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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