output of a lamp is the amount of energy that a
lamp actually converts to light.
Example
• An electric kettle has power rating of 1000 W. it takes this
kettle 3.5 min to heat up 600ml of water from 22.0 °C to
100.0 °C. This required 196 000 J of energy to heat the
water. What is the efficiency of the kettle?
• The energy used by the kettle is E  Pt
P = 1000 W
E  Pt
t = 3.5 min = 210s
E  1000W  210s
E  210000 J
Percent efficiency = (196 000 J/210 000 J) x 100%
Percent efficiency = 93.3 %
Review
• Read p.337 – 339; 342-346; 358
• Questions p.342 (1-6)
»p.348 (1-6)
»p.350 (15, 16; 21-22; 24-30)