Sample Problem: Mesh Analysis (Independ Sources) 11

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Do curso por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

123 classificações

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

123 classificações

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.

Na lição

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

Conheça os instrutores

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 sources first.

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

This is our first example of mesh analysis, so

we’re going to go through a few points of interest and important

steps in performing mesh analysis and coming up with the mesh equations.

First thing we want to know is that mesh analysis is the same thing as loop

analysis.

The terms are used interchangeably.

In this problem, we can identify several meshes.

We have a mesh on the left-hand side of the circuit, which is the left loop,

if you will, but also the mesh on the right-hand side, which is the right mesh,

or right loop of the circuit.

We also have mesh which is around the outside of the circuit, and

so we have three different meshes.

They're not independent of one another because

the outside mesh composes the left and the right meshes, so

we're going to write mesh equations for two different meshes.

Mesh 1 and Mesh 2 for this problem.

So, first of all, we're going to identify the meshes.

This is our Mesh 1 and this is the mesh current associated with Mesh 1,

and this is Mesh 2, the right-hand side mesh, and

we're going to assign currents for Mesh 2 to I sub 2.

So, we have two meshes, Mesh 1 and Mesh 2, and mesh currents I1 and I2.

So, the first step in writing the Mesh equation is to

use the passive sign convention to assign polarity is across the different elements.

Our voltage sources already have polarities assigned so

we don't have to do that.

But we have to identify polarities for the resistors and

the voltage drop across the resistors.

So using the passive sign convention, if we start with the left most loop and

we're looking at the current I1 It flow through R1.

And positive sign convention tells us that the positive current flows

into the positive polarity of the voltage drop across those resistors.

So this would be identified like this, as V R1.

So V R1. We also have when

we continue around that loop, past resistor one,

we have resistor three So we have a voltage drop across it, V R3.

And if we continue around that mesh, we have R2 at the bottom of that loop.

And the voltage drop, again using the passive sign convention is V R2

across the resistor at the bottom of the left hand loop.

Okay.

So now we're going to turn to the right hand loop and

we're going to assign polarities for it.

So if we look at that, and starting at the lower left hand side of that loop, and

going around the loop.

We already have a polarity assigned for R3 so we don't need to do that.

We have our voltage drop at the top VS2 As a polarity already assigned.

And then we come around to the right-hand side of the circuit, we have resistor R4.

Using the pass of we know that the positive direction of the current

meets the positive polarity of the voltage drop across R4 first.

So he assigned.

That voltage drop, VR4, as shown.

And then we go around that loop, continuing around the loop to R5.

Same technique, we have a voltage drop VR5 across resistor 5.

So that's what our polarities and

our voltage drop assignments Are based on the passive sign convention.

The next thing we're going to do is we're going to add up,

The voltages around each loop to come up with the mesh equation.

Because we know that we're going to use Kirchhoff's voltage law

as a way of solving These problems in mesh analysis.

And Kirchoff’s voltage law tells us that, the sum of the voltage

around any closed loop is equal to zero at any instant in time.

So whether we take loop one,

loop two, or if we even choose the outside loop around the outside of the circuit.

If we add up those voltage drops,

the sum of them is going to be zero at any instant of time.

So let's do that, let's start mesh one and let's add up the voltage Around that loop.

Starting with the lower left hand corner of that mesh,

travelling upward in a clockwise direction.

The first thing we come across is voltage source VS1 and

we, in fact hit the negative polarity of that voltage source first.

So we have minus VS1 For our first voltage rock.

We continue on that circuit and we encounter resister R 1 with

voltage drop V R 1 and counting the positive polarity of it first.

We have plus V R1 and we have

continuing around that circuit the voltage drop across R3 so we have a plus VR3.

And we continue around that loop and

we encounter resistor R2 and the drop across it.

And that's V R2.

We hit the positive polarity of each one of those resistor

voltage drops first, so they're positive quantities.

And so that's our first KVL equation.

That's our first mesh equation for loop 1.

We also have loop 2.

Again, doing the same thing with loop two, starting at the lower left and

corner of loop two, traveling clockwise around the loop.

The first thing we encounter as we start that journey is we encounter the voltage

up across resister R three.

We get the negative polarity first, so it's minus VR3.

We then continue and we encounter the voltage source VS2,

the positive polarity first, so it's + VS2.

This is a 3.

Continue your round that loop, we encounter resistance R4 and

the voltage drop across it.

Positive polarity first so it's plus VR4 and

travelling onward the bottom part of that loop, we encounter R5 Positive

polarity of the voltage drop across that first and so we have VR5 and

the sum of those four voltage drops around that loop is equal to zero.

Okay.

So now we have these two equations and ultimately these will allow us to

solve for The loop currents I1 and I2.

We can make some exchanges here and they're easy in some cases.

We know that Ohm's Law V = IR can be used wherever

we have a resistor we can replace The voltage,

for instance, VR1, we can replace that by

equal to I1 times R1.

We know that voltage drop

VR2 The R2 is equal

to the same current I1 times R2.

I skip R3 for a minute and move on to R4 and R5.

We know that for Resistor R4 if we want the voltage

drop across it, it's going to be I2 times R4.

And similarly for R5 the voltage drop is going to be I2 times R5.

Now let's look at R3 again.

R3 has A current I1 flowing through it, and it also has a current

I2 flowing through it, but they're flowing in opposite directions.

So if you remember, using the passive sign convention, we assign the positive

polarity across R3, the voltage drop across R3 to be at the top of R3,

based on A positive current I one flowing in to the top of R three.

Okay so we have that current flowing in and we also have

the current I two which is flowing the opposite direction through R three.

So the voltage drop VR three, VR three.

Is going to be equal to I1

minus I2 times R3.

And we chose I1 as the positive current because using the passive sign convention

That was our positive current flowing into that voltage drop across R3.

And the I2 is the opposite direction, so it has the negative sign in front of it.

So now we have R1, R2, R3, R4, and R5 identified, we see that

if we plug those into equation one and equation two, That we have two unknowns.

Our only unknowns would be I1 and I2.

So we'd have two equations which are independent of each other, and

we'd have two unknowns, I1 and I2.

We could solve for I1 and I2 if we knew values for the resistors and for

the voltage sources.

And if we solve for I1 and I2,

then we have everything else that we would need for the circuit.

We could solve for the power, absorb the power supply.

We could solve for the voltage drops across any element in the circuit.

So, that's all we need in order to able to solve for all the values of interest for

this circuit.

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