Max power

[Femme_physics]

So in a question I have to find out what's the "max power" that can be spread around a circuit without causing any damage to the resistor. So, unsure what "max power" means, I googled "max power" and ended up with many images and resources of Homer Simpson, half naked girls and cars, with no referrence to the electrical engineering term. Where do I find some educational referrences to calculating the max power in the sense I need? I'm not looking for a full electronics guide, just an article with a specific referrence to that.

[Xarren]

To "the" resistor? Where is the resistor? What is the circuit?

Power is current x voltage. Max power usually the amount of power the component can dissipate as heat. If more power is applied, the component will melt/break.

Resistors have "power ratings" which tell you how much power can be allowed through it. Lets say you have a 10ohm resistor, with a power rating of 10w (watts, unit of power).

Let V = 5

V=IR

I=V/R

P=IV

P=(V/R)V

P=V^2/R

PR^1/2 = Vmax

This way you can find the maximum Voltage that can be applied to the resistor without damaging it.

[Femme_physics]

Well, this is the circuit (attached). There are 5 resistors. So if say each resistor has a 1W power, a total power of more than 5 will definitely cause damage?

[Studiot]

a total power of more than 5 will definitely cause damage?


Not necessarily.

Start by labelling the outer two terminals A and B.
Then ask yourself, "How could the power be applied?"

[MATLABdude]

Courtesy of IMDB:
Homer Simpson: Kids: there's three ways to do things; the right way, the wrong way and the Max Power way!
Bart: Isn't that the wrong way?
Homer Simpson: Yeah, but faster!
http://www.imdb.com/title/tt0701118/quotes?qt0229108

For junior electrical engineering students, max power refers to the optimal source impedance that leads to maximum power transfer to a load resistor:
http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

However in your case, this is probably referring to the maximum power (P = I * V) that can be dissipated in a resistor before it burns up / explodes / vapourizes. All resistors are rated for a maximum power dissipation (the most commonly encountered are 0.25 W, and they're usually darn toasty when you put this much power through them). Note also that this makes certain assumptions about the ability for heat to be carried away to keep the casing under 60 C (if I recall correctly).

Usually, the higher the power dissipation, the bigger the casing (for some reason, my Google-fu fails to find an image of various power-rating resistors lined up in a row):
http://www.curtpalme.com/forum/viewtopic.php?p=256721

Make sure your resistors dissipate under the specified maximum, and you should be good. Note that if a 100 ohm resistor has 250 mA going through it, you have 250 mW. However, if you have two 50 ohm resistors in series with 250 mA going through them, they each dissipate only 125 mW.

EDIT: Beaten to the chase! Exceeding 5W total is only a problem if you exceed 1W in any of the component resistors.

[Femme_physics]

Let V = 5
But if you don't know V?

Not necessarily.

Start by labelling the outer two terminals A and B.
Then ask yourself, "How could the power be applied?"
Hmm...I guess it could be applied in 3 ways Horizontally. In an arch over R1 and R2, or in an arch below R5. Does this affect the result? Because as far as I understand, there's only 1 answer.

[Femme_physics]

But if you don't know V?
I just realized I can extract V from the formulas for each of the R's.

Courtesy of IMDB:
Homer Simpson: Kids: there's three ways to do things; the right way, the wrong way and the Max Power way!
Bart: Isn't that the wrong way?
Homer Simpson: Yeah, but faster!
http://www.imdb.com/title/tt0701118/quotes?qt0229108

I'll have to watch this episode hehe

[Studiot]

In order to supply power to the circuit I presume they meant you to apply a voltage between terminals A & B. Call this voltage V. The power drawn will increase as V increases.

This allows you to calculate the currents in each resistor in terms of V
The total power is the sum of all the individual resistor powers.
Each individual resistor power is given by the square of the current (a function of V) and the individual resistance.
So you then need to let V increase until one of the individual resistor powers reaches 5 watts.

You can then use this value to calculate the total power.

[Xarren]

Well the battery voltage is the V in case of one resistor. If you have more than one resistor, Vs are distributed in ratios according to their resistances. Eg two 1ohm resistors and 5V supply = 2.5V across each resistor.

[Femme_physics]

I've been staring at this problem for a while now and I keep running into the same loop of confusion. I don't have V. The only thing I can do is define P = 1 at each at the resistor, and since I need Pmax I figured I can just add them up together. Apparently I can't.

This way you can find the maximum voltage that can be applied to the resistor without damaging it.

Yes, but I don't need maximum voltage, I need maximum power.

In order to supply power to the circuit I presume they meant you to apply a voltage between terminals A & B. Call this voltage V. The power drawn will increase as V increases.

Makes sense.

This allows you to calculate the currents in each resistor in terms of V

Right, but each resistor has its own V. I don't know what it is unless V is defined to me.

The total power is the sum of all the individual resistor powers.
Yes, that I do know, which is why I thought to just add 1+1+1+1+1 = 5 [W], and bang, problem solved, but it seemed too easy.


So you then need to let V increase until one of the individual resistor powers reaches 5 watts.

You can then use this value to calculate the total power.

But you've just defined Watt to be 5, why do I need to calculate it?

If you think I'm currently too low a level of an electronic student to understand this, I'll say fair enough. Sorry if I'm showing some n00bish elements. Pretty new to electronics. Thanks for all your help.

[I like Serena]

It seems to me your problem statement is incomplete.
Does it say for instance, how much power a resistor can take before blowing up?

[Femme_physics]

Yes, 1 [W]

[I like Serena]

All right!

Now let's say we put a voltage V on the left side (call it A), and set the right side (call it B) to earth (that is a voltage of 0 V).

A current will start flowing that we will call Ia.

Let's take a look at resistor R5 which is 50 ohm.
Since there is a voltage V, there will be a current I5 = V / R5 = V / 50.
The dissipated power in R5 will be P5 = V2 / R5 = V2 / 50.

So we can say that if V is such that P5 = 1 W, then R5 will blow.
This would happen if V2 / 50 = 1 W, which means V = √50 = 7.07 V

Now if we would calculate the Req of the circuit, we know that the total power dissipated in the circuit would be Pmax = V2 / Req which you could already calculate.

Note that you would have to check the dissipated power in the other resistors as well, to see if they would already blow with a lower V.

[Studiot]

But you've just defined Watt to be 5, why do I need to calculate it?



Sorry I din't look properly, I thought the max was 5 watts for any resistor. Now I realise it is 1 watt.

:blushing:

Right, but each resistor has its own V. I don't know what it is unless V is defined to me.



V is applied between A and B

So the current in R5 is V/50

Now R1+R2 is 20,
put this in parallel with R3 the effective resistance is 10
add this to R4 to get 40

So the current in R3 is V/40

I will leave you to split this to calculate the currents in R1 and R2.

[Femme_physics]

I like Serena, I followed your lead verbatim. Attached. Do I get your seal? (btw come to think about it, this should've been posted in the HW section. I just didn't expect you all to be so nice and lead up to my question)



Note that you would have to check the dissipated power in the other resistors as well, to see if they would already blow with a lower V.

I will leave you to split this to calculate the currents in R1 and R2.

But I can just do Rtotal without having to calculate each separately, right?

[Studiot]

The voltage across R5 is V
The voltage across R4 is V times 30/40
The voltage across R3 is V times 10/40
The voltage across R2 is V times 10/40 times 15/20
The voltage across R1 is V times 10/40 times 5/20

Edit

I think it more important that you can follow the above, rather than that you can find a short cut, because this is really a test of your understanding of combined series and parallel circuits.

[Femme_physics]

The voltage across R5 is V
The voltage across R4 is V times 30/40
The voltage across R3 is V times 10/40
The voltage across R2 is V times 10/40 times 15/20
The voltage across R1 is V times 10/40 times 5/20

Edit

I think it more important that you can follow the above, rather than that you can find a short cut, because this is really a test of your understanding of combined series and parallel circuits.

Fair enough, but if they're asking me for Pmax why do I need to provide them results in terms of voltage across each R? :confused:

[I like Serena]

I like Serena, I followed your lead verbatim. Attached. Do I get your seal? (btw come to think about it, this should've been posted in the HW section. I just didn't expect you all to be so nice and lead up to my question)

But I can just do Rtotal without having to calculate each separately, right?

Sorry, no seal yet ;)

In your total resistance you were right in your calculation of R1,2,3. But then you added the other resistors. This would only be correct if they were all in series, which they are not.

And in your second sheet you write that if Pmax=1, that RT will blow.
But that's not correct either.
What you would have, is that with Pmax=V2/RT=(√50)2/22.22=2.25 W, that R5 will blow.

The resistor RT as such does not exist - it is only an "equivalent" resistance.
If anything blows, it will be an individual resistor.
I have (randomly) picked the resistor R5 to see when it would blow.
But any of the others may blow as well.

That's where Studiot's list might come in handy, because he's given you the voltages across each resistor. With each voltage and the corresponding resistance, you can calculate the power dissipated in each resistor with the formula P=V2/R

[Femme_physics]

I think this is too much for me. Thanks for trying. I've been trying to figure it out for a while but I'm only getting more confused. I don't understand how Studiot's list make sense, he just decided out of thin air that R5 equals = V? Maybe I have some gaps in my knowledge?...

In your total resistance you were right in your calculation of R1,2,3. But then you added the other resistors. This would only be correct if they were all in series, which they are not.

I added them in parallel where they're in parallel , then they became in series. That's how I've been doing it in class to get the right figures. Now I'm wrong with that, too? Great, I'm back to just knowing ohm's law, and even there my confidence is shaken :P

[I like Serena]

I added them in parallel where they're in parallel , then they became in series. That's how I've been doing it in class to get the right figures. Now I'm wrong with that, too? Great, I'm back to just knowing ohm's law, and even there my confidence is shaken :P

You can firm up your own beliefs again, because between your 2nd and 3rd drawing you did apply the law correct for parallel resistors! :smile:

In your third drawing you have R1,2,3 in series with R4, so those add up.
Then I guess you should have made a 4th drawing with R1,2,3,4 which is in parallel with R5.....

I think this is too much for me. Thanks for trying. I've been trying to figure it out for a while but I'm only getting more confused. I don't understand how Studiot's list make sense, he just decided out of thin air that R5 equals = V? Maybe I have some gaps in my knowledge?...

You appear to be missing some knowledge about how voltages vary in a circuit, and how currents add up in a circuit.
This will come back when you get more knowledge about for instance Kirchhoff's circuit laws.
I had the impression you already knew about those.
My bad! :frown:

As to Studiot's list, he did not say that R5 equals V, but he wrote that the voltage "across" R5 is V.
This means that ohm's law is applicable, and you can find for instance the current "through" R5 with the formula I5=V/R5 (Ohm's law :smile:).

[Femme_physics]

In your third drawing you have R1,2,3 in series with R4, so those add up.
Then I guess you should have made a 4th drawing with R1,2,3,4 which is in parallel with R5.....

I went with the conclusion that the power source is wherever I want it to be. If the power sources are between the two points then yes, I'm aware that my drawings would be wrong. Shouldn't the power source be defined a location?

You appear to be missing some knowledge about how voltages vary in a circuit
I do know that each R has its own V. Based on the formula V = IxR
and how currents add up in a circuit.

I do know that in a series circuit I is constant, in a parallel circuit it splits, but only if we know V can we know I everywhere. In this problem we have neither I or V!

Kirchhoff's circuit laws
We went through
Sum of all I going in equals to sum of all I going out
Sum of all series V equals to Vtotal.

And I think that's it.

As to Studiot's list, he did not say that R5 equals V, but he wrote that the voltage "across" R5 is V.

The voltage across R5 is R5 x I, as far as I know.

[I like Serena]

I do know that each R has its own V. Based on the formula V = IxR

I think we're onto something here.
It's not that each R has its own voltage V, but each R has its own "voltage difference", usually denoted ΔV.

Imho it's usually easiest to assign each point "between" resistors a "voltage amount".
The "voltage difference" of a resistor is then the difference between the "voltage amounts" on either side of the resistor.

A more precise version of the formula is ΔV=IxR.

I do know that in a series circuit I is constant, in a parallel circuit it splits, but only if we know V can we know I everywhere. In this problem we have neither I or V!

Yes, that is correct. That is Kirchhoff's current law.

We went through
Sum of all I going in equals to sum of all I going out
Sum of all series V equals to Vtotal.

And I think that's it.

Yes, you seem to understand Kirchhoff's current law perfectly. :)

Kirchhoff's voltage law is a little more nuanced.
What is more precisely says, is that the sum of all "voltage differences" in a closed loop is zero.
This works out to what you say, if we're just talking about resistors in series.

The key however, is that we're not talking about a "voltage amount" but about a "voltage difference".
So we talk about a current "through" a resistor, which is the current in the resistor.
And we talk about the voltage "across" a resistor, which is the "voltage difference" over the resistor.

The voltage across R5 is R5 x I, as far as I know.

That's entirely right! :)

I went with the conclusion that the power source is wherever I want it to be. If the power sources are between the two points then yes, I'm aware that my drawings would be wrong. Shouldn't the power source be defined a location?

Confused :confused:

The power source would be for instance a battery with the positive pole on the left side of the circuit and the negative pole on the right side of the circuit. It is not part of the circuit.

This would cause a "voltage difference" across the entire circuit. And because it is a battery, this voltage would be constant, regardless of the current flowing.

Typically we would assign the left side of the circuit a "voltage amount" V, and the right side of the circuit a "voltage amount" 0.

[Femme_physics]

The power source would be for instance a battery with the positive pole on the left side of the circuit and the negative pole on the right side of the circuit. It is not part of the circuit.

Yes, I know, this was the circuit in my mind (attached). But I guess I'm not allowed to define the source wherever I want. It's just that it wasn't defined on the question or in the drawing, so I've taken the liberty, heh.

Thank you so much for the detailed reply and the encouragement Serena, and everyone. I'll try reading on some guides and going over my notes again to see how to solve this question. If I ever do solve it I'll post back here. Will definitely use all the replies here as referrence when I try. It might take time. But, anyway, even while having failures after failures, it's always good to try :) What doesn't kill you makes you stronger...even though in electricity it often just kills you. heh.

[Studiot]

I'm glad you know Kirchoff's laws.
However what I did is much more fundamental and can be done by inspection.
Since it seems like magic I will walk you through it (It is actually called a walking through analysis in electronics)

Firstly label the terminals A & B as I said. These are the small circles to the left and right of your diagram.
Apply any voltage V across these terminals. So V volts appears across AB.

Now R5 is connected directly across AB so the voltage across R5 is V.

In the other leg of the circuit you have a complicated arrangement containing R1 through R4.
However can you see that this leg is also connected directly across AB?
So the voltage across the whole leg is V.

R1 & R2 are in series and add up to 5+15 = 20
This series pair in in parallel with R3 and so is 20 in parallel with 20 = 10

So the second leg has 10 in series with R4 ie 10 in series with 30 a total of 10 + 30 = 40
As already noted the voltage across this is V
In a series circuit the total voltage (V in this case) is distributed in proportion to the resistances in series to the total.

So the voltage across R4 is 30/40 times V

And the voltage across the combination of R1, R2 & R3 is 10/40 times V.

This voltage (10V/40) is directly across R3 so is the voltage across R3.

It is also the voltage across the series combination of R1 and R2, again distributed in proportion

So voltage across R2 is 10V/40 times 15/20
and voltage across R1 is 10V/40 times 5/20

Now that we have all the voltages and resistances we can assemble an equation for the total power

Powe{r_{total}} = \frac{{V_{R1}^2}}{{R1}} + \frac{{V_{R2}^2}}{{R2}} + \frac{{V_{R3}^2}}{{R3}} + \frac{{V_{R4}^2}}{{R4}} + \frac{{V_{R5}^2}}{{R5}}

Substituting

= \frac{{{V^2}}}{{50}} + \frac{{{{\left( {\frac{{30V}}{{40}}} \right)}^2}}}{{30}} + \frac{{{{\left( {\frac{{10V}}{{40}}} \right)}^2}}}{{20}} + \frac{{{{\left( {\frac{{10V}}{{40}}*\frac{{15}}{{20}}} \right)}^2}}}{{15}} + \frac{{{{\left( {\frac{{10V}}{{40}}*\frac{5}{{20}}} \right)}^2}}}{5}

Now if we examine each of the 5 terms in this expression the one with the largest numerical coefficient of V2 will be the one to reach the 1 watt limit first.

Do you follow this reasoning here?

If we set this equal to 1 watt we can solve for V

Then we can back substitute and determine the total power.

The arithmetic is not really so fearsome as it looks - really.

[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.

go well

[Femme_physics]

You're incredible, Studiot. Absolutely incredible. I will definitely print this post and add to my notebook. I WAS able to follow you, actually, and fathom the concept thoroughly with this analysis and see exactly what you did. This is perfect. Exactly what I needed.

I'm not sure if I got my arithmatic right in the calculation I attached now but I understand how you got there. I'm not actually used to seeing a voltage source where the + is on one end and the - at the other-- it threw me off. Thank you for caring enough to post it, Studiot.


(The amount of care and help in this forum amazes me every time anew. It works because there are people like you around! What an educational treasure on the net. I never expected that.)

[Femme_physics]

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.

go well

I will definitely do that as well.

[Studiot]

Whilst I agree with multiplying the equation through by 300, I don't make the second and later terms the same.

your 10 etc should appear ouside the brackets, not inside where you have them.

that is

\frac{{{{\left( {\frac{{30V}}{{40}}} \right)}^2}}}{{30}}*300 = 10*{\left( {\frac{{30V}}{{40}}} \right)^2}not{\left( {\frac{{300V}}{{40}}} \right)^2}

etc

Looking again at my 2 equations I see I have reversed the order of terms between the first and second, although both are correct. Sorry if that caused confusion.

[Femme_physics]

Right, algebra kinks. I will work it out better, thanks.

[Femme_physics]

= \frac{{{V^2}}}{{50}} + \frac{{{{\left( {\frac{{30V}}{{40}}} \right)}^2}}}{{30}} + \frac{{{{\left( {\frac{{10V}}{{40}}} \right)}^2}}}{{20}} + \frac{{{{\left( {\frac{{10V}}{{40}}*\frac{{15}}{{20}}} \right)}^2}}}{{15}} + \frac{{{{\left( {\frac{{10V}}{{40}}*\frac{5}{{20}}} \right)}^2}}}{5}
.

Shouldn't the first term -> V^2/50 have brackets separating the ^2 from the V/50?

[Studiot]

This is where I inadvertanly reversed the term order, relative to the first equation.

R5 = 50 so power in R5 is

V2/50

R5 is the only resistor that experiences the full voltage, V.
All the others see V reduced by some factor.
However you still need to check that R5 reaches 1 watt first, since we also divide by the resistor value to get the power.

I am off now for some parvus et circum in the shape of dinner and Alien v Predator 2.

go well

[Femme_physics]

I see. It's also a bit late here. I'll post tomorrow the numbers hopefully I'd get someone's seal of approval. Thank you Studiot, Serena.

[Femme_physics]

1) I've scanned my second attempt of the solution

2) I've also tried to calculate for I, is my equation correct?

Also, as far as how you've gotten the the formula for each of the resistor, it's basically:

"the potential difference of the resistor (or the entire parallel region), divided by the potential difference across the entire closed loop, divided by the potential difference of the resistor."

I've tried to derive that from your formula. I don't recall it being explained to me that way.

[Studiot]

Oh dear, you really are better at mechanics.

I will post some more after lunch. Meanwhile perhaps you would like to think about this.

Why does your 'solution' equate the total power to 1 and then 300?

The current method is not really derivable from the voltage method - V does not actually appear anywhere.

Kirchoffs laws, loops and such won't help here. I'm just using the basic definition of series and parallel connections, ohm's law and the equation for power.

[Femme_physics]

Oh dear, you really are better at mechanics.

LOL. Appreciate the recognition though.

I do know how to find Rt and easily break down a parallel+series circuit to an only series circuit, which I initially thought was challenging but I understand the simple principles of it.

Why does your 'solution' equate the total power to 1 and then 300?

Well, I solved for V, so since I had a fraction and 300 was my common denominator, I multiplied both sides...

So quote is not true? ->

"the potential difference of the resistor (or the entire parallel region), divided by the potential difference across the entire closed loop, divided by the potential difference of the resistor."


It seems to be true to how you've decided the formula for each R in your P equation.

[Studiot]

There are two separate issues here.

One is arithmetic and I will sort that first by finishing the solution (in the attachment).

Note that in order to back substitute and calculate the total power I only need to calculate V2.
I don't need to take the square root and find V.

I have lined up the calculations so the individual contributions from each resistor is shown term by term.

Is there anything you don't follow?

[I like Serena]

1) I've scanned my second attempt of the solution

I'll leave the electrical part of the discussion to Studiot, because I think he's doing a good job at it, and he's obviously taking the time and energy to try to explain it (I think he's improving on that too :smile:).

So I'd like to draw your attention to an arithmetic part (which he ignored for now).

In your scan you write one of the terms (2nd line) as:

10 \cdot \left(\frac {30V} {40}\right)^2

Then in the 3rd line you have turned this into:

10 \cdot \left(\frac {30V^2} {40^2}\right)

And in the 4th line this is:

\left(\frac {300V^2} {1600}\right)

This is not quite right. ;)

Can you see what this should be?

[Femme_physics]

Great catch, Serena. You two are really too kind. I appreciate the effort. I also re-read the walking through analysis in page 2 just to make sure it's all clear.

This is not quite right. ;)

I'm plugging it into my calc and I now see it doesn't turn up quite right. I seem to have broken a simple rule of parathesis BEFORE exponents.

Apparently, what's right is what Studiot did, but I'm not able to follow him. Did he just use the properties of exponents?

Not sure how he got from this to that

http://img717.imageshack.us/img717/992/howhow.jpg (http://img717.imageshack.us/i/howhow.jpg/)

He seems to have used the exponent on both the numerical denominator and numerator in the fraction and then just pull V out and leave it to the power of 2. I wasn't aware you can do that. He also took the 30 on the lowest denominator and multiplied it by the upper fraction. But at this page I already seemed to have lost Studiot on too many fronts.

Is this all confusion due to my lack of understanding about simplifying exponential fractions containing unknowns? If it is, I'll go back and review it.

[I like Serena]

Great catch, Serena. You two are really too kind. I appreciate the effort. I also re-read the walking through analysis in page 2 just to make sure it's all clear.

I'm plugging it into my calc and I now see it doesn't turn up quite right. I seem to have broken a simple rule of parathesis BEFORE exponents.

Apparently, what's right is what Studiot did, but I'm not able to follow him. Did he just use the properties of exponents?

Not sure how he got from this to that

http://img717.imageshack.us/img717/992/howhow.jpg (http://img717.imageshack.us/i/howhow.jpg/)

He seems to have used the exponent on both the numerical denominator and numerator in the fraction and then just pull V out and leave it to the power of 2. I wasn't aware you can do that. He also took the 30 on the lowest denominator and multiplied it by the upper fraction. But at this page I already seemed to have lost Studiot on too many fronts.

Is this all confusion due to my lack of understanding about simplifying exponential fractions containing unknowns? If it is, I'll go back and review it.

I'm focusing on arithmetic now, because you will definitely need that too with calculus and dynamics.

Studiot skipped a few steps, so I'll write them out for you for this specific instance.

He used the following rules for exponents and fractions:

(2 \cdot 3)^2 = 2^2 \cdot 3^2

\left(\frac 2 3\right)^2 = \frac {2^2} {3^2}

3 \cdot \frac 2 3 = 2

\frac {\left( \frac 2 3 \right)} 5 =\frac {3 \cdot \frac 2 3} {3 \cdot 5} = \frac 2 {3 \cdot 5}

\frac {2 \cdot 3} 5=\frac 2 5 \cdot 3


So taking you step by step, applying the rules above, we have:

\frac {\left(\frac {30} {40}V\right)^2} {30}
=\frac {\left(\frac {30} {40} \cdot V\right)^2} {30}
=\frac {\left(\frac {30} {40}\right)^2 \cdot V^2} {30}
= \frac {\frac {30^2} {40^2} \cdot V^2} {30}
= \frac {40^2 \cdot \frac {30^2} {40^2} \cdot V^2} {40^2 \cdot 30}
= \frac {30^2 \cdot V^2} {40^2 \cdot 30}
= \frac {30^2} {40^2 \cdot 30} \cdot V^2
= \left( \frac {30 \cdot 30} {40 \cdot 40 \cdot 30} \right) \cdot V^2

If you do this a lot, you can do it all at once, or otherwise when you write it down, you leave it to the reader to figure out the intermediary steps.

[Studiot]

Well I'm glad I didn't have to type all that out.

Thank you ILS!

[Femme_physics]

He's incredible! :)

I did, however, try to solve it on my own using the properties of exponent you have written out to me, I Like Serena. I swear, I didn't even look at your final big solution! (attached is my draft attempt and success [there was a moment of desperation which might show in the page]... )

I'll look back at some of the earlier posts now, thanks :)

[Femme_physics]

Okay, I am looking back at Studiot post. You've found Ptotal, they're asking for Pmax. They didn't ask "which resistor will be damaged first", they asked "what is the max power you can spread around the circuit without causing damage to any of the resistors"..shouldn't the answer be in terms of Pmax = ?

[Studiot]

Yes I should have said Ptotalmax not just Ptotal when I put in the value for V2

The maximum safe power in that network is 2.25 Watts.

At that power input R5 is at its rated limit and if you look carefully R4 is pretty close as well.

Are we all agreed now?

[I like Serena]

He's incredible! :)

I did, however, try to solve it on my own using the properties of exponent you have written out to me, I Like Serena. I swear, I didn't even look at your final big solution! (attached is my draft attempt and success [there was a moment of desperation which might show in the page]... )

I'll look back at some of the earlier posts now, thanks :)

I'm proud of you! Your calculation is entirely correct :smile:

(However, Studiot had 0.01875 V2. How could that be?)

[Femme_physics]

I'm proud of you! Your calculation is entirely correct

:) Great.

I always make sure to plug a number to see it makes sense!

Yes I should have said Ptotalmax not just Ptotal when I put in the value for V2

The maximum safe power in that network is 2.25 Watts.

At that power input R5 is at its rated limit and if you look carefully R4 is pretty close as well.

Are we all agreed now?


If you can confirm my own verbal-only walk through analysis

In order to find Pmax we first write up the circuit's formula for total power, by doing that we are able to easily detect which resistor will have the greater reaction to upping of the power dosage, i.e. which resistor will burn first.


So after discovering the culprit, we set it to be LESS THAN or EQUAL to 1. i.e. we use the formula for power again, just this time looking at an individual naughty resistor and setting it equal or less than 1.

That allows us to determine MAX voltage [to the power of 2].

(hmm, interesting. What if the voltage was given to us?)

Having found Vmax^2, we just plug it into the Ptotalmax formula and solve. Voila!



(However, Studiot had 0.01875 V2. How could that be?)

*glances at Studiot...whistles, averts gaze*

[I like Serena]

*glances at Studiot...whistles, averts gaze*

:rofl:
I'm afraid you were both right in your calculations.
Can you spot the difference and how it matters?

[Studiot]

*glances at Studiot...whistles, averts gaze*

I don't seem to get the joke here.

FF, in post#43 you posted some working where you started with a fraction in brackets and squared it.

Yes your arithmetic in that post was correct.

Unfortunately you had not copied down the fraction correctly so the bottom 30 appeared inside your bracket, but was outside in my original.

??

[Femme_physics]

Oh, yea, I put the paranthesis all over, whereas the parenthesis just should've been over the upper fraction, so that's what raised to the power of 2, not the lower part

Edit: I wrote it before Studiot posted!

[Femme_physics]

I don't seem to get the joke here.

FF, in post#43 you posted some working where you started with a fraction in brackets and squared it.

Yes your arithmetic in that post was correct.

Unfortunately you had not copied down the fraction correctly so the bottom 30 appeared inside your bracket, but was outside in my original.

??

Yes, my bad, it's just that I have a thing with experts and solution manuals writing the wrong figures so I keep looking for it even when it's not there. Anyway, you're a world of help! :) Can we go back to -->
"
If you can confirm my own verbal-only walk through analysis

In order to find Pmax we first write up the circuit's formula for total power, by doing that we are able to easily detect which resistor will have the greater reaction to upping of the power dosage, i.e. which resistor will burn first.


So after discovering the culprit, we set it to be LESS THAN or EQUAL to 1. i.e. we use the formula for power again, just this time looking at an individual naughty resistor and setting it equal or less than 1.

That allows us to determine MAX voltage [to the power of 2].

(hmm, interesting. What if the voltage was given to us?)

Having found Vmax^2, we just plug it into the Ptotalmax formula and solve. Voila!
"

[I like Serena]

Oh, yea, I put the paranthesis all over, whereas the parenthesis just should've been over the upper fraction, so that's what raised to the power of 2, not the lower part

Edit: I wrote it before Studiot posted!

@Studiot: Sorry, I just thought it funny that FP thought you had it wrong, but of course you were right all along.

I did hope that FP would have found that out for herself ;).

[Studiot]

I really hate to miss a joke, even one at my expense.

Folks can call me anything they like, just so long as it is clearly in jest.
Laughter is almost as good as a good dinner.

[Studiot]

(hmm, interesting. What if the voltage was given to us?)


Surely if the voltage is fixed the power is fixed?

The rest of your summary is good.

:approve:

How do you feel about the circuit analysis now?

[Femme_physics]

Surely if the voltage is fixed the power is fixed?
I get it, so the max voltage is always the max power! Brilliant. Absolutely brillant.

Somebody tell Homer Simpson that! :)


The rest of your summary is good.

:approve:

How do you feel about the circuit analysis now?


Not...that bad frankly. It's kinda lovely, and makes sense. It's, again, one of those subjects you must love to succeed in it. I love :) I hope to get as good as mechanics in it. I'm glad to have great mentors like you two (if you don't mind me calling you that :) ), that don't even mind going back to the basic groundworks to get up the scale back to electronics.

By the way, do you know how much self-restraint it took not to answer your question with "electrifying"?? lol

Oh, btw, as long as you and ILS are here, I have another coursework question that I... *see them both bolt for their lives* ;)

[I like Serena]

*bolting*

--I like ILSe

[wikipedia: "249 Ilse - an asteroid named for the legendary German princess of the Harz Mountains"]

[Studiot]

If you are comfortable with basic circuit analysis, you might like to look at this.

http://www.physicsforums.com/showthread.php?t=493068

[Femme_physics]

If you are comfortable with basic circuit analysis, you might like to look at this.

http://www.physicsforums.com/showthread.php?t=493068


I was gonna solve his problem for him just for practice but even the problem bolted and ran away from me. Don't believe me? Here, look!

[I like Serena]

I was gonna solve his problem for him just for practice but even the problem bolted and ran away from me. Don't believe me? Here, look!

:rofl:

[I like Serena]

I was gonna solve his problem for him just for practice but even the problem bolted and ran away from me. Don't believe me? Here, look!

Seriously.
You can't treat R1 and R2 as parallel resistors, since there is an R3 in between.*

However, if you deform the circuit a bit and bring the end points of R5 and R6 together to one point, which you can do because they are short circuited, it becomes a problem I think you can already solve.

In the new deformed circuit the resistors R5 and R6, for instance, are parallel.

----
* Note that I followed your numbering of the resistors, and not the numbering of the thread.

[Femme_physics]

Seeing how you can deform the circuits is indeed crucial in solving these types of problems, I was rather shortsighted this time. Thanks, I'll probably won't post in trying to help him or anyone with electronics problem anytime soon.

[I like Serena]

Seeing how you can deform the circuits is indeed crucial in solving these types of problems, I was rather shortsighted this time. Thanks, I'll probably won't post in trying to help him or anyone with electronics problem anytime soon.

If by "anytime soon" you mean "after a couple of days", I can agree. :)

And anyway, I'm pretty sure any contribution you make would be appreciated.
It shows how enthusiastic you are, and you already know how well people like that and respond to that.
Imho ppl respond much less to a "know it all" professor, even if he/she is right.

(Just try not to hijack any such thread :wink:)

[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.

:rofl:

[Femme_physics]

I've enroled in a class of handwriting lessons.

:rofl:

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.

http://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.

34951

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://imageshack.us/photo/my-images/15/rt1i.jpg/)

http://img860.imageshack.us/img860/9381/i1i2.jpg (http://imageshack.us/photo/my-images/860/i1i2.jpg/)

http://img801.imageshack.us/img801/424/vrsc.jpg (http://imageshack.us/photo/my-images/801/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 VR4 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! :smile:

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! :wink:

[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