Sample Problem: Series/Parallel (Independ Sources) 2

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En provenance du cours de Georgia Institute of Technology

Linear Circuits 1: DC Analysis

192 notes

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

192 notes

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.

À partir de la leçon

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

Rencontrer les enseignants

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 series and parallel independent sources.

In this problem, we want to determine the current I3 and

the voltage associated with the node

which has been designated as node 1 which is the node at the top of the circuit.

And so to do this, the first thing we're going to do is we're going to start

combining some of our elements.

We see that first of all,

that we have a parallel combination of several elements in this circuit.

But in particular we have a parallel combination of current sources.

We have a 4 amp current source on the right hand side of the circuit which is

downward current source.

And we have a 6 amp current source on the right hand side which is the upward

current source.

So we can just think of those first of all as two sources which

are in parallel with one another.

Two current sources so

they look kind of like this if we re draw just those two components of our circuit.

We have a 6A upwards, 4A downward and

ultimately what we would end up in this case is we'd end up

with 2A from the combination of those two sources.

So we're going to redraw our circuit with those

two sources combined with each other, and there's our 2A source from doing that.

We also noticed that we have these resistors

that are a resistor network on the right-hand side of our circuit.

We have two 6 ohm resistors in parallel, and

they're in series with a 15 ohm source.

So if we start with the two 6 ohm resistors that we have,

we see that the combination of those two 6 ohm resistors,

6 and parallel with 6 is going to equal 3 ohms.

That gives us a 3 ohm resistance from the combination of those two resistors in

parallel.

We also know that those two resistors which are in parallel and

series with a 15 ohm.

So we have a 15 ohm plus a 3 ohm and

that gives us 18 ohms equivalent for those two,

for those three resistors and series in parallel.

And so we're going to re-draw our circuit with those resistors but

we're going to take it one step further.

We're going to take those resistors which combine to be 18 ohms and

we're going to also notice that they are in parallel with the 9 ohm resister,

so we have an 18 ohm resistor in parallel with a 9 ohm resistor.

The parallel combination of the 18 ohm resistor and

the 9 ohm resistor gives us a 6 ohm equivalent resistance.

And so we're going to redraw our circuit with those changes to our circuit.

So we still have our 3 ohm resistor and the reason we didn't combine that with

other resistors is because we're looking for the current through that resistance.

So we don't want to combine it and lose our ability to solve for i 3 by doing so.

So we're going to take those different elements that we have left after we do our

combination of resistors and we've already done our combination of current sources.

We're going to leave our current controlled current source, intact.

So, 0.9 I3 going upward and

we also have the combined resistances

left which is a 6 ohm resistance.

So that's our redrawn circuit.

Now its a little bit easier for us to go back and to,

write our Norton equation for node one at the top of our circuit.

So we're going to use Kirchhoff's current law to solve this circuit

since it's now a two node circuit, a node at the top, and

our ground node at the bottom of the circuit.

Kirchhoff's current law can be either,

used either by combining the currents that are flowing into

a node or complimentary you can combine the currents out of that node.

So in this problem we're going to combine the currents or

add the currents that are flowing into the top node.

First of all we know that we have a two amp source which is flowing into the node.

We have I3, which is flowing out of the node.

We have point nine I3 flowing into the node and

we also have the current which is going down through

the 6 ohm resistor on the right hand side of the circuit.

And that current, we know, is going to be equal to, and

it's flowing out of the top node so we're going to put a negative sign.

It's going to be equal to V sub 1,

which is the voltage associated with the top node.

Divided by 6 ohms using Ohm's law and that's equal to 0.

So we've used Kirchhoff's current law which is the sum of the currents in to any

node equal 0 at any instant in time to come up with our first equation.

So that's equation number one.

So we notice that equation number one has two unknowns.

It has an unknown I sub 3, and it has V sub 1.

So we need another independent equation so

that we'll have two equations along with our two unknowns to solve this circuit.

So, to do this, we notice that we have a relationship between the nodal voltage and

I sub 3 through the 3 ohm resistor.

We know, using Ohm's law,

that V1 = I3 times the 3 ohms.

So that gives us a second equation which is independent from our first equation.

V1 = I sub 3(3).

Now we have two equations and two unknowns, so if we use matrix analysis or

elimination of the variables, however you want to solve the circuit.

We find I sub 3 is equal to 2/0.6 amps and

once we known I sub 3, we could find V sub 1,

using our equation too that we've already come up with.

So we know that V sub 1 using Ohm's law

is equal to 2/0.6 amps times 3 ohms, and so

that gives us 10 volts 4 V sub 1.

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