Sample Problem: Dependent Sources 2

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

97 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

97 calificaciones

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.

De la lección

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

Conoce a los instructores

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 circuits with dependent sources and

the problem is to determine the current in the circuit and

the output on load voltage V sub 0.

So if you look at the circuit that we’ve drawn,

we have a 12 volt source at the left hand side of the circuit and

we have our load resistance at the right hand side of the circuit.

So we have a 5 kilo ohm load resistance on the right hand side of the circuit.

We also notice that we have a dependent source.

The dependent source is at the top of the circuit, and it's a current controlled

voltage source where the voltage associated with that dependent source

is controlled by the current that's flowing through this single loop circuit.

So we're going to again determine the current and

the output voltage associated with this circuit.

So in order to do this, we want to use Kirchhoff's Voltage Law and

we can do that by summing up the voltages as we go around the circuit and

we could choose either a clockwise or a counterclockwise direction to

add these voltages up, the voltage drop associated with each element.

It really doesn't matter whether we go clockwise or counter clockwise.

In this case, we're going to start with the lower left hand

corner of the circuit and we're going to go clockwise around the circuit.

We see that if we do that,

the first thing we encounter is a minus 12 volts associated with the source.

So again, we're doing Kirchhoff's Voltage Law and it's clockwise around the circuit.

So it's minus 12 volts associated with the voltage source on the left hand side of

the circuit and there we have the voltage drop associated

with the 3 kilo ohm resistor and that's going to be I1

times 3 kilo ohms.

And using the passive sign convention, we designate the voltage drop as plus

the minus across the 3 kilo ohm source, 3 kilo ohm resistance sorry.

So we continue going around and we run into the dependent source and

for the dependent source, we have a negative voltage associated with that.

Since we hit the negative polarity of the voltage source first,

so it’s minus V sub A and we continue around.

We also have using passive sign convention,

a plus to minus voltage drop across a 5 kilo ohm load resistance.

And so that is going to be plus

I1 times 5 kilo ohms is equal to 0.

So that's our loop equation using Kirchhoff's voltage law for

this a single loop circuit.

So minus 12 volts plus I1 times 3K, the voltage

drop across the 3 kilo ohm resistor minus V sub A since we hit the negative polarity

of the dependent source first, as we travel clockwise around the circuit, and

then we have the voltage drop I sub 1 5K across the 5 kilo ohm resistance.

And so that gives us an equation with Two unknowns I1 and V sub A.

We also have our relationship so that's our first equation.

We also know the relationship between V sub A and I1 simply through our dependent

source and that gives us our second equation that V sub A is

equal to 2000 which is the amplification factor times I1.

So that's our second independent equation relating VA and I1.

So if we use those two equations and we solve for

I1, we end up with an I1 equal to 2 milli amps.

And using that, we can then solve for our output voltage.

Our output voltage is going to be I1 times 5 kilo ohms.

And we've solved for I1 already,

and so we end up with an output voltage

which is equal to 10 volts.

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