Electrical Overview
Ref: Woud 2.3
I=
Q
C = 1C
Q = charge
N.B. this is a long note and
repeats much of what is is the text
C = 1 coul
(2.50)
t
min = 60 s
t = time
s = 1s
A = 1A
I = current
work done per unit charge = potential difference two points
aka electromotive force (EMF)
Power = U( t) ⋅ I( t)
V = 1V
U = volts
1V⋅ 1 A = 1 W
1V⋅ 1 A = 1 watt
(2.51)
source, resistance, inductance, capacitance
resistance
resistance = R
Ω = 1Ω
ohm = 1 Ω
Ohm's law
U( t) = I( t) ⋅ R
power in a resistor ...
Power = U( t) ⋅ I( t) = I( t) ⋅R
friction in mechanical system
(2.52)
2
2
1Ω ⋅ (1A) = 1 W
inductance
(2.53)
mass of inertia in mechanical system
H = 1H
inductance = L
henry = 1 H
t
d
U( t) = L⋅ I( t)
dt
H⋅
A
s
= 1V
⌠ U( t)
dt
I( t) = ⎮
⎮ L
⌡
or ...
V⋅ s
H
= 1A
(2.54)
0
⎛d ⎞
P = U⋅ I = L⋅ I⋅ ⎜ I ⎟
⎝
dt ⎠
H⋅ A⋅
A
s
(2.55)
= 1W
t
t
I
⌠
⌠
⌠
⎛
d ⎞
⎮
inductive_energy_stored = Eind = ⎮ P( t) dt =
L⋅ I⋅ ⎜ I ⎟ dt = ⎮ L⋅ I dI
⌡
⎮
⌡
⎝
dt ⎠
0
0
⌡
I
⌠
1 2
⎮ L⋅ I dI → ⋅ I ⋅L
⌡
2
0
0
2
A ⋅H = 1 J
capacitance
spring in mechanical system
F = 1F
capacitance = C
farad = 1 F
t
d
I( t) = C⋅ U( t)
dt
(2.56)
F⋅
V
s
= 1A
⌠ I( t)
dt
U( t) = ⎮
⎮ C
⌡
or ...
A⋅ s
F
= 1V
(2.57)
0
d
P = U⋅ I = C⋅ U⋅ U( t)
dt
F⋅ V⋅
t
t
U
⌠
⌠
⌠
d
⎮
capacitive_energy_stored = Ecap = ⎮ P( t) dt =
C⋅ U⋅ U( t) dt = ⎮ C⋅ U dU
⌡
⌡
⎮
dt
0
0
⌡
0
V
s
= 1W
⌠
⎮
⌡
U
0
C⋅ U dU →
1
2
2
⋅ U ⋅C
(2.58), (2.59)
2
V ⋅F = 1 J
1
11/13/2006
Kirchhoff's laws
first ...
number_of_currents
∑
sum_of_currents_towards_node = 0
⎡⎣Ii( t)⎤⎦ = 0
i= 1
second ...
(2.60)
direction specified
sum_of_voltages_around_closed_path = 0
number_of_voltages
∑
⎡⎣Ui( t)⎤⎦ = 0
i= 1
(2.61)
series connection of resistance and inductance ...
U( t) := Um⋅cos( ω⋅t)
imposed ... external
Um = amplitude_of_voltage
V = 1V
ω = frequency
Hz = 1
t = time
min = 60 s
I( t) := Im⋅cos( ω⋅t − φ)
resulting current assumed also harmonic
Im = amplitude_of_current
(2.62)
1
s
A = 1 amp
(2.63)
φ = phase_lag_angle
it is useful to represent this parameters as vectors using complex notation, where the values are represented by the real
parts
Uz( t) := Um⋅cos( ω⋅t) + Um⋅sin( ω⋅t) ⋅i
Iz( t) := Im⋅cos( ω⋅t − φ) + Im⋅cos( ω⋅t − φ) ⋅i
Imaginary parts of Uz(t), Iz(t)
plotting set up
Uz(t)
Iz(t)
1
0.5
0
0
0.5
1
Real parts of Uz(t), Iz(t) = U(t), I(t)
over R voltage drop will be ... U ( t) := R⋅ I( t) → R⋅ I ⋅ cos⎣⎡( −ω ) ⋅t + φ⎦⎤
R
m
over L voltage drop will be ...
UR( t) = R⋅ Im⋅ cos( ω⋅t − φ)
cos( α ) = cos( −α )
d
UL( t) := L⋅ I( t) → L⋅ Im⋅sin⎡⎣( −ω ) ⋅t + φ⎦⎤ ⋅ω
dt
d
⎛π
⎞
L⋅ I( t) = −Im⋅ω⋅L⋅sin( ω⋅t − φ) = Im⋅ω⋅L⋅cos⎜ + ω⋅t − φ⎟
2
dt
⎝
⎠
2
11/13/2006 ⎛ π + α ⎞ → −sin( α ) ⎟
⎝2
⎠
cos⎜
⎛ π + α ⎞ = cos⎛ π ⎞ ⋅cos( α ) − sin⎛ π ⎞ ⋅sin( α ) = 0 ⋅ cos( α ) − 1⋅ sin( α )
⎟
⎜ ⎟
⎜ ⎟
⎝2
⎠
⎝2⎠
⎝2⎠
cos⎜
or ...
UzR( t) := R⋅ Im⋅cos( ω⋅t − φ) + R⋅ Im⋅sin( ω⋅t − φ) ⋅i
in complex (vector) notation ...
⎛π
⎞
⎛π
⎞
UzL( t) := Im⋅ω⋅L⋅cos⎜ + ω⋅t − φ⎟ + Im⋅ω⋅L⋅sin⎜ + ω⋅t − φ⎟ ⋅i
2
2
⎝
⎠
⎝
⎠
Imaginary parts of Uz(t), UzR(t), UzL(t)
plotting set up
Uz(t)
UzR(t)
UzL(t)
1
0.8
0.6
0.4
0.2
0
0.2
0
0.2
0.4
0.6
0.8
1
Real parts of Uz(t), UzR(t), UzL(t) = U(t), UR(t), UL(t)
at this point these vectors are shown with two unknowns included I m and φ
i.e. directions are correct relatively given φ and magnitudes arbitrary given I m
Kirchoff's second law ...
U( t) := UR( t) + UL( t) → R⋅ Im⋅cos⎡⎣( −ω ) ⋅ t + φ⎤⎦ + L⋅ Im⋅ sin⎡⎣( −ω ) ⋅ t + φ⎤⎦ ⋅ ω
⎛π
⎞
Um⋅cos( ω⋅t) = R⋅ Im⋅cos( ω⋅t − φ) + L⋅ Im⋅ω⋅cos⎜ + ω⋅t − φ⎟
2
⎝
⎠
this can be solved for φ and Im after expanding the rhs into sines and cosines and setting cos = cos and sin = sin
easier if think in terms of vectors
3
11/13/2006 Imaginary parts of Uz(t), UzR(t), UzL(t)
Uz(t)
UzR(t)
UzL(t)
UzL(t) rel. to UzR(t)
UzR(t)+UzL(t)
1
0.8
0.6
0.4
0.2
0
0.2
0
0.2
0.4
0.6
0.8
1
Real parts of Uz(t), UzR(t), UzL(t) = U(t), UR(t), UL(t)
for UzR(t) + zL(t) to = Uz(t) magnitude and angle must be =
Um =
(R⋅ Im)
2
(
+ L⋅ Im⋅ω
)2 =
UzR( t) → R⋅ Im⋅ cos⎡⎣( −ω ) ⋅t + φ⎤⎦ − i⋅ R⋅ Im⋅sin⎡⎣( −ω ) ⋅t + φ
R + ( L⋅ω ) ⋅Im
2
2
Uz( t) → Um⋅ cos( ω⋅t) + i⋅ Um⋅sin( ω⋅t)
UzL( t) → L⋅ Im⋅sin⎡⎣(−ω ) ⋅t + φ⎤⎦ ⋅ω + i ⋅ Im⋅ ω ⋅ L⋅ cos⎡⎣( −ω ) ⋅t + φ⎦⎤
or ...
and ...
Im =
Um
R + ( L⋅ω )
2
2
⎛ L⋅ω ⎞
⎟
⎝ R ⎠
φ = atan⎜
using these relationships in the plot ...
plotting set up
4
11/13/2006 Uz(t)
UzR(t)
UzL(t)
UzL(t) rel. to UzR(t)
UzR(t)+UzL(t)
Imaginary parts of Uz(t), UzR(t), UzL(t), etc.
1
0.8
0.6
0.4
0.2
0
0.2
0
0.2
0.4
0.6
0.8
1
Real parts of Uz(t), UzR(t), UzL(t) etc. = U(t), UR(t), UL(t), etc.
N.B. angle may not appear as right angle due to scales
φ shown as lag (positive value with negative sign)
capacitor lead approach (text)
similar for Capacitance
imposed ... external
U( t) := Um⋅cos( ω⋅t)
Um = amplitude_of_voltage
V = 1V
ω = frequency
Hz = 1
1
s
(2.62)
min = 60 s
t = time
this is different from text: lag phase angle vs. lead angle used
resulting current assumed also harmonic
I( t) := Im⋅cos( ω⋅t − φ)
current assumed to have lag angle. this approach taken
to allow common treatment of L and C in circuits
Im = amplitude_of_current
V = 1V
φ = phase_lag_angle
complex (vector) representation, set up with real part expressed as cos
Uz( t) = Um⋅cos( ω⋅t) + Um⋅sin( ω⋅t) ⋅i
Iz( t) = Im⋅cos( ω⋅t − φ) + Im⋅sin( ω⋅t − φ) ⋅i
plotting set up
5
11/13/2006 Imaginary parts of Uz(t), Iz(t)
voltage and current at omega*t positive lag phase angle
Uz(t)
Iz(t)
1
0.5
0
0
0.2
0.4
0.6
0.8
1
Real parts of Uz(t), Iz(t) = U(t), I(t)
voltage across capacitor (from above)
t
(2.57)
t
⌠ I ⋅cos( ω⋅t − φ)
Im⋅sin( ω⋅t − φ)
⌠ I( t)
Im
π⎞
⎮ m
⎛
⎮
dt =
dt =
UC( t) =
=
⋅cos⎜ ω⋅t − φ − ⎟
⎮
⎮ C
C
2⎠
C⋅ω
C⋅ω
⎝
⌡
⌡
0
0
using complex (vector) notation
Uz( t) := Um⋅cos( ω⋅t) + Um⋅sin( ω⋅t) ⋅i
UzC( t) :=
Im
C⋅ω
Iz( t) := Im⋅cos( ω⋅t − φ) + Im⋅cos( ω⋅t − φ) ⋅i
⎛
⎝
⋅cos⎜ ω⋅t − φ −
π⎞
Im
π⎞
⎛
⋅sin⎜ ω⋅t − φ − ⎟ ⋅i
⎟+
2 ⎠ C⋅ω
2⎠
⎝
UzR( t) := R⋅ Im⋅cos( ω⋅t − φ) + R⋅ Im⋅sin( ω⋅t − φ) ⋅i
Kirchoff's second law for resistor with capacitor...
Uz( t) := UzR( t) + UzC( t) → Ω⋅Im⋅cos⎡⎣( −ω ) ⋅t + φ⎤⎦ − i⋅Ω⋅Im⋅sin⎡⎣( −ω ) ⋅t + φ⎦⎤ −
Im
C⋅ω
⋅sin⎡⎣( −ω ) ⋅t + φ⎦⎤ − i⋅
Im
C⋅ω
⋅cos⎡⎣( −ω ) ⋅t + φ⎤⎦
plotting set up
6
11/13/2006 Voltages with phase angle = - 40 deg
Imaginary parts of Uz(t), UzR(t), UzC(t), etc.
1
0.8
N.B. angle is distorted due
to scales; angle between
UzR(t) and UzC(t) is π/2)
0.6
0.4
0.2
Uz(t)
UzR(t)
UzC(t)
UzC(t) rel. to UzR(t)
UzR(t)+UzC(t)
0
0.2
0.4
0
0.2
0.4
0.6
0.8
1
1.2
Real parts of Uz(t), UzR(t), UzC(t) etc. = U(t), UR(t), UC(t), etc.
Uz( t) = Um⋅ cos( ω⋅t) + Um⋅sin( ω⋅t) = R⋅ Im⋅cos( ω⋅t − φ) + i⋅ R⋅ Im⋅
sin( ω⋅t − φ) +
Im
C⋅ω
⋅sin( ω⋅t − φ) − i⋅
Im
⋅cos( ω⋅t − φ)
C⋅ω
look at solution plotted
plotting set up
7
11/13/2006 Imaginary parts of Uz(t), UzR(t), UzC(t), etc.
Uz(t)
UzR(t)
UzC(t)
UzC(t) rel. to UzR(t)
UzR(t)+UzC(t)
0.8
0.6
0.4
0.2
0
0.2
0.4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Real parts of Uz(t), UzR(t), UzC(t) etc. = U(t), UR(t), UC(t), etc.
Uz( t) = Um⋅ cos( ω⋅t) + Um⋅sin( ω⋅t) = R⋅ Im⋅cos( ω⋅t − φ) + i⋅ R⋅ Im⋅ sin( ω⋅t − φ) +
magnitude similar to
above ...
Im =
R +
1
( ω⋅C) 2
N.B. angle may not appear as right angle due to scales
φ shown as lead (negative value with negative sign)
so with both L and C
C⋅ω
phase angle is negative;
hence referred to as
lead angle
Um
2
Im
⋅sin( ω⋅t + φ) − i⋅
Im
⋅cos( ω⋅t + φ)
C⋅ω
⎞
⎟
⎝ ω⋅C⋅R ⎠
φ = −atan⎛⎜
1
φ1 = −26.565 deg in this numerical example
⎛ ω⋅L − 1 ⎞
⎟
ω⋅C⋅R ⎠
⎝ R
φ = atan⎜
8
11/13/2006 Section 2.3.4
2
P = U⋅ = I ⋅R
Direct current (DC)
P( t) = U( t) ⋅ I( t)
Single phase alternating current (AC)
U( t) := Um⋅cos( ω⋅t)
typically sinusoidal ...
Pa :=
average power ...
lim
T→∞
I( t) := Im⋅cos( ω⋅t − φ)
lag phase angle used
⎛⎜ 1 ⌠ T
⎞⎟
1
⋅⎮ U( t) ⋅ I( t) dt → ⋅Um⋅Im⋅cos( φ
) ⎜ T ⌡0
⎟
2
⎝
⎠
in practice effective values are used
Ue :=
Ie :=
lim
T→∞
lim
T→∞
1 ⌠
⎮
⋅
T ⎮
⌡
T
(
Um⋅cos( ω⋅t)
)
2
dt →
0
1 ⌠
⎮
⋅
T ⎮
⌡
T
1
1
1
2
⋅2 ⋅ ⎛ Um
2
2⎞
⎝
2
⎠
→
0
⋅2 ⋅ ⎛ Im
⎝
2
1
2
so average power becomes ...
2
Ue = effective_voltage
1
1
(Im⋅cos(ω⋅t − φ))2 dt
Um
Ue :=
2⎞
2
Ie :=
⎠
Im
2
Ie = effective_current
cos( φ) = power_factor
Pa := Ue⋅Ie⋅cos( φ)
what is current required in two systems with same effective voltage but larger phase lag?
here forward e subscript dropped and U == U e , I == Ie
some power and current definitions
apparent_power = V⋅ A = U⋅ I
real_power = U⋅ ⋅cos( φ)
W = 1W
reactive_power = U⋅ I⋅ sin( φ)
same for current
I = current
A = 1 amp
load_current = I⋅ cos( φ)
A = 1 amp
reactive_current = I⋅ sin( φ) A = 1 amp
V⋅ A
three phase alternating current
⎛ 0 ⎞
⎜
⎟
⎜ 2⋅ π ⎟
α := ⎜ 3 ⎟
⎜ π⎟
⎜ 4⋅ ⎟
⎝ 3⎠
ORIGIN := 1
i := 1 .. 3
(
(
and ...
3
∑
i= 1
)
Up := Um⋅cos ω⋅t − α i
i
phase angle for respective phases
Ip := Im⋅cos ω⋅t − α i − φ
i
)
3
Up expand → 0
i
∑
i= 1
Ip expand → 0
i
9
11/13/2006 IL = Ip
1
1
star connection ...
i := 1 .. 2
UL := Up − Up
i
i
i+
1
(
i := 1 .. 3
e.g.
magnitude is ...
)
(
IL = Ip
3
3
UL := Up − Up
3
3
31
)
Uzp := Um⋅cos ω⋅t − α i + Um⋅sin ω⋅t − α i ⋅i
i
i := 1 .. 2
i := 1 .. 3
IL = Ip
2
2
UzL := Uzp − Uzp
i
i
i+ 1
UzL := Uzp − Uzp
3
3
1
1
1
UzL simplify → Um⋅ cos( ω⋅t) + i⋅ Um⋅sin( ω⋅t) + Um⋅ cos⎛⎜ ω⋅t + ⋅π ⎞⎟ + i⋅ Um⋅sin⎛⎜ ω⋅t + ⋅π ⎟⎞
1
3
3 ⎠
⎝
⎠
⎝
from trigonometry...
1
2
(
2
)
⎛ cos( ω⋅t) + cos⎛ ω⋅t + π ⎞⎞ + ⎛ sin( ω⋅t) + sin⎛ ω⋅t + π ⎞⎞ expand → 3 ⋅ cos( ω⋅t) 2 + 3 ⋅ sin( ω⋅t) 2
⎜
⎜
⎟⎟ ⎜
⎜
⎟⎟
3 ⎠⎠
3 ⎠⎠
⎝
⎝
⎝
⎝
Um *
2
magnitude := Um⋅ 3
(2.85)
angle relative to ω*t (set ω*t = 0)
1
1
→ Um⋅cos( 0 ) + i⋅ Um⋅sin( 0 ) + Um⋅cos⎛⎜ ⋅π ⎞⎟ + i⋅ Um⋅sin⎛⎜ ⋅π ⎞⎟
substitute , t = 0
⎝3 ⎠
⎝3 ⎠
simplify
UzL
1
for plotting ...
i := 1 .. 3
⎛ sin⎛ π ⎞ ⎞
⎜ ⎟ ⎟
⎜
⎝3⎠
angleωt = atan⎜
⎟
⎜ 1 + cos⎛⎜ π ⎞⎟ ⎟
ω1 := 1 t1 := 0.79
⎝
⎝3⎠⎠
(
φ1 := 1
)
U1m := 1 U1p := U1m⋅ cos ω1 ⋅ t1 − α i
i
(
⎛ sin⎛ π ⎞ ⎞
⎜ ⎟ ⎟
⎜
⎝3⎠
atan⎜
⎟ = 30 deg
⎜ 1 + cos⎛⎜ π ⎟⎞ ⎟
⎝
⎝3⎠⎠
)
(
)
U1zp := U1m⋅ cos ω1 ⋅ t1 − α i + U1m⋅ sin ω1 ⋅ t1 − α i ⋅i
i
2
Up1
Up2
Up3
-Up2 ref Up1
Up1-Up2
1.5
1
0.5
0
0.5
1
1
0.5
0
0.5
1
similarly in a delta connection ... current has same geometry
UL = Up
(2.86)
IL = Ip ⋅ 3
(2.87)
10
11/13/2006
2.3.5 Magnetic induction
B = μ⋅
I
T = 1 tesla
B = flux_density
2⋅π⋅r
T=1
2
H
μ = permeability_of_medium
m
=1
m⋅kg
H
2 2
m
A ⋅s
μ 0 = permeability_of_vacuum
μ R = permeability_of_medium_relative_to_vacuum
=1
I
kg
T=1
2
amp⋅ s
m
(2.90)
μ = μ 0 ⋅μ R
Wb
henry
r
m
μ 0 := 4⋅π⋅10
B
7H
m
unitless
magnetic field around wire carrying current
derived from Biot-Savart law
e.g. magnetic field at point P results from motion of charged particle at velocity V in vacuum
→ →
V × ar
B=
⋅q⋅
T
2
4⋅π
r
μ0
parameters
units and equivalents
T = 1 tesla
B = flux_density
T=1
1
2
Wb
m
μ 0 = permeability_in_vacuum
μ 0 := 4⋅π⋅10
C = 1 coul
q = charge
→
V = velocity_vector_of_charge
−7H
H
m
m
=1
N
H
2
m
A
=1
newton
2
amp
C = 1 A⋅ s
m
s
→
ar = unit_vector_from_charge_q_to_point_P
r = distance_from_P_to_charge
m
units check
H
m
⋅C⋅
m 1
⋅
= 1T
2
s
m
→ →
V × ar
dB =
⋅dq⋅
2
4⋅π
r
→
→
q ⋅ V = I⋅ dl
μ0
differential
form
line currents ...
so ..
4
T = 1 × 10 gauss
→ →
dl × ar
dB =
⋅I⋅
2
4⋅π
r
μ0
⌠
⎮
B=⎮
⎮
⌡
μ0
I→ →
dl × ar
4⋅π 2
r
⋅
11
11/13/2006 e.g. long straight wire with current I
I
→ →
dl × ar
μ 0 dl⋅ sin( θ
)
dB =
⋅I⋅
=
⋅I⋅
2
2
4⋅π
4⋅π
r
r
μ0
see figure at right
θ
dl⋅ sin( θ ) = r⋅ dα
dl⋅ sin( θ )
r⋅
=
dα
sin( θ )
=
dα
cos( α )
r
dα
r*dα
α
geometry for solution set up
r
dα
=
cos( α ) ⋅ dα
R
π
dα
dl
⋅ sin( θ )
r
R
r
θ
θ
2
r
R
r
dl
2
r=
dB into
paper
R
sin( θ ) =
r⋅ dα
cos( α ) =
dl
R
r
→
→
dl × ar
μ 0 dl⋅ sin( θ )
μ 0 cos( α ) ⋅ dα
dB =
⋅I⋅
=
⋅I⋅
=
⋅I⋅
2
2
R
4⋅π
4⋅π
4⋅π
r
r
μ0
π
⌠2
μ 0 cos( α )
⌠
⎮
B = ⎮ 1 dB = ⎮
⋅I⋅
dα
R
4⋅π
⌡
⎮
⌡− π
2
⌠2
μ 0 cos( α )
⎮
1 μ0 I
⋅I⋅
dα
→
⋅
⋅
⎮
R
2 π Ω
4⋅π
⎮
⌡− π
B=
μ 0 ⋅I
Q.E.D.
2⋅π⋅R
2
if area not vacuum, substitute μ =μ r*μ 0 ...
magnetic flux density over an area AA
⌠
Φ := ⎮ B dAA
⌡
2
AA = enclosed_area
to distinguish from A
(ampere)
Wb = 1 weber
Wb = 1
kg⋅ m
2
A = 1 amp
A⋅ s
12
11/13/2006 Lorentz force
I
a force will act on a current carrying conductor when it
is placed in a magnetic field
FL
FL = B⋅ I⋅len
(2.92)
→
⎯
FL = I⋅ len × B
FL = Lorentz_force
N = 1 newton
B = flux_density
T = 1 tesla
I = current_thru_conductor
A = 1 amp
len = length
m
l=len
B into paper
T⋅ A⋅ m = 1 N
B⋅ I⋅ sin( angle)
where x is vector cross product and magnitude is
right hand rule applies
view of single coil in magnetic field (B) with current (I) (slightly revised from text; len*sin(β)
I
I into paper
β
h
F into
paper
F
h
len
F
len*sin(β)
side view
force on one segment (h) of coil
torque on coil depends on β
B
top view
F = I × B⋅h
N.B. I is perpendicular to B =>
M = F⋅ len⋅sin( β )
M = F⋅ len⋅sin( β ) = I⋅ B⋅h⋅len⋅sin( β ) = I⋅ B⋅AA⋅sin( β ) = I⋅ Φ ⋅sin( β )
account for multiple windings (turns) (N)
F
len⋅ sin( β ) = distance_between_couple_of_force_F
AA = area_of_coil = enclosed_area
M = N⋅ I⋅ Φ ⋅ sin( β )
general relationship recognizing proportionality to I* Φ
F = I⋅ B⋅h
M = Km⋅Φ⋅I
AA to distinguish
from A (ampere)
Km = constant_for_given_motor
(2.93)
13
11/13/2006 Faraday's Law
Voltage is generated in conductor when moving in magnetic field
E = −B⋅len⋅v
E = induction_potential = electromotive_force
V = 1 volt
units check
T = 1 tesla
B = flux_density
m
v = velocity
T⋅
s
len = length_of_conductor
m
s
⋅m = 1 V
m
E
E as shown is positive value and direction
minus sign is consistent with observation that E as shown
would produce a current in the same direction which in
turn would produce a force opposite to velocity.
v l=len
E = −(B × v)⋅len
vector form ...
consistent with text ... multiply by sin(α) where α is
angle between B and v
B into paper
d
E=− Φ
dt
may also be expressed as ...
since ...
Φ = B⋅ Area
substituting ...
E = −N
d
Φ
dt
Wb
s
d
d
d
d
E = − Φ = − (B⋅ Area) = −B Area = −B⋅len x
dt
dt
dt
dt
in coil rotating in constant magnetic field B
and with N turns ...
for single turn and ..
for N turns
= 1V
as ...
Φ = B⋅ Area⋅ cos( β ) = B⋅ Area⋅cos( ω⋅t)
(2.95)
Area = len⋅ x
where ...
Area = area_enclosed_in_coil
d (
E = −N B⋅ Area⋅cos( ω⋅t) ) → E = N⋅ B⋅ Area⋅ sin( ω⋅t) ⋅ ω
dt
rad
ω = 2⋅π⋅n
n = rpm
rpm = 6.283
E = N⋅ 2⋅ π ⋅n⋅B⋅ Area⋅sin( ω⋅t)
min
as abve for motor constant ... express ...
E = KE⋅Φ⋅n
KE = constant_for_given_motor
E = induced_electromotive_force
V = 1 volt
Φ = magnetic_flux
Wb = 1 weber
n = rotation_speed
rpm = 0.105
1
sec
1Wb⋅ 60rpm = 6.283 V
14
11/13/2006