Rth of Norton's Theorem for circuit analysis

[niceboar]

Homework Statement

I've deleted all current sources and shorted the voltage sources.
[PLAIN]http://img121.imageshack.us/img121/5064/62767285.png
How would I reduce this circuit to an equivalent resistance? What's throwing me off is the location of b and the fact it is a broken wire.

The Attempt at a Solution

I can see it multiple ways
3k and 2k share one node thus in series making it 5k
the 5k resistor is || to the 1k resistor making 833
then add it to 1k?

but that really doesn't make sense because 3k is also in series with the rightmost 1k making 4k
then 4k would be || with 2k making 1.3k
then add it to 1k?

or the path of least resistance 3k?

I'm really confused. Thanks for any help.Edit: I'm thinking it would be better if I gave the full problem.
could someone explain where to make the cuts and the reasoning? where CAN I cut? does it have to be a straight line across? we haven't done anything like this in class and the textbook has no good examples of a problem like this either.
[PLAIN]http://img169.imageshack.us/img169/4356/69377576.png

current of RL is -5.36mA

Edit:
http://engr.astate.edu/jdg/Circuits/Lecture1/TheveninEqExample.html
made everything much more clear

leaving in RL is the key then you can easily see 3k is in series with 1k making 4k
4k is then in parallel with 2k making 1.3k
and 1.3k is in series with 1k making 2.3k in parallel with RL

 

[willem2]

To Find Rab, you can just reduce the circuit by replacing parallel and series resistances,
until you have only one resistance left that connects A and B.
Replacing The 2k and the 3k resistance with a 5k resistance doesn't make sense, because point A lies in between and will disappear from the resultant circuit. This will make it impossible to end up with one resistance that connects A and B

 

[niceboar]

To Find Rab, you can just reduce the circuit by replacing parallel and series resistances,
until you have only one resistance left that connects A and B.
Replacing The 2k and the 3k resistance with a 5k resistance doesn't make sense, because point A lies in between and will disappear from the resultant circuit. This will make it impossible to end up with one resistance that connects A and B

Yes I see that now.

 

[staticd]

Right, you had to leave RL anyway. Your first assumption, that you had to take it out, was incorrect.

I do think that you need to rethink your solution. At the end you should have 4k || (RL + 1k) || 2k.

 

[niceboar]

Right, you had to leave RL anyway. Your first assumption, that you had to take it out, was incorrect.

I do think that you need to rethink your solution. At the end you should have 4k || (RL + 1k) || 2k.

I got the same answer that was on the handout (he gives us all the answers).

why did you make combine RL and 1k? isn't the point trying to find the equivalent resistance outside RL?

 

[staticd]

why did you make combine RL and 1k? isn't the point trying to find the equivalent resistance outside RL?

I didn't read where you were trying to find the Req from RL's perspective.

After you short the Vsource and open the Isources, you are left with two loops where 4k || (RL + 1k) || 2k.

I added the 1k to RL because it is in series with RL.

 

[staticd]

You're correct. My bad. I couldn't tell that you added your work at the bottom...very busy post.

 

[niceboar]

[PLAIN]http://img808.imageshack.us/img808/8651/81511642.png
okay on this one, how would I reduce the resistance?

If I use Norton's to find Rth to the inductor, I break the wire so 5k is not connected so do I ignore it?
Like this?
[PLAIN]http://img301.imageshack.us/img301/6452/35371845.png
[PLAIN]http://img139.imageshack.us/img139/8302/59770342.png
Rth = 3K/4?

But what if I wanted to find it normally i.e. without Norton's or Thevenin's?
How would 5K be connected to 4K/3? Series? Where would I place the 5.75K resistor? In parallel with the inductor? I understand how you can reduce components but how do you reduce it with power sources between them? It is even possible?

 

[staticd]

Rth = 3K/4?

As far as I can tell, yes that's Rth. However, in your drawings, after you open up the current source, just drop the resistors that "go away" after you open or short a source.

But what if I wanted to find it normally i.e. without Norton's or Thevenin's?
How would 5K be connected to 4K/3? Series? Where would I place the 5.75K resistor? In parallel with the inductor? I understand how you can reduce components but how do you reduce it with power sources between them? It is even possible?

You would have source --> 5k --> (3/4)k || L

So, you can't combine the .75k resistor with the 5k resistor because they are not in series or parallel.

You ask a very good question, because, very soon, you will be working with dependent source. Those, I'm afraid, you can not short or open to find Rth.

Therefore, Rth = Voc/Isc.

In this case, you would find the current in the loop with "source --> 5k --> (3/4)k", multiply that current by .75k to get Voc.

Then, you would find Isc by shorting across the inductor, which would give you "source --> 5k", find that current and you've got Isc.

Rth = Voc/Isc!

:]

 

[niceboar]

As far as I can tell, yes that's Rth. However, in your drawings, after you open up the current source, just drop the resistors that "go away" after you open or short a source.
You would have source --> 5k --> (3/4)k || L

So, you can't combine the .75k resistor with the 5k resistor because they are not in series or parallel.

You ask a very good question, because, very soon, you will be working with dependent source. Those, I'm afraid, you can not short or open to find Rth.

Therefore, Rth = Voc/Isc.

In this case, you would find the current in the loop with "source --> 5k --> (3/4)k", multiply that current by .75k to get Voc.

Then, you would find Isc by shorting across the inductor, which would give you "source --> 5k", find that current and you've got Isc.

Rth = Voc/Isc!

:]

Thanks for clearing that up. So voltage and current divider is the only way? Yeah we've been over dependent sources but still shaky on some of the fundamentals. I actually prefer finding Isc and Voc.

Also can't you simply short the inductor anyway in a DC circuit even not finding T/Norton's? Isc would always be 5mA right assuming it is non transitive.

 

[staticd]

So voltage and current divider is the only way?

In this case, you could just use the voltage divider rule to find Voc, if there was a voltage source instead of a current source.

The Isc would just be the current from the source...

 

[niceboar]

In this case, you could just use the voltage divider rule to find Voc, if there was a voltage source instead of a current source.

The Isc would just be the current from the source...

Right.

Another question...
[PLAIN]http://img534.imageshack.us/img534/3921/34174456.png
In this circuit, the break is initially closed (don't know how to draw switches in this program)

at t=0 the switch is thrown open.

My question is, is there a specific node I have to chose for the ground using KCL to find V(t) of the capacitor? I used node A to make things easy. V(t) is simply the differential equation. Anyway I got the right answer but I was unable to duplicate it by picking C as my ground node. Do I have to use A here (or B)? Sorry my teacher isn't that good (but he tries), and the book is much much worse. Unfortunately he teaches directly from the book.

oh this is using the differential equation technique as my book calls it

I figure I would have to use A or B since the capacitor basically becomes a independent voltage source. And since you can't find the current from an IVS directly for KCL, it fails, right? So I have to pick A or B as ground for it to be the special case that allows me to find V directly right?

 

[staticd]

I don't know why you aren't using KVL, since you end up with a resistor in series with the capacitor. Once you find the current, you just plug it into the equation for voltage of the capacitor...

To answer your question, so long as you use the SAME reference for all nodes when you perform KCL, you can select which ever node you want. However, it is almost ALWAYS best to use a node that is common to all of the components for which you want to find the voltage across.

In this case it is either A or B (after you break the circuit down).

 

[niceboar]

I don't know why you aren't using KVL, since you end up with a resistor in series with the capacitor. Once you find the current, you just plug it into the equation for voltage of the capacitor...

To answer your question, so long as you use the SAME reference for all nodes when you perform KCL, you can select which ever node you want. However, it is almost ALWAYS best to use a node that is common to all of the components for which you want to find the voltage across.

In this case it is either A or B (after you break the circuit down).

I don't believe current is a function with a transient capacitor, it's discontinuous

 

[staticd]

Well, how can you use KCL then?

 

[niceboar]

Well, how can you use KCL then?

C * dv/dt = i(t)

i.e. the continuous function is the voltage

 

[staticd]

Very good, now, how about KVL.

 

[niceboar]

Very good, now, how about KVL.

Well to find KVL I need dv/dt...

 

[staticd]

What about V_c(t) = 1/C integral(i(t)dt) + V_c(0) ?

 

[niceboar]

But yeah the only way I know how to solve it is with KCL solving the differential equation.

dv/dt + a*v(t) = A

 

[niceboar]

What about V_c(t) = 1/C integral(i(t)dt) + V_c(0) ?

i(t) is discontinuous, you can't integrate it. Or at least I don't know of a way. It's discontinuous since dv/dt at the flip of the switch goes positive to negative or vice versa.

 

[staticd]

--i(t) is pieceswise continuous, you integrate over the continuous regimes

--if you learn how to do it both ways now, you will save yourself a whole 'lot of *** pain later...i'm just sayin.

 

[niceboar]

--i(t) is pieceswise continuous, you integrate over the continuous regimes

--if you learn how to do it both ways now, you will save yourself a whole 'lot of *** pain later...i'm just sayin.

Okay makes sense. Why not just use diffeq?

 

[staticd]

You end up with a differential equation...

Try it out, you'll see.

 

[niceboar]

You end up with a differential equation...

Try it out, you'll see.

Yeah I didn't really see how you could get around that. So it's just to use KVL?

 

[staticd]

Yeah, and you have to differentiate to solve it.

 

[niceboar]

Yeah, and you have to differentiate to solve it.

Yeah I'll try it out next homework problem

 

[staticd]

I will admit that your approach is correct. It is much easier to use KCL in a series RC circuit. There are many more intermediate steps if you use KVL. This will come in handy when you are dealing with higher order circuits (RLCs). In that case you have to be very familiar with the use of both current and voltage through an inductor and capacitor.