elec130 electrical engineering 1

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ELEC130 Electrical Engineering 1

Week 3Module 2 DC Circuit Tools

5 March, 1999 Lecture 3 2

Administration Items Tutorials - Will be held in ES 210 this week.

Answers tutorial 1 will be revised Introduction to Electronic Workbench - Revised document Faculty PC’s Rm. ES210 - Go to Diomedes

Login: cstudentnumber Password: access keys on students card + daymonth (ddmm) of birth

Use Drive u: to save your work

Laboratory - THIS WEEK in EE 103(a)

Allocation of Laboratory and Tutorial Times NO more changes after Friday 12 March 1999 4 pm If you cannot make your time, please ask for alternative

Quiz 1 - THIS WEEK


Survey Results

Subject Home Page: - through Dept. Pages http://www.ee.newcastle.edu.au/

Then to Undergraduate studies Then to Course Information/Syllabus Then to Subject Web Pages

From the web site you have the option to save the file in power point

You are expected to read the specified text references to build the background information to the topic areas we are covering. You should think of the lecture as an opportunity to reflect on your reading and clarify difficult concepts.


Survey Results (cont.)

Current Sources DC power supply, transistors

Conductance - Parallel Resistance's

Voltage and Current Division

Why - Delta - tutorial 1 Question 19 part 4 Floyd pg. 309 Example 8-19 pg. 312

Superposition


Conductance

Sometimes easier to use inverse of resistance called conductance G = R-1

Symbol: G Units:Siemens S (mhos)

NB: Useful when resistors are connected in parallelGeq = G1 + G2 +... +Gn

1/Req = 1/R1 + 1/R2 +... +1/Rn

Case of two parallel resistance's: Req = R1R2 /(R1 + R2)

Vs R1

I1

+

-R2

I2


Week 2 Summary cont.

Voltage Division Current Division

vR

R RVs1

1

1 2

iR

R RIs1

2

1 2

R2R1

I1 I2

Is

+

-

I R1

R2

V1

+

_

V2

+

_

Vs


Survey (cont.)





Superposition


Wye Delta Transformations

Need to find equivalent resistance to determine current. HOW?(They are not in series, not in parallel)

Use Y to transformation


Survey





Superposition


Week 2 Summary (cont.)

Superposition:

If a linear circuit is excited by more that one independent source, then the total response is simply the sum of the responses of the individual sources.

Voltage sources - short circuit

Current source - open circuit


Power Calculations

Power is not linear! Superposition will not work

directly! With 2 A source opened P’1 = 25

W With 10 V Source shorted P’’1 = 1

W Total P = P’ + P’’ = 26 W

(incorrect)

Must calculate current by superposition and then work out power I’ = 5 A & I’’ = -1 A Total I = I’ + I’’ = 4 A Power P = 42 R = 16 W


Example Week 3

Find I ?

Determine VBC ?

What power is delivered by 4V source ?

I

VBC

C


Week 3

How does the current in the load change if RL is (say) doubled?


Thevenin’s Theorem

Any linear network with a pair of terminals can be replaced by a circuit comprised of a voltage source in series with a resistor.

The observed voltages and currents in the load will be the same using the “Thevenin equivalent” circuit as would be seen using the original circuit.


Thevenin’s Components

VTh Thevenin Voltage

‘open circuit’ voltage

VTh is the voltage which would appear across the terminals of the original and equivalent circuit if those terminals are open circuited.

RTh Thevenin Resistance

Independent sources inactivated

RTh is the total resistance seen when looking into the original circuit with sources inactivated

Can also be obtained by observing the short circuit current. RTh = VTh / Isc.


Steps to finding the Thevenin Equivalent

Step 1 Determine the two points from which the Thevenin is to be found. NB:Polarity

–

Step 2 Find open circuit voltage across these two points by removing the Load (resistance)

VTh = Vo/c

Step 3 Find RTh by looking from the two points into the circuit after replacing all independent

sources

Step 4 Draw the Thevenin Equivalent– Voltage source in series with a resistor


Example Week 3

Find I ?

Determine VBC ?


What is the Thevenin Equivalent circuit between A & B ?

I

VBC

C


Norton’s Theorem

Any linear network with a pair of terminals can be replaced by a circuit comprised of a current source in parallel with a resistor.

The observed voltages and currents in the load will be the same using the “Norton equivalent” circuit as would be seen using the original circuit.


Norton’s Components

IN Norton Current

‘short circuit’ current

IN is the current which would appear through the terminals of the original and equivalent circuit if those terminals are short circuited.

RN Norton Resistance

independent sources inactivated

RN is the total resistance seen when looking into the original circuit with sources inactivated

Can also be obtained by observing the open circuit voltage. RN = Voc / IN .


Steps to finding the Norton Equivalent

Step 1 Determine the two points from which the Norton is to be found. NB:Polarity

–

Step 2 Find the short circuit current through these two points by putting a short across them IN = Is/c

Step 3 Find RN by looking from the two points into the circuit after replacing all independent sources

Step 4 Draw the Norton Equivalent– Current source in parallel with a resistor


Example Week 3 Find I ?

Determine VBC ?


What is the Thevenin Equivalent circuit between A & B ?

What is the Norton Equivalent circuit between A & B ?

I

VBC

C


Relationship between Thevenin & Norton

A particular circuit can be represented by Thevenin or Norton equivalent. Therefore Thevenin and Norton equivalent circuits must be the same.

Hence Req = Rth = RN

RTh = VTh / Isc = VTh / IN VTh = RN IN

RN = Voc / IN = VTh / IN IN = VTh / RTh

elec130 electrical engineering 1

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