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ELE130 Electrical Engineering 1. Week 5 Module 3 AC (Alternating Current) Circuits. Administration. Quiz - next week in lecture material from weeks 4, 5 & 6 Laboratory Test postponed from next week to the week following Easter break (from week 6 to week 7)
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ELE130 Electrical Engineering 1 Week 5 Module 3 AC (Alternating Current) Circuits
Administration • Quiz - next week in lecture • material from weeks 4, 5 & 6 • Laboratory Test • postponed from next week to the week following Easter break (from week 6 to week 7) • will include work from laboratory 1, 2 & 3 • laboratory no. 4 will be done in week 6 instead of week 7 • Help Desk
Why use ac? • In power applications • easier to generate (no brushes or rectifiers) • easier to transform voltages (reduced losses) • easier to distribute (easier switching) • easy utilization (brushless motors, switching) • In communications • ac required to reproduce audio signals, to transmit over distance (radio, microwave, etc.) • DC circuits are special case of ac circuits (when frequency is zero)
Parameters of an ac wave • vs = Vm sin(t + ) • Vm : amplitude of the wave • : radian frequency = 2f, f in Hz, in rad/sec • t : time in seconds • t : (in radians) • : relative angle (in degrees or radians, but don’t MIX them up!)
Note • f (cycles / sec) and (radians/sec) • 1 cycle = 2 radians = 360o • angle can be specified in degrees or radians • 90° = /2 radians • generally, frequencies will be the same (if not big hassles!) • generally waves will have different amplitudes • is relative to reference. Of more importance however is relative phase, or phase difference
Consider two waves • Phase difference (1- 2) • (if frequency different, then relative phase changes with time) • Leading and lagging waves • using time line, the wave whose peak is occurs first is leading • Phase ‘wraps’ around i.e. 20° = 380 ° • usually work in range -180 ° to +180 °
Example - what is the phase difference? • x1(t) = X1 sin(t - 30°) & x2(t) = X2 cos (t - 40°) • convert sin to cos : sin(x) = cos (x - 90o) • x1(t) = X1 cos ( t - 30o - 90o) = X1 cos (t - 120°) • Phase difference is |-120° - (-40°)|= 80° • x2(t) leads x1(t)
Inductor response to an ac source • The current iL is: • as vL = vs: • by Ohm’s Law: |ZL|= L • current lags voltage by 90o • therefore angle of ZL = +90o • multiplying by j • ZL = jL
Capacitor response to an ac source • The current iC is: • by Ohm’s Law: ZC • current leads voltage by 90o • angle of ZC = -90o • multiplying by -j
Example - (solving circuits with an ac source) • RL circuit • solve in dc. - steady state + transient response (revision) • solve in time domain • solve in frequency domain (need complex numbers) • real (time domain ) circuits can be solved by adding an imaginary part, do the calculations in the complex domain, and then simply take the magnitude part of the complex answer as the solution to the real circuit!
Revision of DC Transients -RC • Vs is given • Time constant: • iC may change instantaneously • vC can not change instantaneously • also: • Which implies that capacitor “appears” like an open circuit to DC
Revision of DC Transients-RL • Vs is given • Time constant • iL can not change instantaneously • vL may change instantaneously • also: • Which implies that an inductor “appears” like a short circuit to DC
i(t) = ? vs= Vmcos(t) Time Domain Analysis Assume: i(t) = Imcos(t - )
Solution • KVL : vs = vR + v = i(t) R + L di/dt Vmcos(t) = RImcos(t - ) + L d(Imcos(t - ))/dt = RImcos(t - ) - LIm sin(t - ) = RIm [ cos(t)cos + sin(t)sin ] - LIm [sin(t) cos - sin cos(t)] = cos(t)[RImcos + LIm sin ] + sin(t) [RImsin - LIm cos ] • cos (t) & sin (t) are othoganal functions ( ie no multiples of cos (t) will ever equate to sin (t) ) • Hence, the 2 terms may be considered separately • Vm = RImcos + LIm sin ………………... (1) • 0 = RImsin - LIm cos …………………(2)
Solution (cont.) • From (2) • sin / cos = LIm/ RIm • tan = L / R • = tan -1L/R • Followes that sin = L / R2 + (L)2& cos = R / R2 + (L)2 • From (1) • Vm = RImcos + LIm sin = RImR / R2 + (L)2 + LIm L / R2 + (L)2 = Im [R2 + (L)2 / R2 + (L)2 ] = Im [R2 + (L)2 ] • Im = Vm / R2 + (L)2 • i(t) = Vm / R2 + (L)2 cos (t - tan -1L/R )
Complex number revision • Electrical Engineers already use i therefore use ‘j’ • What is j2 ? • if multiplication by j is a rotation by 90°, then j2 must be a rotation by 180° • j2 = -1 j = -1 (which cannot really exist so must be imaginary) • Complex numbers can be expressed in • cartesian co-ordinates (addition & subtraction) or • polar co-ordinates (multiplication & division) • Euler’s Identity : z=a+jb = M[costjsin t] = Mej t • where M = magnitude
X = Xm where is assumed or stated as a side note Phasor notation - omits the time depend part and allows us to write the magnitude and phase only Polar: X = Xm Cartesian: X = XR + jXI XR = Xmcos XI = Xm sin Xm = (X2R + X2I) = tan -1 (XI / XR) jXI XR Phasor notation Xm
Complex Impedance's • the ratio of voltage to current across and element is known as IMPEDANCE • Impedance is frequency complex, containing both real and imaginary components • Symbol is Z (which is complex) : Z = R + jX • R & X are both real • R is always 0 or positive • X is reactive component, can be positive or negative
Impedance • ZR = R • Inductance gives a positive reactance ZL = j L • Capacitance gives negative reactance
Steps in solve AC Problems 1. Confirm sources are of the same frequency 2. Convert to Phasor notation 3. Treat as a DC problem (with complex numbers) 4. Convert Back to Time domain
Example (cont. in phasor notation) • Total impedance : ZT = R + jL = R2 + (L)2 tan -1 (L/R) • Ohms Law : Vs = IT . ZT • IT = Vm0° / R2 + (L)2 tan -1 (L/R) = [Vm / R2 + (L)2 ] 0° - tan -1 (L/R) = Vm / R2 + (L)2 cos ( t - tan -1 (L/R)) i(t) = ? vs= Vmcos(t) jL Vm0°