Ch. 7 - Electricity SPS10. Students will investigate the properties of electricity and magnetism. a....
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Transcript of Ch. 7 - Electricity SPS10. Students will investigate the properties of electricity and magnetism. a....
Ch. 7 - Electricity
Ch. 7 - ElectricitySPS10.
Students will investigate the properties of electricity and magnetism. a. Investigate static electricity in terms of friction,
induction, conduction
b. Explain the flow of electrons in terms of alternating and direct current. the relationship among voltage, resistance and current. simple series and parallel circuits.
Ch. 7 - Electricity
Ch. 7 - Electricity
I. Electric Charge I. Electric Charge Static ElectricityConductorsInsulatorsElectroscope
A. Static ElectricityA. Static Electricity
Static Electricity the net accumulation of electric
charges on an object
Electric Field force exerted by an e- on anything that
has an electric charge opposite charges attract like charges repel
A. Static ElectricityA. Static Electricity
Static Discharge the movement of
electrons to relieve a separation in charge
A. Static ElectricityA. Static ElectricityThree ways to achieve static electricity
Friction • Rubbing together two objects
Induction• Inducing a positive or negative charge
Conduction• Transferring a positive or negative charge
Demo on Static Electricity
B. ConductorsB. Conductors
Conductor material that allows electrons to move
through it easily e- are loosely held ex: metals like copper and silver
C. InsulatorsC. Insulators
Insulator material that doesn’t allow electrons to
move through it easily e- are tightly held ex: plastic, wood, rubber, glass
D. ElectroscopeD. Electroscope
Electroscope instrument that
detects the presence of electrical charges
leaves separate when they gain either a + or - charge
Ch. 7 - Electricity
Ch. 7 - Electricity
II. Electric CurrentII. Electric CurrentCircuitPotential DifferenceCurrentResistanceOhm’s Law
A. CircuitA. Circuit
Circuit closed path through
which electrons can flow
A. Potential DifferenceA. Potential Difference
Potential Difference (voltage) difference in electrical potential
between two places large separation of charge creates
high voltage the “push” that causes e- to move
from - to + measured in volts (V)
B. CurrentB. Current
Current flow of electrons through a conductor depends on # of e- passing a point in
a given time measured in amperes (A)
C. ResistanceC. Resistance
Resistance opposition the flow of electrons electrical energy is converted to
thermal energy & light measured in ohms ()
Copper - low resistance Tungsten - high resistance
C. ResistanceC. Resistance
Resistance depends on… the conductor
wire thickness• less resistance
in thicker wires
wire length • less resistance in shorter wires
temp - less resistance at low temps
E. Ohm’s LawE. Ohm’s Law
Ohm’s Law
V = I × RV: potential difference (V)
I: current (A)
R: resistance ()
• Voltage increases when current increases.• Voltage decreases when resistance increases.
E. Ohm’s LawE. Ohm’s LawA light bulb with a resistance of 160 is
plugged into a 120-V outlet. What is the current flowing through the bulb?
GIVEN:
R = 160 V = 120 V
I = ?
WORK:
V= I * RDivide both sides by R
I = V ÷ R
I = (120 V) ÷ (160 )
I = 0.75 A
E. Ohm’s LawE. Ohm’s LawA cell phone with a resistance of 50 is plugged into a 110-V outlet. What is the current flowing through the bulb?
GIVEN:
R = 50 V = 110 V
I = ?
WORK:
V= I * RDivide both sides by R
I = V ÷ R
I = (110 V) ÷ (50 )
I = 2.2 A
E. Ohm’s LawE. Ohm’s LawA Television is plugged into a 120-V outlet. How much resistance does it have if the current is measured to be 4.25 A?
GIVEN:
R = ? V = 120 V
I = 4.25
WORK:
V= I * RDivide both sides by R
R = V ÷ I
R = (120 V) ÷ (4.25 A)
R = 28.2
Ch. 7 - Electricity
Ch. 7 - ElectricityIII. Electrical CircuitsIII. Electrical Circuits
Circuit componentsSeries circuitsParallel circuitsHousehold circuits
A. Circuit ComponentsA. Circuit Components
A - battery C - light bulb
B - switch D - resistor
B. Series CircuitsB. Series Circuits
Series Circuit current travels in a single path
• one break stops the flow of current current is the same throughout circuit
• lights are equal brightness each device receives a fraction of the total
voltage • get dimmer as lights are added
C. Parallel CircuitsC. Parallel Circuits
Parallel Circuits current travels in multiple paths
• one break doesn’t stop flow current varies in different branches
• takes path of least resistance• “bigger” light would be dimmer
each device receives the total voltage• no change when lights are added
D. Household CircuitsD. Household Circuits
Combination of parallel circuits too many devices can cause wires to
overheat
Safety Features: fuse - metal melts, breaking circuit circuit breaker - bimetallic strip bends
when hot, breaking circuit
Ch. 7 - Electricity
Ch. 7 - Electricity
IV. Measuring ElectricityIV. Measuring Electricity
Electrical PowerElectrical Energy
A. Electrical PowerA. Electrical Power
Electrical Power rate at which electrical energy is converted to another form of energy
P = I × VP: power (W)
I: current (A)
V: potential difference (V)
A. Electrical PowerA. Electrical Power A calculator has a 0.01-A current flowing through it.
It operates with a potential difference of 9 V. How much power does it use?
GIVEN:
I = 0.01 A
V = 9 V
P = ?
WORK:
P = I · V
P = (0.01 A) (9 V)
P = 0.09 W
A. Electrical PowerA. Electrical Power A flashlight has a 0.05-A current flowing through it.
It operates with a potential difference of 8 V. How much power does it use?
GIVEN:
I = 0.05 A
V = 8 V
P = ?
WORK:
P = I · V
P = (0.05 A) (8 V)
P = 0.4 W
B. Electrical EnergyB. Electrical Energy
Electrical Energy energy use of an appliance depends on power required and time used
E = P × tE: energy (kWh)
P: power (kW)
t: time (h)
B. Electrical EnergyB. Electrical Energy A refrigerator is a major user of electrical power. If
it uses 700 W and runs 10 hours each day, how much energy (in kWh) is used in one day?
GIVEN:
P = 700 W
= 0.7 kW
t = 10 h
E = ?
WORK:
E = P · t
E = (0.7 kW) (10 h)
E = 7 kWh
B. Electrical EnergyB. Electrical Energy A refrigerator is a major user of electrical power. If
it uses 700 W and runs 10 hours each day, how much energy (in kWh) is used in one day?
GIVEN:
P = 900 W
= 0.9 kW
t = 14 h
E = ?
WORK:
E = P · t
E = (0.9 kW) (14 h)
E = 12.6 kWh