PHY
PHY 1214
1214
General
General Physics
Physics II
II
Lecture 21
Generators and Transformers
July 7 & 11, 2005
Weldon J. Wilson
Professor of Physics & Engineering
Howell Hall 221H
wwilson@ucok.edu
PHY 1214: Lecture 21, Slide 1
Lecture Schedule (Weeks 4-6)
We are here.
PHY 1214: Lecture 21, Slide 2
Review: Two uses of RHR’s
+ + + +v
I
– Thumb: v (or I)
– Fingers: B
– Palm: F on + charge
B
F
• Force on moving charge in
Magnetic field
Palm: out of page.
• Magnetic field produced by
moving charges
– Thumb: I (or v for + charges)
– Fingers: where you want to know B
– Palm: B
PHY 1214: Lecture 21, Slide 3
Thumb: out
Fingers: up
Palm: left.
x
•
Review: Two uses of RHR’s
• Magnetic field produced by
moving charges
– Thumb: I (or v for + charges)
– Fingers: curl along B field
PHY 1214: Lecture 21, Slide 4
+ + + +v
F
– Thumb: v (or I)
– Fingers: B
– Palm: F on + charge
B
I
• Force on moving
charge in Magnetic
field
Palm: out of page.
I
Review: Induction
• Lenz’s Law
– If the magnetic flux (ΦB) through a loop changes, an
EMF will be created in the loop to oppose the change
in flux
– EMF
current (V=IR)
additional B-field.
• Flux decreasing => B-field in same direction as original
• Flux increasing => B-field in opposite direction of original
• Faraday’s Law
– Magnitude of induced EMF given by:
∆Φ
Φf − Φi
ε = −N
= −N
∆t
tf − ti
PHY 1214: Lecture 21, Slide 5
Review: Rotation Variables
v, ω, f, T
ω
• Velocity (v):
– How fast a point moves.
– Units: usually m/s
r
• Angular Frequency (ω):
– How fast something rotates.
– Units: radians / sec
v
v
v= ωr
• Frequency ( f ):
– How fast something rotates.
– Units: rotations / sec = Hz
f = ω / 2π
• Period (T):
– How much time one full rotation takes.
– Units: usually seconds
PHY 1214: Lecture 21, Slide 6
T = 1 / f = 2π
/ω
Generators
• Alternating Current (AC) generator
– Converts mechanical energy to electrical energy
– Consists of a wire loop rotated by some external
means
– There are a variety of sources that can supply the
energy to rotate the loop
• For example, these may include falling water or
heat by burning coal to produce steam
PHY 1214: Lecture 21, Slide 7
AC Generators, cont.
• Basic operation of the generator
– As the loop rotates, the magnetic
flux through it changes with
time
– This induces an emf and a
current in the external circuit
– The ends of the loop are
connected to slip rings that rotate
with the loop
– Connections to the external
circuit are made by stationary
brushes in contact with the slip
rings
PHY 1214: Lecture 21, Slide 8
AC Generators, cont.
Area A=ℓa
The emf generated in wire BC is Bℓv⊥ where ℓ is the
length of the wire and v⊥ is the velocity component
perpendicular to the B field (v has no effect on the
charges in the wire ). An emf of Bℓv⊥ is also generated
in the wire DA with the same sense as in BC. Because
v⊥=v sinθ, the total emf is ε =2Bℓv sinθ.
PHY 1214: Lecture 21, Slide 9
AC Generators, cont.
Since v=rω (tangential speed=radius times
angular speed), it follows v=(a/2)ω and
ε = 2Bℓv sinθ = 2Bℓ (a/2)ω sinθ
ωt
Therefore, ε=Bℓaω sinωt and with A=ℓa
ε=NBAω sinωt
PHY 1214: Lecture 21, Slide 10
For a coil with N
turns
Generator equation from
Faraday’s law
ωt
ε=-N∆ΦB/∆t
ε=-NBA[∆(cosθ)/∆t]
From calculus: ∆(cosωt)/ ∆ t=-ωsinωt
ε=NBAωsinωt
εmax = NBAω (maximum value of the emf)
ε=εmaxsinωt=εmaxsin2πf
PHY 1214: Lecture 21, Slide 11
2π f
AC Generators, final
• The emf generated by the
rotating loop can be found by
ε =2Bℓv⊥=2Bℓv sinθ
• If the loop rotates with a
constant angular speed, ω,
and N turns
ε=NBAω sinωt
• ε = εmax when loop is parallel
to the field
• ε = 0 when when the loop is
perpendicular to the field
PHY 1214: Lecture 21, Slide 12
Generators and Torque
ε = ω A B sin(θ)
Voltage!
Connect loop to resistance R use I=V/R:
I = ω A B sin(θ) / R
Recall:
τ = A B I sin(θ)
= ω A2 B2 sin2(θ)/R
•
ω
v
x
r
Torque, due to current and B field, tries to slow
spinning loop down. Must supply external
torque to keep it spinning at constant ω
PHY 1214: Lecture 21, Slide 13
θ
v
Generator
A generator consists of a square coil of wire with 40 turns, each side
is 0.2 meters long, and it is spinning with angular velocity ω = 2.5
radians/second in a uniform magnetic field B=0.15 T. Determine the
direction of the induced current at instant shown. Calculate the
maximum emf and torque if the resistive load is 4Ω.
ε = NA B ω sin(θ
θ)
= (40) (0.2)2 (0.15) (2.5)
= 0.6 Volts
τ = NI A B sin(θ
θ)
= N2 ω A2 B2 sin2(θ)/R
= (40)2 (2.5) (0.2)4 (0.15)2/4
= 0.036 Newton-meters
PHY 1214: Lecture 21, Slide 14
ω
•
v
v
x
θ
Note: Emf is
maximum at θ=90
Note: Torque is
maximum at θ=90
Power Transmission
A generator produces 400 kW of power, which it
transmits to a town 10 km away through copper
power lines. The resistance in the power lines is
200 ohms. How much power is lost if (a)
V=20,000 V (b) 500,000 V?
P = I V I = P/V = (400 kW)/(20 kV) = 20 A
Ploss = I2R =(20 A)2 (200 ohm) = 80 kW
(20% lost!)
P = I V I = P/V = (400 kW)/(500 kV) = 0.8 A
Ploss = I2R =(0.8 A)2 (200 ohm) = 0.128 kW
(0.03% lost!)
PHY 1214: Lecture 21, Slide 15
Transformers
Key to efficient power distribution
Increasing current in primary
creates an increase in flux
through primary and secondary.
∆Φ
∆t
∆Φ
Vs = − N s
∆t
iron
Vp = −N p
ε
∼
PHY 1214: Lecture 21, Slide 16
V
s
Same ∆Φ/∆
∆Φ/∆t
Vs N s
=
Vp N p
Energy conservation!
Vp
IpVp = IsVs
NP
NS
(primary)
(secondary)
R
Question
The good news is you are going on a trip to France. The bad
news is that in France the outlets have 240 volts. You
remember from PHY 1214 that you need a transformer, so
you wrap 100 turns around the primary. How many turns
should you wrap around the secondary if you need 120
volts out to run your hair dryer?
iron
1) 50
2) 100
Vs N s
=
Vp N p
3) 200
ε
 Vs 
120 
N s = N p   = 100
 = 50
 240 
 Vp 
PHY 1214: Lecture 21, Slide 17
∼
Vp
V
s
NP
NS
(primary)
(secondary)
R
Transformers
iron
Transformers depend on a
change in flux so they only
work for alternating
currents!
A 12 Volt battery is connected to a
transformer that has a 100 turn primary
coil, and 200 turn secondary coil. What
is the voltage across the secondary after
the battery has been connected for a long
time?
1) Vs = 0
2) Vs = 6
PHY 1214: Lecture 21, Slide 18
3) Vs = 12
Vp
V
s
NP
NS
(primary)
(secondary)
4) Vs = 24
R
The Tesla Coil
A special case of a step-up
transformer is the Tesla coil. It uses
no magnetic material, but has a very
high N2/N1 ratio and uses highfrequency electrical current to induce
very high voltages and very high
frequencies in the secondary.
There is a phenomenon called “the
skin effect” that causes high
frequency AC currents to reside
mainly on the outer surfaces of
conductors. Because of the skin
effect, one does not feel (much) the
electrical discharges from a Tesla coil.
PHY 1214: Lecture 21, Slide 19
Transformers
• Key to Modern electrical system
• Starting with 120 volts AC
– Produce arbitrarily small voltages.
– Produce arbitrarily large voltages.
• Nearly 100% efficient
!!!Volt!!
PHY 1214: Lecture 21, Slide 20
Motors
• Motors are devices that convert electrical
energy into mechanical energy
– A motor is a generator run in reverse
• A motor can perform useful mechanical
work when a shaft connected to its rotating
coil is attached to some external device
PHY 1214: Lecture 21, Slide 21
Motors and Back emf
Back emf
The applied voltage V supplies the current I to
drive the motor. The circuit shows V along
with the electrical equivalent of the motor,
including the resistance R of its coil and the
back emf ε.
PHY 1214: Lecture 21, Slide 22
Motors and Back emf
• The phrase back emf is used for an emf that tends to
reduce the current due to an applied voltage → current
through the motor: I=V-εb/R, where V is the line
voltage, εb is the back emf and R is the coil resistance
• When a motor is turned on, there is no back emf initially
• The current is very large because it is limited only by
the resistance of the coil
PHY 1214: Lecture 21, Slide 23
Motors and Back emf, cont.
• As the coil begins to rotate, the induced
back emf opposes the applied voltage
• The current in the coil is reduced
• The power (i.e., current) requirements for
starting a motor and for running it under
heavy loads are greater than those for
running the motor under average loads
PHY 1214: Lecture 21, Slide 24
Example: A motor has a 10 Ω coil. When running at its
maximum speed, the back emf is 70 V. Find the current
(a) when the motor starts and (b) when the motor has
reached its maximum speed.
• (a) I=V/R=120 V/10 Ω
• I=12 A
• (b) I=(V-εb)/R
• I=(120 V-70 V)/10 Ω
• I=50 V/10 Ω=5 A
PHY 1214: Lecture 21, Slide 25
End of Lecture 21
Before the next lecture, read sections 22.7
thru 22.9.
PHY 1214: Lecture 21, Slide 26