SERIES RL CIRCUITS (1)
R0  R  j 0 
et   E p sin ωt
+
 E p 0 V
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R
L
ωL90  0  jωL 
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Circuit above is a series RL network connected to an ac
voltage source
Need to find the phasor form of the total impedance of
this combination
The total impedance of this series combination is
ZT  R  j0  0  jωL  R  jωL  R  j X L ohms
The magnitude and angle of ZT can be found by
converting to polar form:
|ZT| = √[R2+(ωL)2] and θ = tan-1(ωL/R)
The plot of ZT:
Im
ZT  R  jωL  ZT θ 
jωL
ZT  R 2  ωL 
2
θ  tan1 ωL R 
θ
R
Series and Parallel AC
Circuits
Re
1
SERIES RL CIRCUITS (2)
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Example: For a series RL combination circuit, R =
300, L = 0.2H and e(t) = 17sin(2000t) V. Find the total
equivalent impedance in polar form and rectangular
form. Sketch the impedance in the complex plane.
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We can use Ohm’s law to find the total current supplied
by a voltage source: iT = v / ZT
Mathematical operation must be carried out using
phasors since all quantities have both magnitude and
angle
The current iT in a series circuit is the same through
every series-connected component
The ac voltage drop across each component can be
found by multiplying each impedance by the current
v1 = iTZ1, v2 = iTZ2, …
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Example: In a series RL circuit, where e(t) = 300 V,
R = 200, XL = j100. Find the total current in the
circuit. Find the voltage drops across R and L. Verify
KVL around the circuit. Draw a phasor diagram showing
e, vR, vL and iT. Sketch the voltage waveforms.
Series and Parallel AC
Circuits
2
SERIES RC CIRCUITS (1)
R0  R  j 0 
et   E p sin ωt
+
R
1
  90  0  j ωC 
ωC
C
 E p 0 V
-
•
•
Above is a series RC network connected to an ac
voltage
The total impedance of this combination is
ZT  R  j0  0  j ωC   R  j ωC  R  j X C
•
The polar coordinates are
ZT  R 2  1 ωC  ohms
2
  1 ωC 
1
θ  tan1 
  tan  1 ωRC
 R 
•
The phasor diagram of the total impedance is
Im
R
Re
θ
Z T  R 2  1 ωC 
2
θ  tan1  1 ωRC 
-j/ωC
ZT  R  j ωC
 ZT θ 
Series and Parallel AC
Circuits
3
SERIES RC CIRCUITS (2)
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As in the series RL circuit, the total current supplied to
the RC network and the voltage drops can be found by
using Ohm’s Law:
iT = e / ZT
vR = iTR and vC = iTXC
Example: For a series RC circuit where e(t) =
18sin(240t + 45) V, R = 3.3k and C = 2.2µF:
a) Find the total current in the circuit in phasor and
sinusoidal form
b) Find the voltage drops across the resistor and
capacitor in phasor and sinusoidal form
c) Verify KVL around the circuit
d) Draw a phasor diagram showing e, iT, vR and vC
e) Sketch the waveforms of e, vR and vC versus angle
Series and Parallel AC
Circuits
4
SERIES RLC CIRCUITS (1)
R0  R  j 0 
i(t)
R
et   E p sin ωt
 E p 0 V
+
L
ωL90  0  jωL 
-
C
1
  90  0  j ωC 
ωC
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The total impedance of the RLC circuit is
ZT  R  j ωL  1 ωC  ohms
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In terms of magnitudes it is: ZT = R + j(|XL| - |XC|)
Inductive and capacitive reactance have opposite signs
Thus net reactance may be either inductive or
capacitive, depending which is larger
Polar coordinates are
ZT  R 2  ωL  1 ωC   R 2   X L  X C
2
θ  tan1

2
X  XC
ωL  1 ωC
 tan1 L
R
R
Series and Parallel AC
Circuits
5
SERIES RLC CIRCUITS (2)
Im
jωL
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The phasor diagram of
the impedance when
inductive reactance is
greater than the
capacitive reactance,
i.e. when |XL|>|XC|
ZT
ωL – 1/ωC
Net reactance
θ
R
Re
-j/ωC
Im
•
The phasor diagram of
the impedance when
capacitive reactance
is greater than the
inductive reactance,
i.e. when |XC|>|XL|
jωL
R
Re
θ
ωL – 1/ωC
Net reactance
ZT
-j/ωC
Series and Parallel AC
Circuits
6
SERIES RLC CIRCUITS (3)
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When there is more than one resistor, capacitor and/or
inductor in a series circuit, the total impedance has a
resistance component equal to the sum of the
resistance values and a reactive component equal to
the sum of the capacitive reactances subtracted from
the sum of the inductive reactances
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Example: In a series RLC circuit, e(t) = 100sinωt, R =
800, ZL = j1250 and ZC = -j450.
a) Find the current in polar form
b) Find the voltage drops vR, vL, vC
c) Verify KVL around the circuit
d) Draw a phasor diagram showing e, i, vR, vL, vC
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Series and Parallel AC
Circuits
7
ADMITTANCE
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The reciprocal of resistance is conductance, with units
Siemens
The reciprocal of impedance is admittance, denoted Y
Y = 1/Z siemens
Impedance is a measure of the extent to which a
component impedes the flow of ac current through it
Admittance is a measure of how well it admits the flow
of ac current
The greater the admittance, the smaller the impedance,
and vice versa
Resistance R is one form of impedance Z, conductance
G is one form of admittance Y
Phasor form of conductance is:
1
1
 0 siemens
R  0 R
1
G   j0 S
R
G
Series and Parallel AC
Circuits
8
SUSCEPTANCE
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Reactance is another special case of impedance
The reciprocal of reactance is called susceptance, B
B = 1/X siemens
There are two types of susceptance
Inductive susceptance:
BL 
1
1
1


  90 siemens
X L ωL90 ωL
BL  0  j 1 ωL  S
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Capacitive susceptance:
BC 
1
1

 ωC90 siemens
X C 1 ωC   90
BC  0  jωC S
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Z = 1/Y; R = 1/G; XL = 1/BL; XC = 1/BC
Example: Find the admittance of a 5 resistor; a 5mH
inductor at f = 60Hz; a 0.2µF capacitor at ω = 1.25106
rad/s.
Example: Find the admittance Y corresponding to Z =
30 + j40 . Draw a phasor diagram showing Y in the
complex plane.
Series and Parallel AC
Circuits
9
PARALLEL AC CIRCUITS (1)
+
YT
Z1
Z2
-
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Above shows a parallel connected set of impedances to
an ac source
Total admittance of the circuit is the sum of the
admittances of the parallel connected components, i.e.
YT = Y1 + Y2 +… + Yn = 1/Z1 + 1/Z2 +…+ 1/Zn
Example: Find the total admittance of a 20mH inductor,
a 1k resistor and a 0.16µF capacitor connected in
parallel to an ac voltage source of 15sin(25103t) V.
Draw a phasor diagram showing the total admittance in
the complex plane. What is the total admittance if the
frequency of the voltage source is doubled?
Series and Parallel AC
Circuits
10
PARALLEL AC CIRCUITS (2)
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The total impedance of a network is the reciprocal of its
total admittance, ZT = 1/YT
For a parallel network:
ZT 

1
1

YT Y1  Y2    Yn
1
1 Z1  1 Z 2    1 Z n
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Example: Three components (an inductor with
impedance j10, a resistor with impedance 2, and
another inductor with impedance j5) are connected in
parallel. Find the total equivalent impedance of this
network.
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Example: A capacitor with impedance –j100 and an
inductor with impedance j500 are connected in
parallel. Find this network’s total equivalent impedance.
Series and Parallel AC
Circuits
11
PARALLEL AC CIRCUITS (3)
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The same voltage appears across every parallelconnected impedance
So current can be found using Ohm’s law and KCL
iT = i1 + i2 +…+ in
+
e
i1 =
e/Z1
in =
e/Zn
i2 =
e/Z2
Z1
Z2
Zn
-
i1  e Z1  eY1
  
in  e Z n  eYn
iT  i1  i2    in
iT  e Z T  eYT
Series and Parallel AC
Circuits
12
PARALLEL AC CIRCUITS (4)
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Example: For the circuit below, find the current in each
impedance. Show that iT = eYT. Draw a phasor diagram
showing e, iT, i1, i2 and i3.
iT
i1
i2
i3
Z1 =
50
Z2 =
j40
Z3 =
-j80
+
e = 60sin(ωt) V
-
Series and Parallel AC
Circuits
13
AC CURRENT SOURCES (1)
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Like a dc current source, an ideal ac current source
supplies the same constant current to whatever
network is connected across its terminals
For ac, the current is constant in the sense that its peak
value does not change
The same symbol is used for an ac current source as
for a dc current source
The direction of the arrow is just a phase reference,
since current reverses direction every half cycle
Reversing the arrow is the same as multiplying the
current by -1, which is the same as adding or
subtracting 180 from its phase angle
Ipsin(ωt + θ) A
= Ipθ
= -Ipθ
Series and Parallel AC
Circuits
=
Ipθ+180
14
AC CURRENT SOURCES (2)
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Example: For the circuit below, find the voltage vT
across the ac current source. Find the currents i1 and i2.
Show that i = i1 + i2. Draw a phasor diagram showing i,
vT, i1 and i2. Sketch the waveforms of i, i1 and i2 versus
angle.
i(t) =
0.24sin(5106t) A
vT
i1
100
Series and Parallel AC
Circuits
i2
2000pF
15
POWER IN CIRCUITS CONTAINING
REACTANCE (1)
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Recall that the average power dissipated by a
resistance carrying sinusoidal ac current can be found
by
Pavg 
I p2 R

Vp I p
2R
2
Veff2
2
 I eff R 
 Veff I eff
R
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2

V p2
Above can be used to find Pavg dissipated by a resistor
in a circuit containing reactance, provided the values
used in the computations are indeed those of the
voltage across and/or current through the resistor itself
In many practical circuits, it is sometimes necessary to
compute Pavg when the resistance or resistor
voltage/current are not known
Series and Parallel AC
Circuits
16
POWER IN CIRCUITS CONTAINING
REACTANCE (2)
i = (Ep/|Z|)  
→
+
e = Ep

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Above is an ac circuit with a general impedance Z
Assuming that the voltage supplied has angle 
i
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|Z|
E p φ

Ep
φ  θ 
Z θ Z
Angle between voltage and current is  - ( - ) = 
The angle between the voltage applied to a network
and the total current supplied to it equals the angle
of the impedance of the network
Resistive component of impedance Z is R = |Z|cos
ohms
Since only resistance in network dissipates power
Pavg 
I p2 Z cosθ
I p2 R

2
2
2
I p Z  Ep , I p Z  I pEp
Pavg 
I p E p cosθ
2
Series and Parallel AC
Circuits
17
POWER FACTOR
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Equations on previous slide provide a means for
computing the average power in terms of voltage e
across whole network and total supply current i
Cos is called the power factor
For a purely resistive network, voltage and current are
in phase, so power factor = cos0 = 1 → Pavg = EpIp / 2
For a purely reactive network, voltage and current are
separated by ±90, and cos ±90 = 0 → Pavg = 0, i.e no
power is dissipated by purely reactive components
(capacitors and inductors)
Since Vp = 2Veff and Ip = 2Ieff we can derive
Pavg 

2Veff

2 I eff
2
 Veff I eff cosθ
 cosθ
Series and Parallel AC
Circuits
18
EXAMPLE OF POWER IN AC
CIRCUITS
j80
i
→
+
e = 600

100
-j20
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In the circuit above:
Find the power factor
Use the power factor to find the average power
dissipated in the network
Verify that the power computed above is the same as
the power computed by using the voltage across and
current through the resistor
Series and Parallel AC
Circuits
19