principles of technology waxahachie high school
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Principles of Technology Waxahachie High School. Energy in Electrical Systems PIC Chapter 5.3. PT TEKS. Energy in Electrical Systems. Objectives : Describe a capacitor. Explain how a capacitor stores energy. Define capacitance. Calculate the electrical energy stored in a capacitor. - PowerPoint PPT PresentationTRANSCRIPT
Principles of TechnologyWaxahachie High School
Principles of TechnologyWaxahachie High School
Energyin
Electrical Systems
PIC Chapter 5.3
Energyin
Electrical Systems
PIC Chapter 5.3
PT TEKS PT TEKS
Energy in Electrical SystemsEnergy in Electrical Systems
Objectives:
Describe a capacitor.Explain how a capacitor stores energy.Define capacitance.Calculate the electrical energy stored in a capacitor.Describe an inductor.Explain how a conductor stores energy.Calculate the electrical energy stored in an inductor.
Objectives:
Describe a capacitor.Explain how a capacitor stores energy.Define capacitance.Calculate the electrical energy stored in a capacitor.Describe an inductor.Explain how a conductor stores energy.Calculate the electrical energy stored in an inductor.
Energy in Electrical SystemsEnergy in Electrical Systems
A capacitor is an electrical device that stores energy in an electric field.
A power supply removes electrons from one plate in a capacitor and deposits them on a second plate.
A capacitor is an electrical device that stores energy in an electric field.
A power supply removes electrons from one plate in a capacitor and deposits them on a second plate.
Energy in Electrical SystemsEnergy in Electrical Systems
This causes the plate that loses electrons to become positively charged and the plate that gains electrons to become negatively charged.
The work done to create this electric field is equal to the potential energy stored in the field.
This causes the plate that loses electrons to become positively charged and the plate that gains electrons to become negatively charged.
The work done to create this electric field is equal to the potential energy stored in the field.
Energy in Electrical SystemsEnergy in Electrical Systems
Equation for Potential Energy in an electric field =
½ x charge x Voltage
W = ½ q V
Equation for Potential Energy in an electric field =
½ x charge x Voltage
W = ½ q V
Energy in Electrical SystemsEnergy in Electrical Systems
If the charge is 10 C (coulombs) and the change in Voltage is 6 V, what is the work done?
W = ½ q x V
W = .5 x 10 x 6
W = 30 J
If the charge is 10 C (coulombs) and the change in Voltage is 6 V, what is the work done?
W = ½ q x V
W = .5 x 10 x 6
W = 30 J
Energy in Electrical SystemsEnergy in Electrical Systems
The charge on either plate of a capacitor is equal to the Capacitance x voltage
Capacitors have fixed values of capacitance.
The charge on either plate of a capacitor is equal to the Capacitance x voltage
Capacitors have fixed values of capacitance.
Energy in Electrical SystemsEnergy in Electrical Systems
The symbol C is used to represent capacitance.
Charge of a plate in a capacitor = Capacitance x voltage
q = C x V
The unit for capacitance is the farad (F).
1 farad = 1 coulomb per volt, or 1 F = 1 C/V
The symbol C is used to represent capacitance.
Charge of a plate in a capacitor = Capacitance x voltage
q = C x V
The unit for capacitance is the farad (F).
1 farad = 1 coulomb per volt, or 1 F = 1 C/V
Energy in Electrical SystemsEnergy in Electrical Systems
If the Capacitance is .0005 F (500 µF) and the Voltage is 200 V, what is the charge?
q = C x V
q = .0005 F x 200 V
q = .1 C
Most capacitors are measured in microfarads (µF) or picofarads (pF)
1 µF = 10-6 F 1pF = 10-12 F
If the Capacitance is .0005 F (500 µF) and the Voltage is 200 V, what is the charge?
q = C x V
q = .0005 F x 200 V
q = .1 C
Most capacitors are measured in microfarads (µF) or picofarads (pF)
1 µF = 10-6 F 1pF = 10-12 F
Energy in Electrical SystemsEnergy in Electrical Systems
Potential energy in a capacitor =
½ (Capacitance) (Voltage)2
PE = ½CV2
Potential energy in a capacitor =
½ (Capacitance) (Voltage)2
PE = ½CV2
Energy in Electrical SystemsEnergy in Electrical Systems
If the Capacitance is .0005 F and the Voltage is 200 V, what is the Potential Energy stored?
PE = ½CV2
PE = .5 (.0005)(200)2
PE = .5 (.0005)(40,000)
PE = 10 J
If the Capacitance is .0005 F and the Voltage is 200 V, what is the Potential Energy stored?
PE = ½CV2
PE = .5 (.0005)(200)2
PE = .5 (.0005)(40,000)
PE = 10 J
Energy in Electrical SystemsEnergy in Electrical Systems
If an electric charge is moving, it will create a magnetic field.If an electric charge is moving, it will create a magnetic field.
Energy in Electrical SystemsEnergy in Electrical Systems
Magnetic field lines with the electric current in the center:
Magnetic field lines with the electric current in the center:
Energy in Electrical SystemsEnergy in Electrical Systems
If two currents are in the same direction, the wires are attracted.If two currents are in the same direction, the wires are attracted.
Energy in Electrical SystemsEnergy in Electrical Systems
If two currents are in opposite directions, the wires are repelled.If two currents are in opposite directions, the wires are repelled.
Energy in Electrical SystemsEnergy in Electrical Systems
The Earth’s magnetic field is caused by currents in the molten iron core found at the center of the earth.
The Earth’s magnetic field is caused by currents in the molten iron core found at the center of the earth.
Energy in Electrical SystemsEnergy in Electrical Systems
The Earth’s Magnetic Field causes the Northern Lights by attracting electrically charged particles given off by the sun.
The Earth’s Magnetic Field causes the Northern Lights by attracting electrically charged particles given off by the sun.
Energy in Electrical SystemsEnergy in Electrical Systems
A magnetic field can also create an electric field. This is called electromagnetic induction (EMF).
A magnetic field can also create an electric field. This is called electromagnetic induction (EMF).
Energy in Electrical SystemsEnergy in Electrical Systems
If a loop of wires moves through a magnetic field, a current is produced in the wire.
If a loop of wires moves through a magnetic field, a current is produced in the wire.
Energy in Electrical SystemsEnergy in Electrical Systems
To increase the electricity created by a magnetic field:• Use a larger magnet• Use a coil (wire) with a larger diameter• Use more turns of the wire• Move the magnet faster
To increase the electricity created by a magnetic field:• Use a larger magnet• Use a coil (wire) with a larger diameter• Use more turns of the wire• Move the magnet faster
Energy in Electrical SystemsEnergy in Electrical Systems
A capacitor stores energy in an electric field and an inductor is an electrical device that stores energy in a magnetic field.
A capacitor stores energy in an electric field and an inductor is an electrical device that stores energy in a magnetic field.
Energy in Electrical SystemsEnergy in Electrical Systems
The amount of energy that an inductor can hold is called inductance.
Use the symbol L to represent Inductance and the unit is the Henry.
Equation for EMF =
-inductance (current/time)
EMF = -L (I/t)
The amount of energy that an inductor can hold is called inductance.
Use the symbol L to represent Inductance and the unit is the Henry.
Equation for EMF =
-inductance (current/time)
EMF = -L (I/t)
Energy in Electrical SystemsEnergy in Electrical Systems
A coil with an inductance of .0045 Henrys and a current of 1000 amps for 2 seconds. What is the EMF?
EMF = -L (I/t)
EMF = -.0045 (1000/2)
EMF = -.0045 (500)
EMF = -2.25 V
A coil with an inductance of .0045 Henrys and a current of 1000 amps for 2 seconds. What is the EMF?
EMF = -L (I/t)
EMF = -.0045 (1000/2)
EMF = -.0045 (500)
EMF = -2.25 V
Energy in Electrical SystemsEnergy in Electrical Systems
Potential energy in inductors (Equation) =
½ inductance (current)2
PE = ½ L I2
Potential energy in inductors (Equation) =
½ inductance (current)2
PE = ½ L I2
Energy in Electrical SystemsEnergy in Electrical Systems
A coil with an inductance of 8 Henrys and a current of 15 A has how much PE is stored?
PE = ½ L I2
PE = ½ (8) 152
PE = ½ (8)(225)
PE = 900 J
A coil with an inductance of 8 Henrys and a current of 15 A has how much PE is stored?
PE = ½ L I2
PE = ½ (8) 152
PE = ½ (8)(225)
PE = 900 J