Sample Problem: Parallel and Series Resistors 2

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

185 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

185 calificaciones

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.

De la lección

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

Conoce a los instructores

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 resistors and what we're trying

to do in this problem is we want to find the equivalent resistance for

the resistor network that is shown.

So we have a network of resistors that are connected in series and parallel.

And we want to be able to ultimately find the equivalent resistance that if we were

to place a home meter across the leads it would give us that resistance measurement.

So, it's RAB across the leads on the leftmost side of the circuit.

To solve these type of problems, usually the best way to start is

to look at the far side of the circuit opposite to where RAB is.

What the equivalent resistance you're looking for is and if that's not the place

to start, then what you want to do is you want to look for something that looks easy

to combine in terms of either series or a parallel combinations of resisters.

So in our circuit, it's pretty easy to decide what we want to do first.

The first thing we want to do is we want to combine the two resistors on

the right most side of the circuit that are in series with one

another that is the 1k and the 2k resistor.

So if we combine the 1k and 2k resistors together,

they're in series with one another and

that gives us a 3k resistor.

Now, we see that that 3k resistor that

we have is in parallel with a 6K resistor.

And so the combination of this 3k resistance on the right most

side of the circuit, and the 6k resistor which is running diagonal down the socket

from the top to the bottom gives us a parallel combination of the 3k and the 6k.

So the combination of these six kilo ohm resistor,

and its equivalent of three kilo ohm resistor

is a 6k in parallel with 3k amp resistors.

And that gives us a 2 kilo ohm resister, so

they're 2k in combination.

So if we redraw this circuit,

we can redraw it with that equivalent resistance in it.

And so we're going to take some time, and

we're going to redraw that circuit with that resistor in there.

So, we have a 6k coming down.

Top to bottom in two places.

And then across here we have the equivalent resistance switch which was 2K.

The top one is a 10k.

This leg is a 6k ohm resistor,

another 6k ohm resistor, a 4k ohm,

and a 2k ohm, and again RAB is measured,

On the leftmost side of the circuit.

So, now we see that we have more to we can do on the rightmost side of the circuit.

We can combine the 10k and the 2k together because they're in series

with one another and we get a 12k resistors from that.

We then can see that that 12k resistance is in parallel with a 6k.

So the combination of these resistors that circle is drawn around in blue,

is a 12k resistor in parallel with a 6k resistor.

And that parallel

combination of the 12k and

6k resistor will give us

a 4 kilo ohm resistor.

So, if take the circuit and we draw it.

Again, we have the 2 kilo ohm resistor at the top.

We have another 2 kilo ohm that I forgot to mark,

another 2 kilo ohm resistor at the top and

then we have a combination of the 12 and 6k in parallel which gives us 4k.

Like this, we still have our 4 kilo ohm from the top

to the bottom of the circuit and we also have the 6 kilo

ohm From the top to the bottom of the circuit.

So, our RAB remains the same.

On the left hand side of the circuit we have a 4 kilo ohm resistor,

a 6 kilo ohm resistor like this.

So, while we slowly whittling the circuit down to an equivalent resistance.

So again, as we have before, we can see that this series combination

of 2 kilo ohm and 4 kilo ohm will give us an equivalent 6 kilo ohm resistance.

We then can take that equivalent 6 kilo ohm and

we will see it's in parallel with the existing 6 kilo ohm

resistor that's already in the circuit, and so we have a 6 kilo ohm.

And parallel with a six kilo ohm and

that gives us a three kilo ohm equivalent resistance.

And so once again, we're going to redraw the circuit so

we're going to go down here and

redraw it with our two kilo ohm in there, we haven't really modified that yet.

And we have the 4 kilo ohm, we haven't use that,

it combined that into our problem yet.

But we have the rest of the circuit combined and

I guess there's a three kilo ohm on the right-most side of the circuit.

So if we redraw it and we put it in our vise the 2 kilo ohms, 4 kilo ohms,

and 3 kilo ohms for our resistors, and ultimately we want to find RAB.

So, now it's becoming an easier and easier problem to work.

Okay, so in our resistor network,

we also have this 9k resistor at the bottom of the network and

we're going to add that into our network.

On the right most circuit like this and

we're also going to add it to our bottom circuit like this.

So, we have this 9k resistor that needs to be added in to each one of those circuits.

And so if we do that, we can then rewrite our circuit one more time

with the two kilo ohm resistor at the top.

A three kilo ohms resistor at the rightmost side,

combine with a 9k, and we still have the 4k here.

So we have a 2 kilo ohm resistor at the top, 4 kilo ohm top to bottom,

and then a 12 kilo ohm resistor on the right hand side of our circuit.

Now, we notice that the 4 kilo ohm and

the 12 kilo ohm are in parallel with one another.

And those two combinations are 4 in parallel with 12 kilo ohm resistor.

Is equal to 3 kilo ohms.

And so if we rewrite our circuit one more time, it will become obvious what RAB is.

We have two kilo ohm in series

with a three kilo ohm,

and so we end up with an RAB

equal to 5 kilo ohms.

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