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# Alternating Current Generators

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Magnetism
Alternating-Current Circuits
Alternating Current Generators
A coil of area A and N turns rotating with constant angular velocity in a uniform
magnetic field produces a sinusoidal emf
Alternating current motor
Instead of mechanically rotating, we can apply an ac potential difference generated
by other ac generator to the coil. This produces an ac current in the coil, and the
magnetic field exerts forces on the wires producing a torque that rotates the coil.
The magnetic field.
Magnetic forces on moving charges.
Magnetic forces on a current element
Torques on current loops and magnets
The Magnetic Field
Magnets
The Earth is a natural magnet with
magnetic poles near the north and
south geographic poles.
The Magnetic Field
Magnets
Does an isolated magnetic
pole exist?
The SI units of magnetic
field is the tesla [T]
Earth magnetic field is
Powerful laboratories
produce fields of 1 -2 T as a
maximum
1 Gauss [G] = 10-4 Tesla [T]
The Magnetic Field
The Magnetic Field. Magnetic force on a moving charge
A proton is moving in a region of crossed fields E =
2x105 N/C and B = 3000 G, as shown in the figure.
(a) What is the speed of the proton if it is not
deflected. (b) If the electric field is disconnected,
draw the path of the proton
The Magnetic Field. Magnetic force on a current element
In the case of a straight segment
of length L
The Magnetic Field. Torques on Current loops and
Magnets
B
A current-carrying loop
experiences no net force in a
uniform magnetic field, but it does
experience a torque that tends to
twist the loop
A circular loop of radius 2 cm has 10 turns of wire and carries a current of 3 A. The
axis of the loop makes an angle of 30Вє with a magnetic field of 8000 G. Find the
magnitude of the torque on the loop.
Sources of Magnetic Field
The Magnetic Field of Moving Charges
The Magnetic Field of Currents. The BiotSavart Law
The Magnetic Field Due to a Current loop
Sources of Magnetic Field
Moving Point Charges are the source of Magnetic Field
Sources of Magnetic Field
A solenoid
A solenoid is a wire tightly
wounded into a helix of
closely space turns . A
solenoid is used to produce a
strong, uniform magnetic
field in the region surrounded
by the loops
In this figure, the length is ten
For a long
solenoid
n = N/L; N number of turns;
L : length of solenoid
Find the magnetic field at the center of a
solenoid of 600 turns, length 20 cm;
radius 1.4 cm that carries a current of 4 A
Sources of Magnetic Field: Solenoid and magnets
Left: Magnetic field lines of a solenoid; right Magnetic field lines of a bar magnet
Magnetic Induction
Magnetic Flux
LenzВґs Law
MAGNETIC INDUCTION
In 1830, Michel Faraday in England and Joseph Henry in the USA
independently discovered that in a changing magnetic field, a changing
magnetic flux through a surface bounded by a stationary loop of wire induces
a current in the wire: emf induced and induced current. This process is
known as induction.
In a static magnetic field, a changing magnetic flux through a surface bounded
by a moving loop of wire induces an emf in the wire: motional emf
Magnetic flux through a surface bounded by a loop of wire.
The SI unit for magnetic flux is
weber [Wb] 1Wb= 1 Tв€™m2
Find the magnetic flux through a 40 cm long
solenoid with a 2.5 cm radius and 600 turns
carrying a current of 7.5 A.
MAGNETIC INDUCTION
The induced emf is in such direction as to
oppose, or tend to oppose, the change that
produces it.
LenzВґs Law
MAGNETIC INDUCTION
The coil with many turns of wire gives
a large flux for a given current in the
circuit. Thus, when the current
changes, there is a large emf induced
in the coil opposing the change. This
self-induced emf is called a back emf
Eddy Currents
Heat produced by eddy currents constitute a power loss in a
transformer. But eddy currents have some practical applications:
damping mechanical oscillations, magnetic braking system
Inductance
Self-inductance
The SI unit of
inductance is
the henry [H]
1 H = 1 Wb/A= 1 T.m2.A-1
Find the self-inductance of a solenoid of length 10 cm,
area 5 cm2, and 100 turns
Magnetic Energy
Alternating Current Generators
A coil of area A and N turns rotating with constant angular velocity in a uniform
magnetic field produces a sinusoidal emf
Alternating current motor
Instead of mechanically rotating, we can apply an ac potential difference generated
by other ac generator to the coil. This produces an ac current in the coil, and the
magnetic field exerts forces on the wires producing a torque that rotaes the coil.
Alternating Current Generators
A coil of area A and N turns rotating with constant angular velocity in a uniform
magnetic field produces a sinusoidal emf
пЃ¦ m пЂЅ NBA cos пЃ±
and пЃ± пЂЅ пЃ· t пЂ« пЃ¤
пЃ¦ m пЂЅ NBA cos( пЃ· t пЂ« пЃ¤ )
пЃҐ пЂЅ пЂ­
dпЃ¦m
пЂЅ NBA пЃ· sin( пЃ· t пЂ« пЃ¤ )
dt
пЃҐ пЂЅпЃҐ
peak
sin( пЃ· t пЂ« пЃ¤ )
frequency
f ; пЃ· пЂЅ 2пЃ° f
Alternating Current Circuits:
Alternating current in a Resistor
Inductors in Alternating Currents
Capacitors in Alternating Currents
Alternating currents in a Resistor
Potential drop across the resistor, VR
Current in the resistor I
Power dissipated in the resistor, P
Average power dissipated in the
resistor Paverage
V R пЂЅ пЃҐ пЂЅ пЃҐ max cos пЃ· t пЂЅ V R , peak cos пЃ· t
I пЂЅ
P пЂЅ I R пЂЅ I peak cos (пЃ· t ) R
2
2
2
Pav пЂЅ ( I ) av R пЂЅ ( I peak cos (пЃ· t )) av R
2
2
2
V R , peak
R
cos пЃ· t пѓћ I peak пЂЅ
V R , peak
R
Root-Mean-Square Values
Inductors in Alternating Current Circuits
The potential drop across the inductor
led the current 90Вє (out of phase)
VL пЂЅ L
dI
dt
пЂЅ
dI
dt
пЂЅ пЃҐ пЂЅ пЃҐ max cos пЃ· t пЂЅ V L , peak cos пЃ· t
V L , peak
L
cos пЃ· t п‚® I пЂЅ
V L , peak
пЃ·L
INDUCTIVE
REACTANCE
sin пЃ· t пЂЅ
I peak пЂЅ
V L , peak
V L , peak
пЃ· L
пЃ·L
;
cos( пЃ· t пЂ­
пЃ°
2
I rms пЂЅ
V L , rms
Instantaneous power delivered by the emf to the inductor is not zero
The average power delivered by the emf to the inductor is zero.
пЃ· L
)
Inductors in Alternating Current Circuits
The potential drop across a 40-mH
inductor is sinusoidal with a peak
potential drop of 120 V. Find the
inductive reactance and the peak current
when the frequency is (a) 60 Hz, and (b)
2000 Hz
I peak пЂЅ
V L , peak
пЃ· L
;
I rms пЂЅ
V L , rms
пЃ· L
INDUCTIVE
REACTANCE
Instantaneous power delivered by the emf to the inductor is not zero
The average power delivered by the emf to the inductor is zero.
Capacitors in Alternating Current Circuits
The
potential
drop lags
the
current
by 90Вє
VC пЂЅ
Q
C
пЂЅ пЃҐ пЂЅ пЃҐ max cos пЃ· t пЂЅ V C , peak cos пЃ· t пѓћ
Q пЂЅ CV C , peak cos пЃ· t
I пЂЅ
dQ
dt
пЂЅ пЂ­ пЃ· CV C , peak sin пЃ· t п‚® I пЂЅ
CAPACITIVE
REACTANCE
I peak пЂЅ
V C , peak
пѓ¦ 1
пѓ¶
пѓ§ пЃ·C пѓ·
пѓЁ
пѓё
V C , peak
пѓ¦ 1
пѓ¶
пѓ§ пЃ·Cпѓ·
пѓЁ
пѓё
cos( пЃ· t пЂ«
пЃ°
)
2
;
I rms пЂЅ
V C , rms
пѓ¦ 1
пѓ¶
пѓ§ пЃ·Cпѓ·
пѓЁ
пѓё
Power delivered by the emf in the capacitor: Instantaneous and average
Driven RLC Circuits
Series RLC circuit
The KirchhoffВґs rules govern
the behavior of potential
drops and current across the
circuit.
(a) When any closed-loop is
traversed, the algebraic
sum of the changes of
potential must equal
zero (loops rule)
(b) At any junction (branch
point) in a circuit where
the current can be
divided, the sum of the
currents into the
junction must equal the
sum of the currents out
of the junction (junction
rule)
Series RLC circuits
V app , peak cos пЃ· t пЂЅ L
2
L
d Q
dt
пЂ«
dQ
dt
RпЂ«
Q
C
dI
dt
пЂ« IR пЂ«
Q
; I пЂЅ
C
пЂЅ V app , peak cos пЃ· t
I пЂЅ I peak cos( пЃ· t пЂ­ пЃ¤ )
dQ
dt
Power delivered to the series RLC circuit
P пЂЅпЃҐ I пЂЅпЃҐ
P av пЂЅ
1
пЃҐ
2
P пЂЅ RI
peak
2
Pav пЂЅ RI
peak
cos пЃ· t I peak cos( пЃ· t пЂ­
)
2
I peak cos пЃ¤ пЂЅ пЃҐ rms I rms cos пЃ¤
dissipated
2
rms
пЃ°
пЂЅ V app , rms
2
in the resistor
R
Z
2
as R / Z пЂЅ cos пЃ¤ and I peak пЂЅ V app , peak / Z
P пЂЅ
1
2
V app , peak I peak cos пЃ¤ пЂЅ V app , rms I rms cos пЃ¤
Power factor:
cosОґ
Series RLC circuits
A series RLC circuit with L = 2 H, C =2 ОјF
and R=20О© is driven by an ideal generator
with a peak emf of 100 V and a frequency of
60 Hz, find (a) the current peak (b) the phase
(c) the power factor, (d) the average power
delivered; (e) the peak potential drop across
each element
Phasors
Potential drop across a resistor can be
represented by a vector VR, which is called a
phasor. Then, the potential drop across the
resistor IR, is the x component of vector VR,
Potential drop across a
series RLC circuit
V app cos пЃ· t пЂЅ L
dI
dt
пЂ« IR пЂ«
Q
C
In the circuit shown in the figure, the
ac generator produces an rms voltage
of 115 V when operated at 60 Hz. (a)
What is the rms current in the circuit
(b) What is the power delivered by
the ac generator (c) What is the rms
voltage across: points AB; points BC;
points CD; points AC; points BD?.
A certain electrical device draws 10 A rms and has an average power of 720 W
when connected to a 120-V rms 60-Hz power line. (a ) What is the impedance
of the device? (b) What series combination of resistance and reactance is this
device equivalent to? (c) If the current leads the emf, is the reactance inductive
or capacitive?
The Transformer
Because of the iron core, there is a large
magnetic flux through each coil, even when
the magnetizing current Im in the primary
circuit is very small .
The primary circuit consists of an ac
generator and a pure inductance (we consider
a negligible resistance for the coil). Then the
average power dissipated in the primary coil
is zero. Why?: The magnetizing current in the
primary coil and the voltage drop across the
primary coil are out of phase by 90Вє
A transformer is a device to raise or lower
the voltage in a circuit without an
appreciable loss of power. Power losses
arise from Joule heating in the small
resistances in both coils, or in currents
loops (eddy currents) within the iron core.
An ideal transformer is that in which these
losses do not occur, 100% efficiency.
Actual transformers reach 90-95%
efficiency
Secondary coil open circuit
The potential drop across the primary coil is
V1 пЂЅ N 1
d пЃ¦ tu rn
dt
If there is no flux leakage out of the iron
core, the flux through each turn is the
same for both coils, and then
V2 пЂЅ N 2
d пЃ¦ tu rn
dt
п‚® V2 пЂЅ
N2
N1
V1
The Transformer
A resistance R, load resistance, in the
secondary circuit
A current I2 will be in the secondary coil, which is
in phase with the potential drop V2 across the
resistance. This current sets up and additional flux
О¦Вґturn through each turn, which is proportional to
N2I2. This flux opposes the original flux sets up by
the original magnetizing current Im in the primary.
However, the potential drop in the primary is determined by the generator emf
According to this, the total flux in the iron core must be the same as when there is no load
in the secondary. The primary coil thus draws an additional current I1 to maintain the
original flux О¦turn. The flux through each turn produced by this additional current is
proportional to N1I1. Since this flux equals вЂ“ О¦Вґturn, the additional current I1 in the
primary is related to the current I2 in the secondary by
N 1 I1 пЂЅ N 2 I 2
These curents are 180 Вє out of phase and produce counteracting fluxes. Since I2 is in
phase with V2, the additional current I1 is in phase with the potential drop across the
primary. Then, if there are no losses
V1 , rms I 1 , rms пЂЅ V 2 , rms I 2 , rms
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