Transcript Slide 1

EE 212
PASSIVE AC CIRCUITS
Instructor : Robert Gander
Email: [email protected]
Office: 2B37
Phone: 966-4729
Class Website: www.engr.usask.ca/classes/ee/212
Text Recommended:
Introduction to Electric Circuits - 5th or Higher Edition
- R.C. Dorf and J.A. Svoboda
2010-2011
EE 212
1
Marking System
Assignment
Midterm Exam
Final Exam
Total
2010-2011
EE 212
15
25
60
100
2
Major Topics
 Introduction
 Phasor diagrams, impedance/admittance, resistance and
reactance in complex plane
 Loop (or Mesh) and Nodal analysis
 Power factor, real and reactive power
 Thevenin's/Norton's theorem, maximum power transfer
theorem, wye-delta transformation
 Superposition theorem, multiple sources with different
frequency, non-sinusoidal sources
2010-2011
EE 212
3
Major Topics (continued)
 Coupled circuits
 Transformer action, equivalent circuit, Losses
 Transformer open and short circuit tests, efficiency and
voltage regulation
 3-phase systems, Y-delta connections/transformations
 Multiple 3-phase loads
 Power Measurement, Wattmeter connections in 1-phase and 3phase balanced/unbalanced systems
 Per Unit system
2010-2011
EE 212
4
Linear Circuit
 obeys Ohm’s Law (i.e. v α i or v = Ri)
 if the i or v in any part of the circuit is sinusoidal, the i and v in every other
part of the circuit is sinusoidal and of the same frequency
Non-linear circuits do not obey Ohm’s Law.
Circuit Elements:
 Active – supply energy: voltage or current source
e.g. battery, function generator, transistor, IC components
 Passive – absorb energy
e.g. resistor, inductor, capacitor
EE 212 deals with
“Steady-state analysis of linear AC circuits” (mainly power circuits)
2010-2011
EE 212
5
Kirchhoff’s Laws
Kirchhoff’s Voltage Law (KVL):
The sum of the instantaneous voltages around any closed loop
is zero.
v1
v2 +
+
V ~
-
+
I
+
v3
-
-
v1 + v2 + v3 - V = 0
Sign convention: For a current going from +ve to –ve, Voltage is +ve
In a voltage source, the polarities are known.
In a passive element (R, L or C), the current always goes from +ve to –ve.
This law can be used to calculate the current in a loop from which the
individual currents in each element can be calculated.
2010-2011
EE 212
6
Kirchhoff’s Laws (continued)
Kirchhoff’s Current Law (KCL):
The sum of the instantaneous currents at any node is zero.
I2
I1
~
I3
- I1 + I 2 + I3 = 0
Sign convention: Current exiting a node is taken as +ve
This law can be used to calculate the voltage at the different nodes in a
circuit.
2010-2011
EE 212
7
Circuit Analysis
When a circuit has more than one element, a circuit analysis is required to
determine circuit parameters (v, i, power, etc.) in different parts of the
circuit. There are different methods for circuit analysis.
Phasor Method
Time Domain Method
- only steady-state circuit analysis
- applicable to both transient and
steady-state circuit analysis
- easy method
- sinusoidal functions represented by
magnitude (usually RMS value) and
phase angle
- Differentiation/integration replaced
by multiplication/division
- very useful for transient analysis
- difficult method (often requires
differentiation & integration of
sinusoidal functions)
Example:
V = Vm /00
2
I = I m /
v = Vm sin wt
i = Im sin (wt + )
2010-2011
EE 212
2
8
Example: Phasor Method
v1 = 50 sin (377 t + 200) volts and
v2 = 10 sin (377 t + 100) volts.
Phasors: V1 = 50 /200 volts, and V2 = 10 /100 volts
V = 50 /200 + 10 /100 volts
Complex Numbers in Polar form
V = 50 (cos 200 + j sin 200)
+ 10 (cos 100 + j sin 100)
Rectangular form
Phasor Diagram
= 46.985 + j 17.101
+ 9.848 + j 1.736
V1
V
= 56.833 + j 18.837
= 59.87
/18.340
V2
volts
v = 59.87 sin (377 t + 18.340) volts.
2010-2011
EE 212
9
Example: Time Domain Method
i
L
R
+
v ~
-
Applying KVL:
C
v = v1 + v2 + v3
100 sin (377t + 300) = L
In Phasor Method:
v = 100 sin (377t + 300)
R = 10 Ω
L = 1/37.7 H
C = 1/7540 F
Find i
100
di + Ri + 1 idt

C
dt
/300
1 1
= L·(jω)· I + R· I + .
·I
C jw
Multiply by jω
2010-2011
perform Laplace Transform
and let s = jω
EE 212
Divide by jω
10
Phasor Representation of a Circuit
i
L
I
R
+
v ~
-
C
Applying KVL:
100
+
V ~
-
V = (jωL) I + R I +
/300
jwL
R
1/(jwC)
1
I
jw C
1
1
= j377 x
·I + 10·I +
·I
1
37.7
j377x
7540
100 /300 = I {j10 + 10 – j20}
I = 5√2 /750 A
i = 5√2 sin (377t + 750) A
2010-2011
EE 212
11
Impedance (Z), Admittance (Y)
I
Z = R + jwL – j/(wC)
jwL
R
+
V ~
-
= R + jXL – jXC
= R + jX
Z (impedance),
R (resistance), X (reactance)
XL (inductive reactance)
XC (capacitive reactance)
Z is a complex number. It can be expressed in
rectangular form, Z = R + jX
or polar form,
Z = |Z| /0 , where
 is the power factor angle
1/(jwC)
Im
jXL
jX
-jXC
|Z|

Re
R
Admittance, Y = 1/Z
Y = G + jB
where, G is conductance and B is susceptance
2010-2011
EE 212
12