Ee201 -revision_contigency_plan

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REVISION Contingency Plan

Transcript of Ee201 -revision_contigency_plan

REVISION

Contingency Plan

Motivation Word

What now?

3

More advanced circuit analysis!

Nodal Analysis Extension of KCL concepts

Loop Analysis Extension of KVL concepts

Thevenin / Norton equivalent circuits Ways to make complex circuits simple

NODAL ANALYSIS

Ref. Node/point,

Datum Node, ground node

NON-Ref. Node/point

Apply kcl at NON-Ref.

Node/point

1

2

SUMMARY

EXAMPLE

What Do You Think?

Test Yourself

`

Nodal Analysis

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How do we tackle a circuit like this?

Not just one loop or one node

Simultaneous equations!

Use node voltages as variables : Nodal Analysis.

If we can find the node voltages, everything else is easy Ohm’s law etc

Nodal Analysis

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Choose one node as the reference.

Usually the node with the most connections.

Commonly called ground.

At ground-zero potential

Might (should!) represent the chassis or ground line in an actual circuit.

Assume all voltages are positive with respect to ground.

For N nodes, we need N – 1 linearly independent KCL equations to solve (with respect to ground).

Nodal AnalysisCircuits with independent current sources

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Consider the network shown.

There are three nodes, so we need two equations.

We can use the bottom node as the reference (ground), so we still need to find v1 and v2.

Use KCL and Ohm’s law to write two nodal equations.

N – 1 linearly independent KCL

Nodal AnalysisCircuits with independent current sources

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Solution 1) Use Gaussian Elimination to find the unknowns, if

IA = 1 mA, R1 = 12 kΩ, R2 = 6 kΩ, IB = 4 mA and R3 = 6 kΩ.

Nodal AnalysisCircuits with independent current sources

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Solution 2) Use Matrix Analysis to find the unknowns, if

IA = 1 mA, R1 = 12 kΩ, R2 = 6 kΩ, IB = 4 mA and R3 = 6 kΩ.

Example

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Oh oh! We can’t write KCL for V1 or V2 (we don’t know the current through the voltage source…)

Example

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But we can write V1 – V2 = 6 V. There are 3 nodes, so we need 2equations: now we have one equation, so we only need one more.

Summary:Problem Solving Strategy

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Step 1)

Determine the number of nodes, select one as the reference (ground). Assign voltage names to the other nodes.

Step 2)

Write an equation for each voltage source (i.e.: V2 = 12 V, or V1 – V4 = -6 V). Each of these equations is one of the linearly independent equations we will use. Identify supernodes when a voltage source connects between two non-reference nodes.

Step 3)

Use KCL to write the remaining equations. For N nodes, you will need a total of N – 1 equations. Solve to find the unknown voltages.

Chap 1 – Steady State Analysis

Sinusoidal Steady State Analysis

• Any steady state voltage or current in a linear circuit with a sinusoidal source is a sinusoid

– All steady state voltages and currents have the same frequency as the source

• In order to find a steady state voltage or current, all we need toknow is its magnitude and its phase relative to the source (wealready know its frequency)

• We do not have to find this differential equation from the circuit, nordo we have to solve it

• Instead, we use the concepts of phasors and complex impedances

• Phasors and complex impedances convert problems involvingdifferential equations into circuit analysis problems

Focus on steady state; Focus on sinusoids.

Characteristics of Sinusoidal

Key Words:

• Period: T , • Frequency: f , Radian frequency • Phase angle• Amplitude: Vm Im

Sinusoidal Steady State Analysis

Characteristics of Sinusoidal

tVv mt sin i

I1

I1

I1 I1 I1 I1

R1

R1

R

5

5

+

_

IS

E

I1

U1

+

- U

I

i

I1 I1 I1 I1

R1

R1

R

5

5

-

+

IS

E

I1

U1

+

- U

I

v、i

tt1 t20

Both the polarity and magnitude of voltage are changing.

Sinusoidal Steady State Analysis

Characteristics of Sinusoidal

Radian frequency(Angular frequency): = 2f = 2/T (rad/s)

Period: T — Time necessary to go through one cycle. (s)

Frequency: f — Cycles per second. (Hz)

f = 1/T

Amplitude: Vm Im

i = Imsint, v =Vmsint

v、i

t 20

Vm、Im

Sinusoidal Steady State Analysis

Phasors

E.g. voltage response

A sinusoidal v/i

Complex transform

Phasor transform

By knowing angular frequency ω rads/s.

Time domain

Frequency domain

eR v tComplex form:

cosmv t V t

Phasor form:

j t

mv t V e

Angular frequency ω is known in the circuit.

Sinusoidal Steady State Analysis

|| mVV

|| mVV

Phasors

Complex Numbers

jbaA — Rectangular Coordinates

sincos jAA

jeAA — Polar Coordinates

jeAAjbaA

conversion: 22 baA

a

barctg

jbaeA j cosAa

sinAb

a

b

real axis

imaginary axis

jjje j 090sin90cos90

Sinusoidal Steady State Analysis

Phasors

Phasors

A phasor is a complex number that represents the magnitude and

phase of a sinusoid:

tim cos mII

Phasor Diagrams

• A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes).

• A phasor diagram helps to visualize the relationships between currents and voltages.

Sinusoidal Steady State Analysis

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.Steady-State Sinusoidal Analysis – Complex Impedances

Phasor Relationships for R, L and C

Resistor:

Q , , R=10,Find i and P。tv 314sin311

VV

V m 2202

311

2

AR

VI 22

10

220

ti 314sin222 WIVP 484022220

Sinusoidal Steady State Analysis

Phasor Relationships for R, L and C

Inductor:

Q,L = 10mH,v = 100sint,Find iL when f = 50Hz and 50kHz.

14.310105022 3fLX L

Atti

AX

VI

L

L

90sin25.22

5.2214.3

2/10050

31401010105022 33fLX L

mAtti

mAX

VI

L

L

k

90sin25.22

5.2214.3

2/10050

Sinusoidal Steady State Analysis

Phasor Relationships for R, L and C

Capacitor:

Q,Suppose C=20F,AC source v=100sint,Find XC and I for f = 50Hz, 50kHz。

1592

11Hz50

fCCXf c

A38.12

c

m

c X

V

X

VI

159.02

11KHz50

fCCXf c

A13802

c

m

c X

V

X

VI

Sinusoidal Steady State Analysis

Phasor Relationships for R, L and C

Review (v-I relationship)

Time domain Frequency domain

iRv IRV

ICj

V

1

ILjV dt

diLvL

dt

dvCiC

CXC

1

LX L ,

,

, v and i are in phase.

, v leads i by 90°.

, v lags i by 90°.

R

C

L

Sinusoidal Steady State Analysis

Phasor Relationships for R, L and C

Summary

R: RX R 0

L: ffLLX L 22

iv

C:ffcc

XC

1

2

11

2

iv

IXV

Frequency characteristics of an Ideal Inductor and Capacitor:A capacitor is an open circuit to DC currents;A Inducter is a short circuit to DC currents.

Sinusoidal Steady State Analysis

Impedance

Complex Impedance

Phasors and complex impedance allow us to use Ohm’s law with complex numbers to compute current from voltage and voltage from current

20k+

-1F10V 0 VC

+

-

= 377

Find VC

Q.

• How do we find VC?

• First compute impedances for resistor and capacitor:

ZR = 20k = 20k 0

ZC = 1/j (377 *1F) = 2.65k -90

Sinusoidal Steady State Analysis

Impedance

Complex Impedance

20k+

-1F10V 0 VC

+

-

= 377

Find VC

Q.

20k 0

+

-

2.65k -9010V 0 VC

+

-

Now use the voltage divider to find VC:

46.82 V31.1

54.717.20

9065.20 10VCV

)0209065.2

9065.2(010

kk

kVVC

Sinusoidal Steady State Analysis

Impedance

Impedance allows us to use the same solution techniquesfor AC steady state as we use for DC steady state.

• All the analysis techniques we have learned for the linear circuits are applicable to compute phasors

– KCL & KVL– node analysis / loop analysis– superposition– Thevenin equivalents / Norton equivalents– source exchange

• The only difference is that now complex numbers are used.

Complex Impedance

Sinusoidal Steady State Analysis

Impedance

Phasor Diagrams

• A phasor diagram is just a graph of several phasors on the complexplane (using real and imaginary axes).

• A phasor diagram helps to visualize the relationships betweencurrents and voltages.

2mA 40

1F VC

+

1k VR

+

+

V

I = 2mA 40, VR = 2V 40

VC = 5.31V -50, V = 5.67V -29.37

Real Axis

Imaginary Axis

VR

VC

V

Sinusoidal Steady State Analysis

Q1

• Find the time-domain voltage over the capacitor using phasor domain techniques. The voltage and current source are both AC (sine) sources with frequency of 1MHz.

ANS:

Q2

• 2. Write the time-domain (differential) node voltage equations for the circuit shown. Do not solve.

ANS:

Q3

• Set-up the phasor-domain node-voltage equations for the following circuit. Do not solve.

ANS:

Q4

• In the circuit shown below, find the current flowing downthrough the inductor. The frequency of the source is 2Mrads/sec and the phase is 0 degrees. You must use phasoranalysis, but your final answer must be converted back totime-domain.

ANS:

Q5

• Write the time-domain (integral-differential) node-voltage equations for the circuit shown, using theground (0) node as your reference. Do not solve.

Ans:

Examples for Sinusoidal Circuits Analysis

v1=120sint v2

i3

i1 i2

Q., Find1I

2I 3I2V in the circuit of the following fig.

Complex Numbers Analysis

Sinusoidal Steady State Analysis

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Additional Example: represent the circuit shown in the

frequency domain using impedances and phasors.

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Additional Example: represent the circuit shown in the frequency domain

using impedances and phasors.

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Example – Steady-State AC Analysis of a Series Circuit

* Find the steady-state current, the phasor voltage across each element, and construct a phasor diagram.

15t500cos707.0ti

15707.0454.141

30100

Z

454.141100j10050j150j100ZZRZ

50jC

1jZ ,150jLjZ ,30100

s

CLeq

CLs

VI

V

1054.3515707.09050C

1j

751.10615707.090150Lj

157.7015707.0100R

C

L

R

IV

IV

IV

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Example – Series/Parallel Combination of Complex Impedances

* Find the voltage across the capacitor, the phasor current through each element, and construct a phasor diagram

901.090100

18010

100j

18010

Z

V

1801.0100

18010

R

V

1351414.050j50

9010

50j50100j

9010

ZZ

t1000cos10180t1000cos10tv

1801050j50100j

4571.709010

ZZ

Z

50j50Z

4571.704501414.0

01

)100j(11001

1

Z1R1

1Z

100jC

1jZ ,100jLjZ ,90-10

C

C

C

CR

RCL

s

C

RCL

RC

sC

RC

C

RC

CLs

I

I

VI

VV

V

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Circuit Analysis

Example – Steady-State AC Node-Voltage Analysis

* Find the voltage at node 1 using nodal analysis

7.29t100cos1.16tv or 7.291.16

: for Solve

5.11.0j2.0j

2j2.0j2.0j1.0

05.15j10j

90-2j5-

-

10

2 node and node1 at KCL Write

11

1

21

21

122

211

V

V

VV

VV

VVV

VVV

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Circuit Analysis

Exercise – Steady-State AC Mesh-Current Analysis

* Solve for the mesh currents

Chap 2 – Mutual Inductance

• Dot Convention

In Practical

• Mutual inductance can exist even in places where we wouldrather it not. Take for instance the situation of a "heavy"(high-current) AC electric load, where each conductor isrouted through its own metal conduit.

• The oscillating magnetic field around each conductor inducescurrents in the metal conduits, causing them to resistivelyheat (Joule's Law, P = I2 R):

• It is standard industry practice to avoid running the conductors of a large AC load in separate metal conduits. Rather, the conductors should be run in the same conduit to avoid inductive heating:

Explain why this wiring technique eliminates inductive heating of the

conduit.• Now, suppose two empty metal conduits stretch between the location of a

large electric motor, and the motor control center (MCC) where the circuit breaker and on/off "contactor" equipment is located.

• Each conduit is too small to hold both motor conductors, but we know we're not supposed to run each conductor in its own conduit, let the conduits heat up from induction.

• What do we do, then?

Answer

• Use terminal blocks to split up" the conductors from one pair into two pairs:

Example 1

• Calculate the phasor currents I1 and I2.

Solution

Example 2

• Determine V0 in the circuit below.

I2= 0.06 -90 degree x 10ohm

Example 3• Determine the phasor currents I1 and I2 in the

circuit below.

I1=0.132 – j2.136 , I2=0.193+j3.214

Ex 4• Determine the phasor currents I1 and I2 in the

circuit below.

-j8 ohm

Ex 5• Calculate Vo.

Ex 5 : Ans

Ex 6

• Calculate I1 and I2.

• (Note that j1 and j2 are reactances and not inductances.

Ex 6:Ans

Chap 3 - Laplace

• Q1. Evaluate the Vo(t) by using Nodal Analysis.

Show all the steps to solve this problem.

Ans:

• Q2. Solve for i(t) for the circuit, given that

V(t) = 10 sin5t V, R = 4 Ω and L = 2 H.

Assume i0=i(0)=0.

i(t)

Ans

• Q3. Evaluate the Vth(s) and Zth(s) by using

Tevenin Theorem. Show all the steps to solve

this problem.

Vth

Ans: