Sample Problem: Series/Parallel (Independ Sources) 4

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來自 Georgia Institute of Technology 的課程

Linear Circuits 1: DC Analysis

125 個評分

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

125 個評分

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.

從本節課中

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

與講師見面

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 series and parallel independent sources.

In this problem we want to determine the power associated with the 80 volts source.

As you can see from the circuit, it has three independent voltage sources

associated with it, it also has four resistors.

We're going to rearrange this circuit using our knowledge of series and

parallel independent sources to make the circuit easier to solve.

Ultimately, what we're looking for, again,

is the power associated with the 80-volt source.

We want to also determine whether that power is absorbed power or supply power.

So the first step in our process is to redraw our circuit.

We're going to take our circuit, we're going to redraw it and

we're going to take it.

And we're going to move the voltage sources so that they look

a little easier as far as the combination of these voltage sources.

The sources are in series of one another.

These are voltage sources.

And we know that when we have voltage sources in series with one another

that we can add them.

We also know, even though it's not relevant in this problem, that if we have

current sources in parallel with one another they can be added as well.

So we have our 80 volt source on the left hand side of the circuit.

We have a 30 volt source at the top, a 20 volt source at the top of the circuit and

then all of our resistances, the 10, the 7, the 5 and the 8 ohm resistors.

So, we've redrawn the circuit so that it becomes more obvious

that these voltage sources can be combined in series with one another.

So we have an 80 volt source.

So if we go around this circuit and we see that we hit the minus 80 so

it's a minus 80 volts and then here we hit another minus 30 volts and

then if we continue we hit a positive 20 volts.

So the net result of that is going to be a 90 volt source.

So we're going to redraw the circuit one more

time with our 90-volt source in there.

And we also know that we have this series combination of resistors as well, and

we're going to combine those as a series combination of resistors, 10, 7, 5 and 8.

So that gives us 30 ohms as a equivalent resistance for those four resistors.

And we also have our current I, drawing it in both of our redrawn circuits.

And so we've reduced our circuit to something that's very easy

to find the current for.

Ultimately if we want to find the power associated with the 80 volt source,

we're going to need that current because we know that the power

is equal to the voltage times the current.

So we know our voltage and

we're also going to find our current using Ohm's Law and our redrawn circuit.

So if we go back to our redrawn circuit, and we're looking for

the current, we know the current is equal to the voltage divided by the resistance.

And so that's going to be equal to 90 volts divided by 30 ohms.

And so that gives us 3 amps for our current source.

We also can use that 3 amps using our power equation

to find the power associated with the 80 volt source.

So if we do that, we have a positive current 3 amps flowing

up into the negative polarity of the 80 volt source.

So we have a minus 80 volts using a passive sign convention where we

have a positive current flowing into the element times the current,

which is 3 amps.

And so, in the end, we end up with a minus 240 watts

of power associated with the 80 volts source.

So since that's a negative power, we know that that is a supplied power.

Remember that using the passive sign convention negative powers

are supply powers and positive would be absorbed powers.

So the 80 volt source is supplying 240 watts of power.

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