ELE130 Electrical Engineering 1
Week 5Module 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
v a va= v a( t) = V a( 0 -p k ) s in ( t+) = V m s in (t+)
t
Vm
-Vm
T
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!)
Generation of an ac wave
t0o
45o
90o
135o
180o
225o270o
315o
360o 0 o 4 5 o 9 0 o 1 3 5 o
1 8 0 o
2 2 5 o 2 7 0 o 3 1 5 o 3 6 0 o
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 °
Lagging and Leading Waveformsv
va
vb
a b
= a - b
Say a = + 30o an d b = -3 0o
th erefore = + 30o - (-30o) = 6 0o
Th erefore vb is lagging va by 6 0o
va is lead in g vref by a (i.e. by +3 0o)
vb is lagging vref by b (i.e. by -3 0o)
vref All vref, va an d vb areof that same freq uency
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
L vL
iL
~vs(t) = V m sin (t)+
-
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
dtvL
idtdiLv LL
LL
1
dttVL
dtvL
i msL )sin(11
)90sin()cos( otLmVt
LmV
)90sin( om tLV
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
C vC
iC
~vs(t) = V m sin (t)+
-
)cos( tCVdtsdvC
dtdv
Ci mC
C
C1
CjCjZC
11
Voltage - Current Relationships
element phase v/ i ratio
R i in phase i proportional v/R
C i leads v by 90o i proportional vC
L i lags v by 90o i proportional v/L
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
+_
R
Vs C vC
iC
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
teIIIi FinitFC
)(
RC
teVVVv FinitFC
)(
dtdv
Ci CC
Revision of DC Transients-RL
+_
R
Vs L vL
iL 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
RL
teFIinitI
FI
Li
)(
teFVinitV
FV
Lv
)(
dtdiLv L
L
Time Domain Analysis
Assume: i(t) = Imcos(t - )
i(t) = ?
vs= Vmcos(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)[RIm cos + LIm sin ] + sin(t) [RIm sin - 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 = RIm cos + LIm sin ………………... (1)
0 = RIm sin - LIm cos …………………(2)
Solution (cont.)
From (2) sin / cos = LIm / RIm tan = L / R = tan -1 L/R Followes that sin = L / R2 + (L)2 & cos = R / R2 + (L)2
From (1) Vm = RIm cos + LIm sin
= RIm R / 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 -1 L/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
Phasor notation 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 + X2
I)
= tan -1 (XI / XR)
XR
jXI 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
CjCjZC
11
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)Vm0°
jL
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