Transcript Power Point

Chapter 6: Electricity
• Section 1: Electric Charge
• Section 2: Electric Current
• Section 3: More Complex Circuits
Section 1: Electric Charge
Defining electricity
• On the atomic level:
 Structurally, the atom consists of
two parts:
 The atomic nucleus which
contains:
protons (+ charge)
neutrons (no charge)
 The electron cloud which
contains:
electrons (- charge)
Nucleus
Protons (+)
Neutrons (0)
Electron Cloud
Electrons (-)
• Both proton and electron are charged particles.
 Although the proton has a much larger mass than the
electron, the magnitude (size) of the positive (+) charge in
a proton is equal to the magnitude of the negative (-)
charge in an electron
• Because of the structure of the atom protons cannot easily
move, so electricity is simply the movement of electrons (or
the (-) charge) from one place to another
Section 2: Electric Current
Electric current – the net movement of electric charges in a
single direction
• An uncontrolled flow of charge is called static discharge
 Examples: the shock you feel when touching a metal
object, and lightning
• Current can be controlled and directed with circuits
 A circuit consists of a conductor, typically copper wire,
separated from the environment by an insulating material
 The insulating material can be cloth, paper, or plastic
• There are two types of circuits:
Flow of current
1. A series circuit which provides
only one path for the current +
to follow
2. A parallel circuit which provides
more than one path for the
current to follow
+
Flow of current
Note: Current always flows from the negative (-) pole to the
positive (+) pole
Section 2: Electric Current
Coulomb’s Law describes the relationship between charge,
capacitance, and potential difference in an electric circuit
• Charge – Atoms can gain or lose electrons
 If the atom acquires an electron, the atom becomes
negatively charged because there are now more electrons
than protons
• When an atom gains an electron it acquires a negative (-)
charge
 Charge is measured in units called coulombs (C)
 1 electron has a charge of 1.60 x 10-19C, and it requires
6.25x1025 electrons to equal a charge of 1C
• Law of Conservation of Charge: an electric charge cannot be
created or destroyed, but can be moved from one object to
another
• Capacitor – a device that can control the flow of current
through a circuit
 Consists of two metal plates separated by an insulator. As
the current flows, electrons “pile” up on one of the plates.
As the negative charges accumulate on the top plate, an
equivalent number of positive charges accumulate on the
bottom plate. Depending on the size of the capacitor,
charges will accumulate until there are enough to jump
the gap.
 Capacitance is the measure of how much charge a
capacitor can store
 The unit for capacitance is the farad (f)
Section 2: Electric Current
Potential difference
• Electrons can be easily moved, while the nucleus of an atom
cannot
 When the electrons are separated from the nuclei they
have stored energy—potential energy
 This electrical potential energy is measured by a value
called potential difference
 Potential difference is created by having extra electrons in
one place and a lack of electrons in another. So, the more
electrons in one place the greater the potential difference
 In a battery there are more electrons on the negative pole
than on the positive terminal
 Potential difference is measured in volts
Coulomb’ Law: Charge = capacitance x potential difference
Q = charge (coulomb) (C)
C = capacitance (farads) (f)
V = potential difference (volts) (V)
Example 1: What is the capacitance
of a capacitor that requires 0.23 C
to charge it to 15 V?
Q = CV
C= Q
V
V= Q
C
Q
V
0.25 C
C=
15 V
C = 0.01667f
C=
Notice that in the solution there is no algebra or dimensional
analysis
Section 2: Electric Current
Example 2: What is the voltage of
a 5f capacitor with a charge of
50 coulombs?
V=
Q
C
50C
5.0x10-6f
V = 1.0x107V
V=
Capacitance in a parallel circuit
A capacitor is able to hold
charge. When connected
in parallel, each capacitor
stores an amount of
charge, and the total
charge is: Qt = Q1 + Q2
Also, the voltage across each capacitor must equal the
total available voltage: Vt = V1 = V2
We know thatQ = CV , so: CtVt = C1V1 + C2V2
Because the voltages are equal, they cancel, and :
Ct = C1 + C2
Example: Determine the total capacitance in the given parallel
circuit
Solution
Ct = C1 + C2
Ct = 15f + 25mf
Ct = 15x10-6f + 25x10-3f
Ct = 2.5x10-2f
Section 2: Electric Current
Capacitance in a series circuit
• Electrons travel to the negative plate at C1
• As they stack up at C1, they repel an
equal number of electrons from the
positive plate of C2
• This happens again at C2
• So, the charge is the same at both capacitors and equal to
the total charge, or: Qt = Q1 + Q2
• The voltage drops across each capacitor, so that: Vt = V1 + V2
• Substituting into
V =Q
Qt
Q1
Q2
=
+
Ct
C1
C2
1
1
1
1
=
+
,or : Ct =
1
1
Ct
C1
C2
+
C1
C2
C , we can say:
• All charges are equal, so:
Example: Find the total capacitance of this series circuit
Solution
Ct =
1
1
1
+
C1 C2
1
Ct =
1
1
+
6f
2.5f
1
Ct =
0.5667
Ct = 1.765f
Section 2: Electric Current
• In a parallel circuit Ct is always greater than any of the
individual capacitors
• In a series circuit, Ct is always less than any of the individual
capacitors
Ohm’s Law describes the relationship between potential
difference, resistance, and current in a circuit
• Resistance – the measure of how much a substance opposes
the flow of electricity
 All substances have the property of electrical resistance.
Some materials—usually metals—have a much lower
resistance than other materials
 The resistance of a conductor depends on its thickness,
composition, length, and temperature
 Resistance varies directly with length, and increase with
temperature
 The unit for resistance is the ohm, and the symbol is the
Greek letter  (omega)
• Current – the flow of electrons through a circuit
 The unit for current is the ampere (amp). The capital letter
I is the symbol for current
Ohm’s Law: potential difference = current x resistance
V = potential difference (volts)(V) V =IR
I = current (amps)
I= V
R
R = resistance (ohms)()
R= V
I
Section 2: Electric Current
Example 1: what is the applied voltage if the current = 14 amps
and resistance = 5?
Solution: V = IR
V = 14amps(5)
V = 70V
Example 2: what is the resistance on a circuit that has a
current of 20 amps and potential diff of 6V?
Solution:
V
R=
I
6V
20amps
R = 0.33
R=
Example 3: What is the current flowing through a circuit when
resistance = 0.25 and V = 9V?
Solution:
V
I=
R
9V
0.25
I = 36amps
I=
Again, notice the solution contains no algebra or dimensional
analysis
Section 2: Electric Current
Resistance in series circuit
• Current flows from negative to
positive
• Same amount of current flows
through R1 and R2, so: It = I1 = I2
• Total voltage is the sum of the voltage across each resistor,
so: Vt = V1 + V2
• Using Ohm’s Law:V1 = I1R1, V2 =I2R2, and Vt = I1R1 +I2R2
 And: Vt = It Rt , so: It Rt = I1 R1 + I2 R2
 And: It = I1 = I2 , so: Rt = R1 +R2
Example: In the circuit shown R1 = 0.05, R2 = 0.10, Rt = ?
Solution: Rt = R1 + R2
Rt = 0.05 + 0.10
Rt = 0.15
In a series circuit, total resistance is always greater than the
resistance any of the individual resistors
Section 2: Electric Current
Resistance in a parallel circuit
• We know that the current will
flow through all paths, so:
It = I1 + I2
• The voltage is equal across both r
resistors, so: Vt = V1 + V2
• We know:
I=
• Substituting:
V
R
, so: It =
Vt
V1
V2
,I1 =
, I2 =
Rt
R1
R2
Vt
V1
V2
=
+
Rt
R1
R2
1
• Canceling the V’s, we get: R
t
=
1
1
1
+
, or: Rt =
1
1
R1
R2
+
R1
R2
Example: In the circuit shown, R1 = 0.05, R2 = 1.0 , Rt = ?
Solution:
1
R1 = 0.05Ω
Rt =
R2 = 1.0Ω
1
1
+
R1 R2
1
Rt =
1
1
+
0.05
1.0
Rt = 0.048
In a parallel circuit, total resistance is always less than any the
resistance of any of the resistors
Section 3: More Complex Circuits
• Power – the rate at which electrical energy is delivered
 The unit for power is the Watt (W)
 Three equations for power:
𝑷 = 𝑰(𝑽)
P = power (watts)
𝟐
𝑷= 𝑰 𝑹
I = current (amps)
𝑽𝟐
V = potential difference (volts)
𝑷=
𝑹
R = resistance (ohms)
Example 1: What is the power if the voltage supplied is 120-V
and the current is 15.25 amps?
Solution V = 120.0 V
𝑃=𝐼 𝑉
I = 115.25 amps
P=?
𝑃 = 15.25 𝑎𝑚𝑝𝑠 120.0 𝑉
𝑷 = 𝟏𝟖𝟑𝟎. 𝟎 𝑾
Example 2: What is the power if the circuit resistance is 75
and the current is 1.5 amps?
Solution R = 75.0Ω
2
I = 1.5 amps
P=?
𝑃= 𝐼 𝑅
𝑃 = 1.5𝑎𝑚𝑝𝑠 2 75.0
𝑷 = 𝟏𝟔𝟖. 𝟕𝟓 𝑾
Example 3: What is the power if the potential difference is 12V
and the total resistance is 0.25?
Solution
V = 12.0 V
𝑉2
𝑃=
R = 0.25 Ω
𝑅
2
12𝑉
P=?
𝑃=
0.25Ω
𝑷 = 𝟓𝟕𝟔. 𝟎 𝑾
Section 3: More Complex Circuits
Circuit Reduction
Example 1: If the charge on the circuit is 15 coulombs, what is
the applied voltage?
Example 2: If the current in the circuit is 2.5 amps, what is the
applied voltage?
The solution to these problems requires two steps:
1. The total capacitance of total resistance is calculated. This
reduces a complex circuit into a simple circuit with only 1
capacitor or resistor.
2. Either Coulomb’s Law or Ohm’s Law is used to find the
required variable.