"Max Power" Calculations in Electrical Engineering

[Studiot]

Yes you can learn as much from your fellow students as you can from the lecturers, if you all cooperate.

 

[Femme_physics]

Yes you can learn as much from your fellow students as you can from the lecturers, if you all cooperate.

I agree.


And FYI, I neatly copied the lessons from this thread into my notebook and it definitely won't leave my mind now.

 

[I like Serena]

I agree.


And FYI, I neatly copied the lessons from this thread into my notebook and it definitely won't leave my mind now.

Looks good! :)

However, there is quite a jump (that is definitely too big) from the circuit to the power formula.

Studiot did explain that in the thread before he posted what you have here.
However that reasoning was not so clearly explained.

So there may be some more work for you there.

And perhaps for Studiot if he's around and willing ;)

 

[Studiot]

I've enroled in a class of handwriting lessons.

 

[Femme_physics]

I've enroled in a class of handwriting lessons.

Heh, not sure what you mean by that, I just wanted to show you your work here is definitely being processed :)

 

[I like Serena]

Heh, not sure what you mean by that, I just wanted to show you your work here is definitely being processed :)

I think he meant that he's around and willing :)

Edit: and do some more handwriting and scanning if you want him to.

 

[Studiot]

Some things to take away from this thead.

The equivalent resistance of two equal resistors in parallel is half of either resistor.

Two equal resistors in series each see half the total voltage across the pair.

A useful development of the parallel resistor formula appears in this thread.

https://www.physicsforums.com/showthread.php?t=485728&highlight=parallel+resistor

 

[Femme_physics]

The equivalent resistance of two equal resistors in parallel is half of either resistor.

That's an interesting fact that actually makes a whole lot of sense :)

Two equal resistors in series each see half the total voltage across the pair.

Yes

A useful development of the parallel resistor formula appears in this thread.

I read it, but it might be digging too deep for me as I'm falling short on some terms "shunt resistor"..." non preferred value"

It could be because my study material is in Hebrew, but probably because I just don't know these terms.
Thanks though.

 

[Studiot]

I read it, but it might be digging too deep for me as I'm falling short on some terms "shunt resistor"..." non preferred value"

Shunt is another word for parallel.

Resistors (and other components) don't come in any old value. They come in a discrete set of values, for example the E12 series has 12 values

10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82 and power of 10 of these

So if you went shopping for E12 resistors (or capacitors or inductors) you could buy

1.8 ohms, 18 ohms, 180 ohms, 1800 ohms, etc etc

but you could not buy 170 ohms or 190 ohms.

So if you needed this value you would have to 'make it up' from two (or more) standard resistors.

Series is easy - you just add them up -so you could get 30 ohms from two 15 ohms in series but parallel is a bit more tricky - unless you use my formula. Sometimes you can get more easily to the required value using a parallel (or shunt) combination, sometimes using a series.

 

[Femme_physics]

I was trying to get back to the concept of doing the same thing with the current. I got stuck at the current for each resistor. The only inkling I got is that R1, R2 and R3 combined equal 10 ohms, so their total can't go over it. Now, once I get into the parallel, how to know the current in each one is rather...daunting. Can you tell me if I'm anywhere near?

 

[Studiot]

Since I have done all the work for you in the votlage analysis, you could also perform a similar analysis with current instead of voltage.

Allow a current I to flow from A to B through your network and calculate the currents in each resistor in terms of I. (no voltage assumptions are required in this case)

Assemble a total power equation as before using power = (curent in each resistor)2 time the resistance.
Solve for I using the term with the largest coefficient.
Back substitute I to get the total power.

This was my original suggestion about current. I have emboldened the important point.

Just as when working in terms of voltage, current did not appear anywhere

When working in terms of current voltage will not appear at all.

---------------------

Here is a start

Let a current I enter the network at A and leave at B.

It is usual to work in terms of conductance (=reciprocal of resistance) for current.

So at the first branch the branch conductances are 0.02 through the 50 ohm and 0.025
Through the other leg.

The current I divides in proportion to the ratio of the conductance of each leg to the total.

So the current in the 50 ohm resistor is

I\left( {\frac{{0.02}}{{0.02 + 0.025}}} \right)

And the current in the other leg is

I\left( {\frac{{0.025}}{{0.02 + 0.025}}} \right)

further subdivision of this leg will yield the other currents

To assemble the power equation we use P = I²R (as we used P = V²/R before).

Can you take it from here?

 

[Femme_physics]

I'm failing to see where did you get the figures for the 50 ohm resistor. At the top, it seems you did V divided by 50, where V = 1. So you've made a presumption that V = 1?

At the bottom presumably should be R ( if you used I = V/R ). So, if V=1 then 0.02 divided by 1 (that's V) divided by the current. Shouldn't this be then 0.02/I?

 

[Studiot]

I don't know how to put it more simply.

I don't know or want to know what the voltage is in this analysis.

When we do current analysis we use - well, current.

Perhaps I jumped a bit too far so here is some preliminary work.

We have already established that the leg containing R1, R2,R3 and R4 has a resistance of 40 ohms. However I have done this again to make sure.

The attachment shows what happens when we replace R1 thro R4 with the 40 ohms.

Note the formulae are pretty fearsome in terms of resistance since a lot of reciprocals and stacked fractions are involved.

But if we note that the reciprocal of resistance is conductance we can avoid all these by calculating the conductances of the resistors to start with. This is what I did in post#

Power engineering where the resistances are low and some AC circuitry is better cariied out in terms of conductivity. It actually makes slightly less work here as well.

 

[I like Serena]

Hi FP,

I find it difficult to understand what Studiot means, so I've made the following diagram which I think should help to explain it.

It shows how the voltage differences are divided over the circuit.

First diagram is the entire circuit.

Second diagram is the circuit with R1, R2, and R3 replaced by their equivalent resistance.
This is needed to determine the voltage differences across (R1,2,3) and R4.

The third diagram shows the original circuit again, but with a further subdivision of the voltage differences.

Now the power that is for instance dissipated in R4 is:

P_4 = \frac {(\Delta V_4)^2} {R_4} = \frac {(\frac {30} {40} V)^2} {30}

And the other power dissipations can be deduced in a similar manner.

This will give you the power formula you already had from Studiot.

 

[Studiot]

I find it difficult to understand what Studiot means, so I've made the following diagram which I think should help to explain it.

Not sure what the problem is. Are you talking about the voltage analysis or the current analysis?

You do one or the other, since this is linear analysis they form dual spaces.

 

[Studiot]

Here is the full calculation using current analysis.

Naturally It gets to the same power (2.25 watts) as the voltage analysis.

Notice I only had to calculate three distinct currents , unlike the 5 different voltages needed for the voltage analysis.

Sorry it is so scruffy.

 

[Femme_physics]

Sorry it took me a bit of time to reply! I was mulling hard about this problem, and I appreciate you two taking so much time following up on me, I hit desperation point a few times I admit, but I never quit!

I spoke to my teacher about this exercise. He said it is easier to do it using current, and that next friday we'll be solving this problem using currents (so I was running ahead of the class in a few weeks!).

And it's not that scruffy, it's readable :) I'll print it before next class to make sure I agree with everything. Thanks. I'll definitely post again here. You two are great, it's my teacher who's terrible for multiple reasons (stupid pointless time-wasting stories is one of them).

 

[Studiot]

Glad it was of some use.

Do you understand what I mean by conductance?

 

[Femme_physics]

Do you understand what I mean by conductance?

I understand it's the opposite of resistance, and Wiki says it's measured by siemens, (though we haven't studied about siemens).

I will say this, IIRC my teacher did tell me that the current in the entire thing will depend on the side with the highest resistance, so in our case it's the R=50 that will determine the current in the circuit. I "think" that's he said.

BTW in your upload I see you've used "sin"! Why do you need a trigonometric function in electronics?!?

 

[Studiot]

since = because.

Conductance is not the opposite of resistance (that would be negative resistance which does occur) it is the reciprocal.

siemens = \frac{1}{{ohms}}

So

I = VS

Use of these quantities really comes into their own with AC circuitry where we have impedance and admittance (symbols Z and Y) instead of resistance and conductance.

 

[Femme_physics]

since = because.

Oh.

Conductance is not the opposite of resistance (that would be negative resistance which does occur) it is the reciprocal.

I like being scientifically spanked :) My favorite kink!

Right. Reciprocal. I'll watch my wording more.

Use of these quantities really comes into their own with AC circuitry where we have impedance and admittance (symbols Z and Y) instead of resistance and conductance.

I'll keep that in check, right now it appears we're still plugging along at the basics.

 

[Studiot]

It is always a good feeling to calculate something by another method and get the same answer both ways.

It is proper engineering practice to conduct an 'independent check' on calculations. Doing the voltage and current methods here is an ideal example of an independent check.

It is all to easy to follow your own or someone else's calculations, nod wisely, and make the same mistake they did.
This cannot happen with a truly independent check since it tests the answer not the method.

 

[Femme_physics]

I'll told you, I'll one day, have what you did figured out :) I want to show you now how much basic electronics knowledge I gained :)

http://img15.imageshack.us/img15/3702/rt1i.jpg

http://img860.imageshack.us/img860/9381/i1i2.jpg

http://img801.imageshack.us/img801/424/vrsc.jpg

I'll carry it on later to find out max power through this :)

I'm now off to help some other students with electronics so I won't be able to reply for a bit. Can I have your seal of approval that I'm allowed to help other students in electronics? :shy:

 

[Studiot]

Your first two pics are OK.

Not sure what you are trying to calculate in your last one though.

If you are trying to calculate the voltage across each resistor by multiplying the current through that resistor by the resistance only V_R4 is correct.

You have shown the current splitting into I₁ and I₂ at the first node. Good.

I₁ splits again at the second node. Say into I₃ and I₄.

However since the resistances in these two branches are equal (20 ohms)

I₃ = I₄ = I₁/2

 

[I like Serena]

I'll told you, I'll one day, have what you did figured out :) I want to show you now how much basic electronics knowledge I gained :)

I'll carry it on later to find out max power through this :)

I'm now off to help some other students with electronics so I won't be able to reply for a bit. Can I have your seal of approval that I'm allowed to help other students in electronics? :shy:

Oh dear, you have become much better at electronics now!

I don't quite get what you did in the third scan though.
And I have yet to see the max power.

So no seal yet (not from me anyway)!
But I'll allow you to help other students in electronics!

 

[Femme_physics]

This is by far a very advanced question when it comes to our course and won't be on the test, most of the students are still stuck at basic circuits finding I's, V's and P's in different parts of it-- which I master now :)

Thanks Studiot, my bad, I do realize what I did wrong since I1 does indeed split in the above wire *slaps forehead*. I'll have to redo it, but right now am really tired. But thanks :)

And thanks for the vote of confidence ILS! Much of it thanks to you, and Studiot! ;)

 

[Studiot]

Here a couple of triangles to help you on your way.

The first is about Ohm's law.

Cover up the quantity you want on the left hand side of the equation and read off the formula.

eg I = V/R; R = V/I ; V = IR

You are hereby qualified to show this to other students to help them.

The second is a test of observation.

What does the message in the triangle say?

go well