Sample Problem: Node Analysis (Depend Sources) 1

Loading...

来自 Georgia Institute of Technology 的课程

Linear Circuits 1: DC Analysis

193 个评分

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

193 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 nodal analysis.

And we're going to work with circuits with dependent voltage sources.

The problem is to find the current i sub 0 in the circuit.

And so i sub 0 is the current which is flowing downward

through through 12 kiloohm resistor on the left-hand side of the circuit.

We've noticed that we have the voltage controlled voltage source in our circuit.

Our voltage is 2v sub x where v sub x

is the voltage drop across the 6 kiloohm resistor in the center leg of the circuit.

We know that with node analysis,

the first step in solving the problem is to identify the nodes in the circuit.

So we'll do that first.

We notice that we have a node one on the left-hand side of the circuit.

We have a node two in the center of the circuit.

We have a node three on the right-hand side of the circuit.

And we have a reference node at zero volts at the bottom of the circuit.

So now we've identified our nodes.

So we have three nodes and we know with nodal analysis

ultimately what we're looking for are the nodal voltages.

Once we have the nodal voltages,

we can find any of the other parameters that we would like to find in the circuit.

So let's solve this circuit.

The first thing we want to do is we want to

sum the currents about each one of the nodes.

So, if we look at node one, we see that we have three currents flowing into node one.

We have the current flowing through the 6 kiloohm resistor right to left.

We have the current flowing up through the 12 kiloohm resistor.

We have the current flowing right to left through our dependent voltage source.

So if we were to write those currents and

add them up we would have a known current based on nodal voltages for

the 12 kiloohm resistor and for the 6 kiloohm resistor.

But we would have two entities, an additional unknown current for

the dependent voltage source.

So we'd have the current flowing up through the 12 kiloohm,

the current flowing downward through the 6 kiloohm and

the current flowing right to left through the voltage source.

This current through the voltage source would introduce an additional

unknown variable, i sub 2, v sub x.

So we'd end up with four unknowns and

three equations and we'd not be able to solve the problems using that approach.

So what we have to do is we have to use the concept

of supernodes to solve this problem.

We use the concept of a supernode anytime that we have a voltage source which is

tied between two nodes floating up in the circuit somewhere and

one of those nodes is not a ground node.

So we don't have our reference node on one side of that voltage source.

So we have no way of determining the current through that voltage source.

So what we're going to do is we going to use the concept of supernode.

And we're going to first identify our supernode as

a supernode which envelopes node one, node two, and

that dependent source, as we see from the dashed lines.

That's our supernode.

So we're going to use Kirchhoff's current laws and

we're going to sum the currents into the nodes starting with the supernode.

And so for the supernode, our equation's going to be the sum of a current

coming through this 6k resistor at the top,

coming up through the 12k resistor on the left-hand side,

coming up through this 6k at the center of the circuit and the current flowing

right to left through the 12 kiloohm resistor in the center of the circuit.

So there's going to be one, two, three, four currents as part of this equation.

Starting with a 12 kiloohm on the left-hand side of the circuit,

the current flowing up through that

is going to be the ground node 0 volts minus v1 divided by 12k.

We also have the current flowing through the 6 kiloohm resistor at

the top of the circuit, right to left, v3 minus v1 divided by 6k.

We have the current through the 12 kiloohm resistor on the right

hand side of the circuit, v3 minus v2 divided by 12k.

And we have a current which is flowing up through the 6

kiloohm resistor and the center of the circuit, 0 for

the ground or reference node minus v2 divided by 6k.

And that's equal to 0.

That's our first independent equation for the supernode.

We look at that equation,

we see that we have three unknown nodal voltages in that equation.

So we're going to need three equations to solve for these three unknowns.

A second equation we're going to write is about node three.

So I'm going to call that three for node three.

And just as we've seen in previous problems, we notice that there is

a 6 volt source between node three and the ground node.

And again, noting that for nodal analysis what we're ultimately looking for

are the nodal voltages, we can find for this problem that we can directly

calculate or find the nodal voltage for

node three from this relationship between

the voltage source and our known ground node.

So in this case, v3 which is tied to the positive polarity of the 6 volt source

minus zero is equal to 6 volts.

So we can find v3 directly is equal to 6 volts.

That's one of our unknowns.

Our last equation is our constrained equation for the supernode.

And we know that for the supernode,

v1, tied to the positive polarity of

our independent source, minus V2,

tied to the negative polarity, is equal to 2 v sub x.

So now we have this additional unknown v sub x that we've introduced

into our set of equations.

Now we have four unknowns and so

we use our second constraining equation for

this problem which relates our

controlling variable v sub x to our nodal voltages v1, v2, and v3.

So that equation is going to be v2,

which is tied to the positive polarity of v sub x

minus 0 is equal to v sub x.

So v sub x is equal to v sub 2.

And so now we have four equations and four nodes that we can solve for.

Remember for this problem ultimate, we're looking for i sub 0.

I sub 0 can be found if we look at the equation for

i of 0 using Ohm's law, it's going to

be v1 minus 0 divide by 12 kiloohms.

So what we really need to find is that nodal voltage for node one.

That's really the only one of interest to us for this problem.

Once we have v1, then we can solve for i0.

So if we use our four equations that we have here and

solve for v1, we end up with a v1 which is equal to nine halves of a volt.

If v1 is equal to nine-halves of a volt, then we can find i0,

and our i0 comes out to be three eighths milliamps,

simply by plugging in our v1 into our equation for i0.

探索我们的目录

免费加入并获得个性化推荐、更新和优惠。

Coursera 致力于普及全世界最好的教育，它与全球一流大学和机构合作提供在线课程。

© 2019 Coursera Inc. 保留所有权利。

通过 App Store 下载

通过 Google Play 获取