2010-2011 EE 212 1

EE 212

PASSIVE AC CIRCUITS

Instructor : Robert Gander Office: 2B37 Email: bob.gander@usask.ca 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

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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

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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)

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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)

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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.

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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.

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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

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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

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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

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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

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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)