2010-2011EE 2121 PASSIVE AC CIRCUITS Instructor : Robert Gander Office: 2B37 Email:...
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Transcript of 2010-2011EE 2121 PASSIVE AC CIRCUITS Instructor : Robert Gander Office: 2B37 Email:...
2010-2011 EE 212 1
EE 212
PASSIVE AC CIRCUITS
Instructor : Robert Gander Office: 2B37 Email: [email protected] 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 2
Marking System
Assignment 15Midterm Exam 25Final Exam 60Total 100
2010-2011 EE 212 3
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
Major Topics
2010-2011 EE 212 4
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 3-phase balanced/unbalanced systems
Per Unit system
Major Topics (continued)
2010-2011 EE 212 5
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.
Linear Circuit
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 6
Kirchhoff’s Voltage Law (KVL):
The sum of the instantaneous voltages around any closed loop is zero.
Kirchhoff’s Laws
v1
V ~+
-
+ +
+
v2
v3
-
- -
I
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 7
Kirchhoff’s Current Law (KCL):
The sum of the instantaneous currents at any node is zero.
Kirchhoff’s Laws (continued)
~
I1 I2
I3 - I1 + I2 + 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 8
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.
Time Domain Method Phasor Method
- applicable to both transient and steady-state circuit analysis
- very useful for transient analysis
- difficult method (often requires differentiation & integration of sinusoidal functions)
- only steady-state circuit analysis
- easy method
- sinusoidal functions represented by magnitude (usually RMS value) and phase angle - Differentiation/integration replaced by multiplication/division
Example:
v = Vm sin t V = /00
2mV
i = Im sin (t + ) I = /2
mI
2010-2011 EE 212 9
v1 = 50 sin (377 t + 200) volts and v2 = 10 sin (377 t + 100) volts.
Example: Phasor Method
Phasors: V1 = 50 /200 volts, and V2 = 10 /100 volts
V = 50 /200 + 10 /100 volts
V = 50 (cos 200 + j sin 200) Rectangular form + 10 (cos 100 + j sin 100)
Complex Numbers in Polar form
v = 59.87 sin (377 t + 18.340) volts.
V1
V2
V
Phasor Diagram = 46.985 + j 17.101 + 9.848 + j 1.736
= 56.833 + j 18.837
= 59.87 /18.340 volts
2010-2011 EE 212 10
Example: Time Domain Method
v ~ C
RL
+
-
iv = 100 sin (377t + 300)R = 10 ΩL = 1/37.7 HC = 1/7540 FFind i
100 sin (377t + 300) = L + Ri + dt
di idtC
1
Applying KVL: v = v1 + v2 + v3
perform Laplace Transform and let s = jω
Multiply by jω Divide by jω
In Phasor Method:jwC
1.
1100 /300 = L·(jω)· I + R· I + · I
2010-2011 EE 212 11
Phasor Representation of a Circuit
v ~ C
RL
+
-
i
V ~ 1/(jC)
RjL
+
-
I
100 /300 = I {j10 + 10 – j20}
I = 5√2 /750 A
i = 5√2 sin (377t + 750) A
Applying KVL: V = (jωL) I + R I + ICj
1
100 /300 = j377 x ·I + 10·I + ·I
75401
j377x
1
37.7
1
2010-2011 EE 212 12
Impedance (Z), Admittance (Y)
V ~ 1/(jC)
RjL
+
-
IZ = R + jwL – j/(wC)
= R + jXL – jXC
= R + jX
Z is a complex number. It can be expressed in rectangular form, Z = R + jXor polar form, Z = |Z| /0 , where
is the power factor angle
|Z|
R
jX
-jXC
jXL
Re
Im
Admittance, Y = 1/Z Y = G + jBwhere, G is conductance and B is susceptance
Z (impedance), R (resistance), X (reactance)XL (inductive reactance)XC (capacitive reactance)