Transcript Source Transformation

Impedance and Admittance
Objective of Lecture
 Demonstrate how to apply Thévenin and Norton
transformations to simplify circuits that contain one or
more ac sources, resistors, capacitors, and/or
inductors.
Source Transformation
 A voltage source plus one impedance in series is said to
be equivalent to a current source plus one impedance
in parallel when the current into the load and the
voltage across the load are the same.
Equivalent Circuits
Thévenin
Vth = In Zn
Norton
In = Vth/Zth
Example 1
First, convert the current source to a cosine function and then to a phasor.
I1 = 5mA sin(400t+50o) = 5mA cos(400t+50o-90o)= 5mA cos(400t-40o)
I1 = 5mA -40o
Example 1 (con’t)
 Determine the impedance of all of the components
when w = 400 rad/s.
In rectangular coordinates
Z C1 =  j / wC1  =  j /400rad / s 1F  =  j 2.5k
Z R1 = R1 = 3k
Z L1 = jwL1 = j (400rad / s )(0.3H ) = j120
Z L2 = jwL2 = j (400rad / s )(2 H ) = j800
Z R2 = R2 = 5k
Z C2 =  j / wC2  =  j /400rad / s 0.7 F  =  j 3.57k
Example 1 (con’t)
 Convert to phasor notation
Z C1 = 2.5k  90o
Z R1 = 3k0 o
Z L1 = 12090o
Z L2 = 80090o
Z R2 = 5k0 o
Z C2 = 3.57k  90o
Example 1 (con’t)


Vt h1 = I1Z C1 = 5m A  40o 2.5k  90o


Vt h1 = (5m A)2.5k   40o   90o
Vt h1 = 7.5V  130o


Example 1 (con’t)
 Find the equivalent impedance for ZC1 and ZR1 in series.
This is best done by using rectangular coordinates for
the impedances. Z eq = Z R  Z C = 3k  j 2.5k
1
Z eq1 =
1
1
3k2   2.5k2  tan1  2.5k
Z eq1 = 3.91k  39.8o
3k 
Example 1 (con’t)
 Perform a Norton
transformation.
In1 = Vt h1 /Zeq1

 3.91k  39.8 
= 7.5V / 3.91k  130   39.8 
In1 = 7.5V  130o
In1
In1 = 1.92m A  90.2o
Z n1 = Z eq1
o
o
o
Example 1 (con’t)
 Since it is easier to combine admittances in parallel
than impedances, convert Zn1 to Yn1 and ZL1 to YL1. As
Yeq2 is equal to YL1 + Yn1, the admittances should be
written in rectangular coordinates, added together,
and then the result should be converted to phasor
notation.
Example 1 (con’t)
Yn1 = 1 Z n1 = 0.256m 139.8o
 


Yn1 = 0.256m 1 cos 39.8o  j sin 39.8o
Yn1 = 0.198 j 0.164m 1

YL1 = 1 Z L1 = 8.33m 1  900
Yn1 =  j8.33m 1
Yeq 2 = 0.198 j 0.164m 1  j8.33m 1
Yeq 2 = 0.198 j8.17m 1
Yeq2 =
0.1982   8.172 m 1 tan1  8.17 0.198
Yeq2 = 0.817m 1  88.60
Example 1 (con’t)
 Next, a Thévenin transformation will allow Yeq2 to be
combined with ZL2.
Example 1 (con’t)
Vt h2 = In1 / Yt h2
Vt h2
Vt h2
1.92m A  90.2o
=
8.17m 1  88.6o
= 0.235V  1.6o
Example 1 (con’t)
Z t h2 = 1 / Yt h2 = 12288.6o
 


Z th2 = 122 cos 88.6o  j sin 88.6o
Z th2 = 2.98  j122

Z L2 = j800
Z eq3 = Z th2  Z L2
Z eq3 = 2.98  j122  j800
Z eq3 = 2.98  j 922
Z eq3 =
2.982  9222 tan1 922 2.98
Z eq3 = 92289.8o
Example 1 (con’t)
 Perform a Norton transformation after which Zeq3 can
be combined with ZR2.
Example 1 (con’t)
In2 = Vt h2 Z eq3
0.235V  1.60
In2 =
92289.80
In2 = 0.255m A  91.40
Example 1 (con’t)
Yeq4 = 1 / Z eq3  1 / Z R 2
Yeq4 = 1.08m 1  89.8o  0.2m 10 o


Yeq 4 = 1.08m 1 cos(89.8o )  j sin(89.8o )  0.2m 1
Yeq 4 = 0.204 j1.08m 1
Yeq4 = 1.10m 1  79.4 o
Z eq4 = 90679.4 o
Example 1 (con’t)
Use the
equation for
current
division to
find the
current
flowing
through ZC2
and Zeq4.
IC 2 =
YC 2
YC 2  Yeq 4
In2
YC 2 = 1 / Z C 2 = 0.280m90o
YC2 = j 0.280m
Yeq4 = 0.204 j1.08m
IC 2
0.280m90o
=
0.255m A  91.40
j 0.280m  0.204 j1.08m
IC 2
0.280m90o
=
0.255m A  91.40
0.204 j 0.8m
IC 2
IC 2



0.280m90o
0
=
0
.
255
m
A


91
.
4
0.826m  75.7 o
= 86.0 A74.3o



Example 1 (con’t)
 Then, use Ohm’s Law to find the voltage across ZC2 and
then the current through Zeq4.


VC 2 = IC 2 Z C 2 = 86.0A74.3o 3.57k  90o
VC 2 = 0.309V  15.7 o
VC 2 = Veq4
Ieq4 =
Veq4
Z eq4
0.309V  15.7 o
=
90679.4o
Ieq4 = 0.341m A  95.1o

Example 1 (con’t)
 Note that the phase angles of In2, Ieq4, and IC2 are all
different because of the imaginary components of Zeq4
and ZC2.
 The current through ZC2 leads the voltage, which is as
expected for a capacitor.
 The voltage through Zeq4 leads the current. Since the
phase angle of Zeq4 is positive, it has an inductive part to
its impedance. Thus, it should be expected that the
voltage would lead the current.
Electronic Response
 Explain why the circuit on the right is the result of a
Norton transformation of the circuit on the left. Also,
calculate the natural frequency wo of the RLC network.
Summary
 Circuits containing resistors, inductors, and/or
capacitors can simplified by applying the Thévenin
and Norton Theorems.
 Transformations can easily be performed using currents,
voltages, impedances, and admittances written in
phasor notation.
 Calculation of equivalent impedances and admittances
requires the conversion of phasors into rectangular
coordinates.
 Use of the current and voltage division equations also
requires the conversion of phasors into rectangular
coordinates.