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Sample Problem: Node (Indep and Dep Sources)

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Linear Circuits 1: DC Analysis

乔治亚理工学院

4.5（198 个评分） | 16K 名学生已注册

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.

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审阅

4.5（198 个评分）

5 stars
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DM

May 09, 2018

Great course, learned a lot of things that they don't teach at my school, Thanks for the opportunity.

YZ

Jan 22, 2018

Excellent course where I get to learn a lot of things that is left from my AP Physics C E&M Course.

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

教学方

Dr. Bonnie H. Ferri
Professor
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
Dr Mary Ann Weitnauer

脚本

The topic of this problem is nodal analysis.

And we're going to solve circuits with current sources in them.

In particular, we have a problem that has both dependent and

independent current sources.

The problem itself is to determine V sub 0.

That's the output voltage across the 1 kilohm

resistor on the right hand side of the circuit.

This problem, again, has two current sources.

It has a dependent source,

which is a current of voltage controlled current source.

It's 2 V sub X divided by 1000.

Where the controlling parameter is a voltage.

V sub X which is dropped across

the 1 kilohm resistor on the left hand side of the circuit.

We also have a two milliamp source, independent source in the circuit.

So you know when we're doing the analysis, the first step in nodal analysis is to

find the equations for the nodes using Kirchhoff's Current Laws.

So we're going to use Kirchhoff's Current Law to sum the currents

either into or out of the nodes of the circuit.

In this circuit we have four nodes we have the node at the upper left hand side

of the circuit which is node 1, we have node 2 kind of the center of the circuit.

And the upper side we have node 3, on the right hand side of the circuit.

Node 3 is the node which encompasses

all the different elements flowing into this particular point in the circuit.

We also have a reference node which we know is at 0 volts at the bottom

of the circuit.

So we have four nodes in the circuit.

We're going to use Kirchhoff's Current Law to sum the currents either into or

out of node 1, node 2 and node 3.

So we're going to choose to solve the problem using Kirchhoff's Current Law and

summing the currents into the different nodes.

So we're going to start with node 1.

If we look at node 1 and we want to sum the currents into node 1, we have

current through the 1 kilohm resistor on the left hand side of the circuit, and

we have current flowing through the 1 kilohm resistor on the top of the circuit.

So the current flowing through the 1 kilohm resistor on the left hand side of

the circuit, if it's flowing into node 1, it originates at the ground node, and

flows up into node 1 through the 1 kilohm resistor, so it's 0- V sub 1,

which is the nodal voltage for node 1, divided by 1 kilohm.

We also have the current through the one kilohm resistor at the top of the circuit,

flowing from node 2 to node 1, since we're summing currents into the nodes.

Starting at 2 flowing into 1 is going to be nodal

voltage of V2- nodal voltage V1 divided by 1K.

Those are the only two currents flowing into node 1.

So the sum of those at any estimate time is equal to 0.

It's important to note that with nodal analysis, what we're solving for

are the nodal voltages, V1, V2, and V3.

If we can find those nodal voltages then we can find any of the associated currents

and any of the branches of the circuit and we can find the voltage drops as well.

So there's our first independent equation.

Our second equation is for node 2 in the center of the circuit.

Again, assuming the currents into node 2 looking at

the 1 kilohm resistor at the top of the circuit first.

The current flowing into node 2 originates at

node 1 flows to node 2 through the 1 kilohm resistor.

So it's V1- V2 divided by 1 kilohm.

We also have a current following up through the 2 milliamp source into node 2.

It's a positive current so it's + 2 milliamps and we have

the current through the 2 kilohm resistor at the top of the circuit as well.

Originating at node 3 to node 2 through the 2 kilohm resistor so

it's + V3- V2 divided by 2k and

that's = 0 that's all the current flowing into node 2.

Now looking at node 3.

Node 3 has again three currents flowing into it.

We have a current flowing through the 2 kilohm from left to right.

It's V2- V3 over 2k,

Plus the current flowing up through the dependent source.

Which is 2V sub x divided by 1000.

And then we have the current flowing through the 1K resistor on the right

hand side of the circuit starting at the ground node,

the reference node 0 volts, flowing up to node 3.

So its + 0- V3 divided by 1K = 0.

So if we look at these three equations we have three equations and three unknowns.

The unknowns are the nodal voltages of V1, V2 and V3.

And everything else is given to us so we can solve for each one of those voltages.

[COUGH] We also need to find an equation

that relates V sub x to the unknown voltages V1, and V2, and V3.

because in fact, the independent current source adds yet

another a invariable V sub x.

So we need to find the relationship between V sub x,

our controlling parameter and our nodal voltages.

And if we look at the circuit, it's quite easy to find that constraint equation.

And we know that V1 = V sub x.

So now we have our four equations and

our four unknowns if we include our unknown from the dependent current source.

So we can solve these simultaneously.

Openly, we see the V sub 0 = V sub 3.

It's the same voltage.

And if we solve these four equations and

four unknowns, then we get a V sub 0 = 4 v.

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