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
ELEMENTS
Atoms - Smallest piece of an element containing all of
the properties of that element
The center portion of
an atom containing the
protons and neutrons
Positively charged
atomic particles
Uncharged atomic
particles
The atomic number is
equal to the number of
protons in the nucleus
of an atom.
The atomic number
identifies the element.
How many
protons are in
this nucleus?
Negatively charged
particles
Orbits in which
electrons move around
the nucleus of an atom
The outermost ring of
electrons in an atom
3D2D
Orbit
Number
Maximum
Electrons
1 2
2
3
4
5
6
Valence
Orbit
2
72
32
8
Orbits closest to the nucleus fill first
18
50
8
Conductors Insulators
Electrons flow easily
between atoms
1-3 valence electrons in
outer orbit
Examples: Silver,
Copper, Gold, Aluminum
Electron flow is difficult
between atoms
5-8 valence electrons in
outer orbit
Examples: Mica, Glass,
Quartz
•
•
•
•
I = q/t
q = I x t
I – current
q = amount of charge flowing past a point
t = time
•
•
1 ampere = 1A =1 C/s
C is the unit of electric charge coulomb. 1 ampere = 6 billion or 6
x 108 C/s electrons flowing along a conductor for each second.
The amount of charge that passes through the filament of a certain light bulb
in 2 seconds is 1.67 C. What is the current in the light bulb? What is the
number of electrons that pass through the filament in one second?
Given: t = 2 seconds; q = 1.67 C; I = ?
Solution:
I =
𝑞
𝑡
=
1.67 𝐶
2 𝑠
I = 0.835 C/s
Number of electrons: = 1 e = 1.6 x 10-19 C
= 0.835 C x
1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
1.6 𝑥 10−19 𝐶
= 5.22 x 1018
Answer:
The current in the light bulb is 0.835 C per second, and the number of electrons
that passes through the filament is 5.22 x1018 per second.
Over the course of an 8 hour day, 3.8x104 C of charge pass through a
typical computer (presuming it is in use the entire time). Determine the
current for such a computer.
Given: q = 3.8 x 104C; t = 8 hr = 28,800 s; I = ?
Solution:
I =
𝑞
𝑡
I =
3.8 𝑥 104𝐶
28,800 𝑠
= 1.32
𝐶
𝑠
𝑜𝑟 1.32 𝐴
Answer:
1.32 A of current does the computer uses.
The large window air conditioner in Anita Breeze's room draws 11 amps
of current. The unit runs for 8.0 hours during the course of a day.
Determine the quantity of charge that passes through Anita's window
AC during these 8.0 hours.
Given: t = 8 hrs = 28, 800 s; I = 11 A; q =?
Solution:
q = I x t
q = 28,800 s x 11 A
q = 316, 800 C or 3.16 x 105
Answer:
The quantity of charge that pass through Anita’s window AC is 3.16 x 105 for 8
hours.

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Electricity

  • 1.
  • 3. ELEMENTS Atoms - Smallest piece of an element containing all of the properties of that element
  • 4. The center portion of an atom containing the protons and neutrons Positively charged atomic particles Uncharged atomic particles
  • 5. The atomic number is equal to the number of protons in the nucleus of an atom. The atomic number identifies the element. How many protons are in this nucleus?
  • 6. Negatively charged particles Orbits in which electrons move around the nucleus of an atom The outermost ring of electrons in an atom 3D2D
  • 8. Conductors Insulators Electrons flow easily between atoms 1-3 valence electrons in outer orbit Examples: Silver, Copper, Gold, Aluminum Electron flow is difficult between atoms 5-8 valence electrons in outer orbit Examples: Mica, Glass, Quartz
  • 9.
  • 10. • • • • I = q/t q = I x t I – current q = amount of charge flowing past a point t = time
  • 11. • • 1 ampere = 1A =1 C/s C is the unit of electric charge coulomb. 1 ampere = 6 billion or 6 x 108 C/s electrons flowing along a conductor for each second.
  • 12. The amount of charge that passes through the filament of a certain light bulb in 2 seconds is 1.67 C. What is the current in the light bulb? What is the number of electrons that pass through the filament in one second? Given: t = 2 seconds; q = 1.67 C; I = ? Solution: I = 𝑞 𝑡 = 1.67 𝐶 2 𝑠 I = 0.835 C/s Number of electrons: = 1 e = 1.6 x 10-19 C = 0.835 C x 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 1.6 𝑥 10−19 𝐶 = 5.22 x 1018 Answer: The current in the light bulb is 0.835 C per second, and the number of electrons that passes through the filament is 5.22 x1018 per second.
  • 13. Over the course of an 8 hour day, 3.8x104 C of charge pass through a typical computer (presuming it is in use the entire time). Determine the current for such a computer. Given: q = 3.8 x 104C; t = 8 hr = 28,800 s; I = ? Solution: I = 𝑞 𝑡 I = 3.8 𝑥 104𝐶 28,800 𝑠 = 1.32 𝐶 𝑠 𝑜𝑟 1.32 𝐴 Answer: 1.32 A of current does the computer uses.
  • 14. The large window air conditioner in Anita Breeze's room draws 11 amps of current. The unit runs for 8.0 hours during the course of a day. Determine the quantity of charge that passes through Anita's window AC during these 8.0 hours. Given: t = 8 hrs = 28, 800 s; I = 11 A; q =? Solution: q = I x t q = 28,800 s x 11 A q = 316, 800 C or 3.16 x 105 Answer: The quantity of charge that pass through Anita’s window AC is 3.16 x 105 for 8 hours.