Sample Problem: Dependent Sources 1

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 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.

The problem is to determine the output or load voltage, V0.

We have the circuit, which is drawn below,

as a 10 milliamp source on the left-hand side of the circuit.

And it has a current source on the right-hand side of the circuit,

which is a dependent source, it's a current controlled current source.

With a current controlling that current source, is the current I sub 0

through the 3 kiloohm resistor, downward through the 3 kiloohm resistor.

The output voltage is the voltage across the load resistance,

which is four kiloohms.

So in this case the output voltage is found in the center of the circuit

across the lower element in the lower left part of the circuit.

So in order to solve this problem we have to use Kirchhoff's laws.

And we notice that this is, essentially, a three node circuit.

We have a node at the top of the circuit, we have a node at the bottom of

the circuit, the ground node, and we also have this node kind of in the center

of the circuit between the 2 and the 4 kiloohm resistances.

So we're going to solve the circuit first of all using Kirchhoff's Current Law.

And we're going to sum the currents into the top node,

which we're going to designate as node 1.

And so if we do that, then we'll end up with.

A falling equation, it's Kirchhoff's Current Law.

And instead of summing the currents into the node one, we're going to sum

the currents out of node one just to show that it doesn't matter whether you sum

the current into or out of the node just so your consistent throughout the problem.

So it's KCL out of node 1 at the top of the circuit.

So first of all we know that we have 10 milliamps flowing out

on the left hand side of the circuit.

We also have the voltage, which is across the 2 kiloohm and

the 4 kiloohm resistances.

That results in a current, which is going to be V1, the nodal

voltage associated with node 1, divided by the 2 kiloohm

plus the 4 kiloohm resistance for that particular part of the circuit.

We also have the voltage, which is flowing through the 3 kiloohm resistance,

which is I not or I sub 0, it's going to be V1 divided by 3 kiloohm.

And on the right hand side of the circuit we have a current

flowing into the top node, which is going to be 4 I sub zero.

And so that's all the currents flowing into node one.

And so the sum of those current in any instance of time is going to be equal

to zero.

That's our first equation.

We notice in that equation we have two unknowns.

We have V sub one, and we have I sub zero.

So we need another equation that relates V sub one and I sub zero,

and that's fairly easy to come up with using Ohm's law.

We know that the voltage, V sub 1, which is a voltage from the top node

to the ground node, is a voltage across the 3 kiloohm resistor.

So V sub 1 is equal to I sub

0 times 3 kiloohms.

So that gives us our second equation,

which is independent of the first equation.

So we have two equations, and we have two unknowns.

So if we go through, and

we solve for the nodal voltage,

then we'll have a nodal voltage V1,

which is equal to 12 volts.

If we know that we can step back and we can use our voltage division that we've

learned early in the course to find the voltage across the 4 kilo ohms resistor.

Namely Vout, V sub 0 the output voltage is equal to the 12 volts,

which is dropped across both the 2k and the 4 k resistances,

and using voltage division we know that the amount of that

voltage drop across the 4k is simply 4K divided by 2K plus 4K.

And so that it gives us 8 volts for

the voltage drop across the 4 kiloohm resistor.

So V sub 0 is equal to 8 volts, and

we could've just as easily I've written that as V sub 4K.

The output across the load resistance 4K.

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