Sample Problem: Max Power (Depend Sources)

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来自 Georgia Institute of Technology 的课程

Linear Circuits 1: DC Analysis

89 个评分

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 个评分

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

从本节课中

Module 2

The module describes some basic principles used in circuit analysis.

Sample Problem: DC Op Amp 19:29

Sample Problem: DC Op Amp 25:33

Sample Problem: DC Op Amp 34:23

Sample Problem: DC Op Amp 46:22

Sample Problem: DC Op Amp 55:33

Sample Problem: Current Division 13:57

Sample Problem: Current Division 25:20

Sample Problem: Parallel and Series Resistors 12:24

Sample Problem: Parallel and Series Resistors 28:07

Sample Problem: KVL6:05

Sample Problem: KVL 25:48

Sample Problem: KVL 32:28

Sample Problem: KCL 12:12

Sample Problem: KCL 24:43

Sample Problem: KCL 32:51

Sample Problem: KCL 41:26

Sample Problem: KCL 52:10

Sample Problem: Op Amps 19:17

Sample Problem: Op Amps 25:29

Sample Problem: Op Amps 34:59

Sample Problem: Nodes, Branches, Paths, and Loops3:19

Sample Problem: Summing Op Amp6:05

Sample Problem: Differential Amp6:47

Sample Problem: Inverting Op Amp5:09

Sample Problem: Non-inverting Op Amp7:54

Sample Problem: Max Power (Depend Sources)4:06

Sample Problem: Dependent Sources 15:10

Sample Problem: Dependent Sources 24:48

Sample Problem: Series/Parallel (Independ Sources) 12:59

Sample Problem: Series/Parallel (Independ Sources) 27:22

Sample Problem: Series/Parallel (Independ Sources) 33:00

Sample Problem: Series/Parallel (Independ Sources) 44:35

Sample Problem: Op Amps 49:10

Sample Problem: Op Amps 59:29

Sample Problem: Op Amps 65:33

Sample Problem: Op Amps 74:23

Sample Problem: Op Amps 86:22

Sample Problem: Op Amps 95:33

Sample Problem: Op Amps 109:17

Sample Problem: Op Amps 115:29

Sample Problem: Op Amps 124:59

与讲师见面

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Maximum Power, and

we're working with circuits with dependent sources.

The problem is to find the Thevenin's equivalent circuit and

use it to determine the load resistance for maximum power and

the value of the maximum power delivered to the load.

So here's our initial circuit from our previous video problem.

So this was the circuit that we started with before our previous problem, and

then what we did was we worked through this problem finding Voc,

finding I short circuit.

So that we could ultimately write the circuit in the form of

a Thevenin's equivalent circuit which contains the Voc, R Thevenin's or

R Thevenin's is equal to Voc divided by I source circuit and the load resistance.

So our initial circuit has everything we need.

And we're going to determine what R sub L is for maximum power delivered to R sub L.

And then we're going to determine the power associated with R sub L.

So we know though the maximum power theorem that maximum power is

delivered to the load when R sub L is equal to R Thevenins.

We found, in this problem, from the previous video solution to it,

that R Thevenin is equal to A kilo ohms.

It's Voc, which is 8 volts, divided by I short circuit, which was 4 milliamps.

And so that ended up giving us 2 kilo ohms for R Thevenin's.

If we want four maximum power delivered to our load,

the condition that R sub L is equal to R Thevenin's.

Then our R sub L in this problem is going to be equal to 2 kilo ohms.

That's our value for R sub L.

Now if we know that and we want to find the maximum

power delivered to that R sub L, then we can

use our simple equation for power, which relates the power to the voltage.

So we have V R sub L divided by V

squared divided by R sub L.

And we know what the voltage is going to be across

R sub L using voltage division of our simplified Thevenin's equivalent circuit.

We have an A volt source.

We have a 8 kilo ohm resistor, we now have identified our load's resistance as 8 kilo

ohms, so our voltage drop across this is going to be the voltage division

of 8 volt source between these two resistors which are the same.

So VRL is going to be 8 volts.

And then R sub L which is 8k divided by 8k + 8k.

That's the amount of the 8 volt source which Is

across the load resistance of 8 kilo ohms.

And we know that number is in fact going to be 4.

So we have a 4 squared at the top, and

we have a load resistance which is equal to 2 kilo ohms.

So the power in this case is going to be

4 squared, which is 16 volts squared divided by 2 kilo ohms.

So that gives us 8 milliwatts for the power.

And that's a power max delivered to the load.

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