Sample Problem: Mesh Analysis (Independ Sources) 9

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来自 Georgia Institute of Technology 的课程

Linear Circuits 1: DC Analysis

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Georgia Institute of Technology

Linear Circuits 1: DC Analysis

125 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 3

This module demonstrates physical resistive circuits and introduces several systematic ways to solve circuit problems.

Sample Problem: Mesh/Node (Supernode)10:38

Sample Problem: Mesh Analysis (Super Mesh)10:17

Sample Problem: Mesh Analysis (Depend Sources) 16:48

Sample Problem: Mesh Analysis (Depend Sources) 27:56

Sample Problem: Mesh Analysis (Depend Sources) 36:52

Sample Problem: Mesh Analysis (Independ Sources) 19:22

Sample Problem: Mesh Anaysis (Independ Sources) 27:13

Sample Problem: Mesh Analysis (Independ Sources) 310:57

Sample Problem: Mesh Analysis (Independ Sources) 46:12

Sample Problem: Mesh Analysis (Independ Sources) 56:09

Sample Problem: Mesh Analysis (Independ Sources) 64:15

Sample Problem: Mesh Analysis (Independ Sources) 77:17

Sample Problem: Mesh Analysis (Independ Sources) 83:42

Sample Problem: Mesh Analysis (Independ Sources) 98:35

Sample Problem: Mesh Analysis (Independ Sources) 109:07

Sample Problem: Mesh Analysis (Independ Sources) 1110:16

Sample Problem: Node Analysis 14:19

Sample Problem: Node Analysis 23:57

Sample Problem: Node Analysis 37:13

Sample Problem: Node (Indep and Dep Sources)7:13

Sample Problem: Node Analysis (Independ Sources) 15:01

Sample Problem: Node Analysis (Independ Sources) 28:10

Sample Problem: Node Analysis (Independ Sources) 35:14

Sample Problem: Node Analysis (Independ Sources) 49:48

Sample Problem: Node Analysis (Independ Sources) 57:48

Sample Problem: Node Analysis (Independ Sources) 65:19

Sample Problem: Node Analysis (Depend Sources) 19:44

Sample Problem: Node Analysis (Depend Sources) 27:43

Sample Problem: Node Analysis (Depend Sources) 35:56

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 Mesh Analysis.

And we're going to work with circuits with independent voltage circuits.

The problem is to write the mesh equations for the circuit shown below.

It's a circuit that has one voltage source,

it's is a 6 volt independent source.

It also has five resistors associated with it in different legs of the circuit.

So in any mesh analysis problem, we recognize, first of all,

that we're going to use Kirchhoff's Voltage Laws to solve the problem.

And we know the first step in this process is to identify the loops associated with

the circuit.

So we need to identify the independent loops.

And the first one is loop 1, and we have loop current 1 or

mesh current 1 in the clockwise direction.

We have another mesh in the lower left-hand side of circuit,

and we're going to assign I2 clockwise for that loop or mesh.

And we have a third mesh on the right-hand side at the bottom.

And we're going to identify a current I sub 3 associated with that mesh.

So we have these three different meshes that we need to write

the mesh equations for.

Ultimately, we'll have three equations for the different meshes,

and we'll have three unknowns, I1, I2, and I3.

We can solve for I1, I2, and I3, and

then we will be able to solve for any voltage drops, or

any powers absorbed or supplied by the elements in the circuit.

So let's go about this and solve, or write them as equations

starting with mesh 1 which is the top mesh in the circuit.

So if we start at the lower left-hand corner of this top mesh and

start proceeding around in a clockwise fashion, we first encounter the 4 volt or

the 4 kiloohm resistor at the top of the circuit.

So the voltage drop there is going to be 4,

this is mesh 1, it's going to be 4 kiloohms times I1.

We continue around that circuit, and if we continue around it,

we are going to encounter the 6 kiloohm resistor next.

And the voltage drop across a 6 kiloohm resistor is the 6K times I1 and

which is flowing clockwise through the 6K resistor.

And we also have I3 which is flowing the opposite direction,

so the current through 6K is going to be I1- I3.

That's the voltage drop across a 6 kiloohm resistor.

And as we continue around that loop, we encounter the 6 volt source,

the positive polarity first.

So we add 6 volts to this, and

that gives us our first mesh equation or loop equation.

Upon inspection, we see that that has two unknowns, it has I1 and

I3 associated with it.

In the end, we want three equations, right, because we have three unknowns,

I1 and I2 and I3.

Let's look at loop 2.

Loop 2 is the lower left-hand loop starting in the lower left-hand corner,

traveling up through the 9 kiloohm resistor.

We have a voltage drop of 9K times I2.

Continuing around this loop,

we have a voltage drop of course associated with our 6V source,

which is a negative polarity first, so it's going to be- 6 volts.

Continuing around this loop,

we encounter the 3 kiloohm resistor in the center leg of the circuit.

And the voltage drop there is going to be 3K, and

then I1 is flowing into the 3K,

into the positive polarity of the 3K resistor.

So it's going to be I2- I3 which is flowing into the negative polarity.

And that's equal to 0.

So that's our second independent equation, and

upon inspection it has two unknowns as well, I2 and I3.

Now our third equation for the right-hand side of the circuit, so

the right-hand lower loop has three voltage drops.

It has first of all starting at the lower left-hand corner,

voltage drop across the 3 kilometer resistor, then the 6, and

then the 12 volt on the right-hand side of the circuit.

So starting with the 3K voltage drop,

we see that we are in the negative

polarity of that voltage drop,

so it's -3K (I2- I1).

Now I'm going up to the 6 kiloohm resistor in the negative

polarity first of that voltage drop, it was 6K (I1- I3).

That was that voltage drop.

We see that from equation 1.

And then we have voltage drop associated with the 12K resistor.

And it's going to be 12K times I sub 3,

and that's equal to 0.

So we have these three equations and three unknowns, so

we would solve these equations for I1, I2, and I3.

And then we could find any other value we wanted for

voltages or currents in the circuit, as well as power.

A point of interest is that [COUGH] these equations while we wrote them in

this fashion, it's perhaps easier to think of these voltage drops

in a little bit different way.

So if we look at this voltage drop in equation 3, it's -3K (I2- I1).

First of all, it should be -3K (I2- I3).

Okay, that should be a 3 instead of a 2 in that first term.

That's the voltage drop across this element, it's this voltage drop here.

In this voltage drop, we have assigned it plus to minus, and

if you look at the second equation here it is, right here.

We're just rewriting it in the third equation because we introduce and

see the negative polarity of the 3K resistor voltage drop first.

So we have this expression.

It's interesting to know that if we write that -3K (I2- I3),

That's equal to 3K (I3- I2).

And so, we could just as easily have, when we're summing up the loops around

this last loop, or summing around the voltages around this last loop,

starting at the lower left-hand corner, we could have said,

the voltage drop here is 3K I3, which is flowing this direction,

positive direction through 3K- I2.

That's what we have down here.

It's exactly the same expression as what we had

before using our sign voltage drops.

So sometimes, and many times it's easier to write these mesh equations,

or loop equations, thinking of each loop current as being a positive and

each associated current with the loops that are adjoining that loop

that we're adding up the voltage drops across as a negative current.

That's what we've done here 3K I3, which is positive in loop 3,

-I2 which is flowing in the opposite direction of the positive I3 current.

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