AC Circuits
Part 10, EE1103
Po-Tai Cheng
Dept. of Electrical Engineering
National Tsing Hua University
Hsin-Chu, TAIWAN
Part 10, EE1103 – p.1/21
Overview
Direct Current (DC) vs. Alternating Current (AC).
Sinusoidal AC;
AC generators;
single-phase and three-phase
Phasor and impedance;
Time domain and phasor domain;
Sinusoidal steady state;
Definition of impedances;
Root Mean Square (RMS) Values and AC power calculations.
Part 10, EE1103 – p.2/21
Direct Current (DC)
DC batteries are first developed by Alessandro Volta in 1799.
DC generators are developed by Thomas A. Edison; First DC distribution
network implemented in Manhanttan, New York, USA.
Volta’s first battery
Pearl Street Power Station in Mahattan, NYC, NY
Part 10, EE1103 – p.3/21
Direct Current (DC)
Edison’s DC generators.
Part 10, EE1103 – p.4/21
Alternating Current (AC)
AC induction machines are developed in 1888 by Nikola Tesla.
AC generation and distribution system developed in 1890s by Tesla and
Westinghouse.
Part 10, EE1103 – p.5/21
AC generator
a1
PSfrag replacements
c2
N
b2
ω
b1
S
c1
a2
A winding placed in the rotor to produce flux, and the rotor rotates at the rate
of 3600 RPM(= 60 × 2π = 377 rad./s)
Three sets of copper windings are distributed around the stator. The flux
linkage of each set of windings are given as:
φa = Φm sin(ωt);
φb = Φm sin(ωt − 2π/3);
φc = Φm sin(ωt + 2π/3);
Part 10, EE1103 – p.6/21
Three-Phase generator
By Farady’s law, the induced voltage across the winding terminals are:
dφa
dφb
= ωΦm cos(ωt); vb =
= ωΦm cos(ωt − 2π/3);
va =
dt
dt
dφc
= ωΦm cos(ωt + 2π/3);
vc =
dt
vb
va
1
vc
0.8
0.6
0.4
0.2
0
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−0.2
−0.4
−0.6
−0.8
−1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Part 10, EE1103 – p.7/21
Sinusoidal Steady State
Assuming the circuit under sinusoidal excitation has reached steady-state:
v(t) = Vm sin (ωt + φ)
where
Vm
=
the amplitude of the sinusoid
ω
=
the angular frequency in rad/sec
φ
= the phase angle of the sinusoid
Periodic signal: v(t) = v(t + nT ), where T =
2π
ω ,
and n is an integer.
Constant frequency: ω does not vary with time.
Part 10, EE1103 – p.8/21
Definition of Phasors
Eular’s identity: ejφ = cos φ + j sin φ
A sinusoidal signal in the time domain can be presented in complex form using
Eular’s identity:
v(t) = Vm cos(ωt + φ) = <[Vm (cos(ωt + φ) + j sin(ωt + φ))] = <[Vm ej(ωt+φ) ]
= <[Vm ejφ ejωt ] = <[Vejωt ]
where V = Vm ejφ .
In sinusoidal steady state, ω is a known constant,
Phasor of v(t) = Vm cos(ωt + φ) −→ V = Vm ejφ
or Vm ∠φ
The phasor analysis is developed by Charles P. Steinmetz in 1900s.
Part 10, EE1103 – p.9/21
Definition of Phasors
For a steady state signal
v(t) = Vm cos(ωt + φ)
dv
= −ωVm sin(ωt + φ)
dt
= ωVm cos(ωt + φ + π/2)
= <[ωVm ejφ ejωt ej
= <[jωVejωt ]
dv
dt
→ jωV
π/2
]
Z
1
Vm sin(ωt + φ)
ω
1
= Vm cos(ωt + φ − π/2)
ω
1
−π
= <[ Vm ejφ ejωt ej( /2) ]
ω
1
= <[ Vejωt ]
jω
v(t)dt =
R
v(t)dt →
1
jω V
Part 10, EE1103 – p.10/21
Impedances
iC
iL
PSfrag replacements
PSfrag replacements
vC
vL
dv
= −ωCVm sin(ωt + φ)
iC = C
dt
= ωCVm cos(ωt + φ + π/2)
= <[ωCVm ejφ ejωt ej
π/2
]
= <[jωCVejωt ]
IC = jωCV
VC
1
=
−→ ZC =
IC
jωC
Z
1
1
v(t)dt =
Vm sin(ωt + φ)
iL =
L
ωL
1
Vm cos(ωt + φ − π/2)
=
ωL
1
−π
Vm ejφ ejωt ej( /2) ]
= <[
ωL
1
Vejωt ]
= <[
jωL
1
VL
V → ZL =
= jωL
IL =
jωL
IL
Part 10, EE1103 – p.11/21
Impedances
vc = Vm cos(ωt + φ)
vL = Vm cos(ωt + φ)
ic = ωCVm cos(ωt + φ + π/2)
VC = Vm ejφ ;
g replacements
φ
ic
IC = jωCV
Ic
φ
PSfrag replacements
vc
ωt
π/2
1
Vm cos(ωt + φ − π/2)
ωL
1
jφ
V
VL = V m e ; I L =
jωL
iL =
ic
vc
ωt
Im
π/2
Vc
φ
Re
Im
Vc
φ
Re
Ic
Part 10, EE1103 – p.12/21
Impedances
Time domain
Phasor domain
v(t) = Vm cos(ωt + θv )
V
V = Vm ejθv or Vm ∠θv
V
i(t) = Im cos(ωt + θi )
A
I = Im ejθi or Im ∠θi
A
dv/dt
R
vdt
V /sec
jωV
V ·rad/sec
V · sec
V/jω
V ·sec/rad
R
Ω
ZR = R
Ω(resistance)
C
F
ZC = 1/jωC
Ω(reactance)
L
H
ZL = jωL
Ω(reactance)
At dc, ω = 0, ZL = 0, ZC → ∞.
At high frequency, ω → ∞, ZL → ∞, ZC → 0.
Part 10, EE1103 – p.13/21
Kirchhoff’s Laws for Phasors
KVL: v1 + v2 + · · · + vn = 0
→
→
→
Vm1 cos(ωt + φ1 ) + Vm2 cos(ωt + φ2 ) + · · · + Vmn cos(ωt + φn ) = 0
<[Vm1 ejφ1 ejωt ] + <[Vm2 ejφ2 ejωt ] + · · · + <[Vmn ejφn ejωt ] = 0
<[(V1 + V2 + · · · + Vn )ejωt ] = 0
→ V1 + V 2 + · · · + V n = 0
KCL: i1 + i2 + · · · + in = 0 . . . →
replacements
PSfrag replacements
Z1
Z2
I 1 + I2 + · · · + I n = 0
Z3
Zpara
Zser
Zser = Z1 + Z2 + Z3 + · · ·
Z1
1/Zpara
Z2
Z3
= 1/Z1 + 1/Z2 + · · ·
Part 10, EE1103 – p.14/21
Example 1
100 Ω
iC (t)
PSfrag replacements
0.001 F
vC (t)
vs (t)
vs (t) = 1000 cos(ωt + 30◦ ), ω = 10 rad/sec,
Find the phasors Ic , Vc , and the time domain representation ic (t), vc (t).
For ω = 100 rad/sec, find the above quantities.
Part 10, EE1103 – p.15/21
Example 1
The phasor representation of vs is Vs = 1000∠(30◦ ) V. For ω = 10 rad/sec, the
reactance of the capacitor is
1
1
=
= −j100 Ω
ZC =
jωC
j × 10 × 10−3
10∠(75◦ )
1000∠(30◦ )
1000∠(30◦ )
Vs
√
√
=
=
=
I=
◦
R + ZC
100 − j100
100 2∠(−45 )
2
10∠(75◦ )
1000∠(−15◦
1
√
√
)=
(−j100) =
VC = I(
jωC
2
2
The corresponding time-domain representation of i and vc :
i(t) = 7.07 cos(10t + 75◦ );
vC (t) = 707 cos(10t − 15◦ )
Part 10, EE1103 – p.16/21
Example 1
The phasor representation of vs is Vs = 1000∠(30◦ ) V. For ω = 100 rad/sec, the
reactance of the capacitor is
1
1
=
= −j10 Ω
ZC =
jωC
j × 100 × 10−3
1000∠(30◦ )
1000∠(30◦ )
Vs
◦
=
=
9.95∠(35.7
=
)
I=
◦
R + ZC
100 − j10
100.5∠(−5.7 )
1
) = 9.95∠(35.7◦ )(−j10) = 99.5∠(−54.3◦ )
jωC
The corresponding time-domain representation of i and vc :
VC = I(
i(t) = 9.95 cos(100t + 35.7◦ );
vC (t) = 99.5 cos(100t − 54.3◦ )
Part 10, EE1103 – p.17/21
Example 2
100 Ω
iL (t)
PSfrag replacements
10 H
vL (t)
vs (t)
vs (t) = 1000 cos(ωt + 30◦ ), ω = 10 rad/sec,
Find the phasors Ic , Vc , and the time domain representation ic (t), vc (t).
For ω = 100 rad/sec, find the above quantities.
Part 10, EE1103 – p.18/21
Example 2
The phasor representation of vs is Vs = 1000∠(30◦ ) V. For ω = 10 rad/sec, the
reactance of the inductor is
ZL = jωL = j × 10 × 10 = j100 Ω
1000∠(30◦ )
1000∠(30◦ )
Vs
√
=
=
= 7.07∠(−15◦ )
I=
R + ZL
100 + j100
100 2∠(45◦ )
VC = I(jωL) = 7.07∠(−15◦ )(j100) = 707∠(75◦ )
The corresponding time-domain representation of i and vc :
i(t) = 7.07 cos(10t − 15◦ );
vC (t) = 707 cos(10t + 75◦ )
Part 10, EE1103 – p.19/21
Root Mean Squre (RMS)
The average power consumption of a resistor R can be calculated as:
Z T 2
Z T
1 1
1
v (t)
dt =
v 2 (t)dt
PAV =
T 0
R
RT 0
q R
1 T 2
Define VRM S =
T 0 v (t)dt . The average power delivered by v(t) can be
expressed as PAV =
2
VRM
S/R
.
For a sinusoidal voltage v(t) = Vm cos(ωt + θv ),
s
s
Z T
Z T
2
1
V
1 + cos(2ωt + θv )
m
2
2
dt
V cos (ωt + θv )dt =
VRM S =
T 0 m
T 0
2
r
Vm
Vm2 T
· = √
=
T
2
2
Part 10, EE1103 – p.20/21
Conclusion
Sinusoidal AC sources;
Phasors: sinusoidal steady-state, periodic signal of a single frequency.
R, L, and C in phasor domain calculation.
KCL and KVL in the phasor domain.
RMS values.
Future Extension:
Circuit Analysis (EE2220): DC circuit analysis; AC circuits analysis.
Signal and Systems (EE3610): Continuous-time and Discrete-time
system theories.
Power Electronics (EE4815): Switch-mode power conversion circuits.
Electric Machinery (EE4840): Transformers and various electric
machines.
Part 10, EE1103 – p.21/21